General solution for missing sequence numbers

I occasionally need to determine a number that I don't have in a sequence,
either the first missing one in a gap in a set of sequential numbers or the
next one in line at the end of a numbered series. Always it meant some
fumbling around, with either VBA at first or later with SQL when I got good
enough at it, establishing the proper join parameters and such. For SQL
experts, this is probably routine and trivial, but for me it was always a
bit of a chore. The last straw came with a database which I recently wrote,
where the converted data had such a numbered series, and the owner wanted to
be able to do both, fill in missing numbers in the gaps AND add new numbers
at the end.

Walking home from a bar last night, I got to thinking about it and realized
that both problems are actually fairly similar and that a simple and general
solution is possible.
I put together a simple table containing one field with the following
entries:

1,2,3,4, 8,9,10, 15,16,17,18, 20, 22,23,24,25, 28,29,30

Missing a


5,6,7, 11,12,13,14, 19, 21, 26,27 and 31 on up.

This is the dataset used for all of the following examples.


Finding the next new number at the end of a series with SQL is trivial; here
is a simplified version of a statement that I found somewhere in the
discussion groups a few years ago:

SELECT Max(MyTable.MySeqFld)+1 FROM MyTable;

This will return a one-record, one-field recordset containing exactly one
value: 31, which is one greater than the largest value so far used in that
field. This is what you would want to use instead of Access's autonumber, if
the field is to contain meaningful sequence numbering, rather than just a
unique identifier.

Locating gaps is a little more complicated: it involves a self-join from N
to N+1 and finding where N+1 doesn't exist, indicating a gap at that point.

SELECT MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL;

This generates a recordset of 5, 11, 19, 21, 26, 31, where each value is the
first missing value in a gap, including the "open gap" at the end, and
that's where the trick to a general solution begins. Since these situations
normally call for either the first (lowest number) gap or last (end of
recordset) gap, you need either the first or last record returned by this
query. Sorting and using the TOP predicate gives you exactly that.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld;

This will again return a one-record, one-field recordset containing exactly
one value: 5, the first missing number in the first gap in the sequence.
Ascending sort order is the default, so the smallest number is the first
returned.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld DESC;

This will return a one-record, one-field recordset containing exactly one
value: 31, the same "one greater than the highest value so far used in that
field" that is returned by the first simple example. Specifying the
descending order here is necessary, since we want the last (greatest) record
from the set and Access SQL does not have a BOTTOM predicate.

Finally, an even more general statement can be used:

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY ((INSTR("LF",[First or Last (F or L)]) *2)-3)*MT1.MySeqFld;

This expects one parameter, F or L and will return either the first missing
number or the next number at the end of the line. The INSTR expression
evaluates to either -1 or 1 (or -3 if the parameter supplied is neither F
nor L, but that has the same effect as -1 in this instance), that is then
used as a multiplier for the sort field, so the sort is either by the field
or by the negative of the field (or 3 times the negative of the field),
giving either ascending or descending order and with the TOP 1 predicate
again returns exactly the one value of interest.


--

Pete

This e-mail address is fake to keep spammers and their auto-harvesters out
of my hair. If you need to get in touch personally, I am 'pdanes' and I use
Yahoo mail. But please use the newsgroups whenever possible, so that all may
benefit from the exchange of ideas.

"Peter Danes" wrote in message
...

snip

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld DESC;

This will return a one-record, one-field recordset containing
exactly one
value: 31, the same "one greater than the highest value so far
used in that
field" that is returned by the first simple example. Specifying
the
descending order here is necessary, since we want the last
(greatest) record
from the set and Access SQL does not have a BOTTOM predicate.

Peter Danes,

I am not sure what the difference is between the above and the
below.

SELECT MAX(MT1.MySeqFld) + 1
FROM MyTable AS MT1;


Sincerely,

Chris O.

You might be interested in the analysis I had in my April, 2004 "Access
Answers" column in Pinnacle Publication's "Smart Access". You can download
the column (and sample database) for free from
http://www.accessmvp.com/djsteele/SmartAccess.html

--
Doug Steele, Microsoft Access MVP
http://I.Am/DougSteele
(no e-mails, please!)


"Peter Danes" wrote in message
...
I occasionally need to determine a number that I don't have in a sequence,
either the first missing one in a gap in a set of sequential numbers or
the
next one in line at the end of a numbered series. Always it meant some
fumbling around, with either VBA at first or later with SQL when I got
good
enough at it, establishing the proper join parameters and such. For SQL
experts, this is probably routine and trivial, but for me it was always a
bit of a chore. The last straw came with a database which I recently
wrote,
where the converted data had such a numbered series, and the owner wanted
to
be able to do both, fill in missing numbers in the gaps AND add new
numbers
at the end.

Walking home from a bar last night, I got to thinking about it and
realized
that both problems are actually fairly similar and that a simple and
general
solution is possible.
I put together a simple table containing one field with the following
entries:

1,2,3,4, 8,9,10, 15,16,17,18, 20, 22,23,24,25, 28,29,30

Missing a


5,6,7, 11,12,13,14, 19, 21, 26,27 and 31 on up.

This is the dataset used for all of the following examples.


Finding the next new number at the end of a series with SQL is trivial;
here
is a simplified version of a statement that I found somewhere in the
discussion groups a few years ago:

SELECT Max(MyTable.MySeqFld)+1 FROM MyTable;

This will return a one-record, one-field recordset containing exactly one
value: 31, which is one greater than the largest value so far used in that
field. This is what you would want to use instead of Access's autonumber,
if
the field is to contain meaningful sequence numbering, rather than just a
unique identifier.

Locating gaps is a little more complicated: it involves a self-join from N
to N+1 and finding where N+1 doesn't exist, indicating a gap at that
point.

SELECT MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL;

This generates a recordset of 5, 11, 19, 21, 26, 31, where each value is
the
first missing value in a gap, including the "open gap" at the end, and
that's where the trick to a general solution begins. Since these
situations
normally call for either the first (lowest number) gap or last (end of
recordset) gap, you need either the first or last record returned by this
query. Sorting and using the TOP predicate gives you exactly that.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld;

This will again return a one-record, one-field recordset containing
exactly
one value: 5, the first missing number in the first gap in the sequence.
Ascending sort order is the default, so the smallest number is the first
returned.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld DESC;

This will return a one-record, one-field recordset containing exactly one
value: 31, the same "one greater than the highest value so far used in
that
field" that is returned by the first simple example. Specifying the
descending order here is necessary, since we want the last (greatest)
record
from the set and Access SQL does not have a BOTTOM predicate.

Finally, an even more general statement can be used:

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY ((INSTR("LF",[First or Last (F or L)]) *2)-3)*MT1.MySeqFld;

This expects one parameter, F or L and will return either the first
missing
number or the next number at the end of the line. The INSTR expression
evaluates to either -1 or 1 (or -3 if the parameter supplied is neither F
nor L, but that has the same effect as -1 in this instance), that is then
used as a multiplier for the sort field, so the sort is either by the
field
or by the negative of the field (or 3 times the negative of the field),
giving either ascending or descending order and with the TOP 1 predicate
again returns exactly the one value of interest.


--

Pete

This e-mail address is fake to keep spammers and their auto-harvesters
out
of my hair. If you need to get in touch personally, I am 'pdanes' and I
use
Yahoo mail. But please use the newsgroups whenever possible, so that all
may
benefit from the exchange of ideas.

What is the specific *NEED* to find the missing numbers?

Peter Danes wrote:
I occasionally need to determine a number that I don't have in a sequence,
either the first missing one in a gap in a set of sequential numbers or the
next one in line at the end of a numbered series. Always it meant some
fumbling around, with either VBA at first or later with SQL when I got good
enough at it, establishing the proper join parameters and such. For SQL
experts, this is probably routine and trivial, but for me it was always a
bit of a chore. The last straw came with a database which I recently wrote,
where the converted data had such a numbered series, and the owner wanted to
be able to do both, fill in missing numbers in the gaps AND add new numbers
at the end.

Walking home from a bar last night, I got to thinking about it and realized
that both problems are actually fairly similar and that a simple and general
solution is possible.
I put together a simple table containing one field with the following
entries:

1,2,3,4, 8,9,10, 15,16,17,18, 20, 22,23,24,25, 28,29,30

Missing a


5,6,7, 11,12,13,14, 19, 21, 26,27 and 31 on up.

This is the dataset used for all of the following examples.


Finding the next new number at the end of a series with SQL is trivial; here
is a simplified version of a statement that I found somewhere in the
discussion groups a few years ago:

SELECT Max(MyTable.MySeqFld)+1 FROM MyTable;

This will return a one-record, one-field recordset containing exactly one
value: 31, which is one greater than the largest value so far used in that
field. This is what you would want to use instead of Access's autonumber, if
the field is to contain meaningful sequence numbering, rather than just a
unique identifier.

Locating gaps is a little more complicated: it involves a self-join from N
to N+1 and finding where N+1 doesn't exist, indicating a gap at that point.

SELECT MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL;

This generates a recordset of 5, 11, 19, 21, 26, 31, where each value is the
first missing value in a gap, including the "open gap" at the end, and
that's where the trick to a general solution begins. Since these situations
normally call for either the first (lowest number) gap or last (end of
recordset) gap, you need either the first or last record returned by this
query. Sorting and using the TOP predicate gives you exactly that.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld;

This will again return a one-record, one-field recordset containing exactly
one value: 5, the first missing number in the first gap in the sequence.
Ascending sort order is the default, so the smallest number is the first
returned.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld DESC;

This will return a one-record, one-field recordset containing exactly one
value: 31, the same "one greater than the highest value so far used in that
field" that is returned by the first simple example. Specifying the
descending order here is necessary, since we want the last (greatest) record
from the set and Access SQL does not have a BOTTOM predicate.

Finally, an even more general statement can be used:

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY ((INSTR("LF",[First or Last (F or L)]) *2)-3)*MT1.MySeqFld;

This expects one parameter, F or L and will return either the first missing
number or the next number at the end of the line. The INSTR expression
evaluates to either -1 or 1 (or -3 if the parameter supplied is neither F
nor L, but that has the same effect as -1 in this instance), that is then
used as a multiplier for the sort field, so the sort is either by the field
or by the negative of the field (or 3 times the negative of the field),
giving either ascending or descending order and with the TOP 1 predicate
again returns exactly the one value of interest.

There are three differences:

1. Your example is the same as my first example which returns only the
"greatest +1", except that you additionally include an alias to the table,
the "AS MT1" at the end of the statement. It doesn't hurt anything, but
isn't really necessary.

2. Youe example doesn't call for a parameter, mine does, to determine the
sort order and so whether you get the first missing number or the next in
line greater than all numbers used so far.

3. Obviously, the example you posted is considerably simpler, and if you
only need what it returns, simpler is preferable. The point of my 'lecture'
was simply that a general solution to these related problems is possible
with a single SQL statement. I do not claim that it is preferable in all
situations, or even any particular situation.

Pete


"Chris2" pí¹e v diskusním
pøíspìvku ...

"Peter Danes" wrote in message
...

snip

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld DESC;

This will return a one-record, one-field recordset containing
exactly one
value: 31, the same "one greater than the highest value so far
used in that
field" that is returned by the first simple example. Specifying
the
descending order here is necessary, since we want the last
(greatest) record
from the set and Access SQL does not have a BOTTOM predicate.

Peter Danes,

I am not sure what the difference is between the above and the
below.

SELECT MAX(MT1.MySeqFld) + 1
FROM MyTable AS MT1;


Sincerely,

Chris O.

Thank you Doug, interesting article. I like your addition of the range, I
think I'll be able to use that in something I'm working on now. And many of
the other titles look intriguing as well - time to do some reading.

(BTW, the description for invoice 11 says how about sending me an e-mail,
but your signature says no e-mails, please. I'm feeling schizophrenic. Maybe
if I write you one but don't send it...?)

Pete


"Douglas J Steele" pí¹e v diskusním
pøíspìvku ...
You might be interested in the analysis I had in my April, 2004 "Access
Answers" column in Pinnacle Publication's "Smart Access". You can download
the column (and sample database) for free from
http://www.accessmvp.com/djsteele/SmartAccess.html

--
Doug Steele, Microsoft Access MVP
http://I.Am/DougSteele
(no e-mails, please!)


"Peter Danes" wrote in message
...
I occasionally need to determine a number that I don't have in a
sequence,
either the first missing one in a gap in a set of sequential numbers or
the
next one in line at the end of a numbered series. Always it meant some
fumbling around, with either VBA at first or later with SQL when I got
good
enough at it, establishing the proper join parameters and such. For SQL
experts, this is probably routine and trivial, but for me it was always a
bit of a chore. The last straw came with a database which I recently
wrote,
where the converted data had such a numbered series, and the owner wanted
to
be able to do both, fill in missing numbers in the gaps AND add new
numbers
at the end.

Walking home from a bar last night, I got to thinking about it and
realized
that both problems are actually fairly similar and that a simple and
general
solution is possible.
I put together a simple table containing one field with the following
entries:

1,2,3,4, 8,9,10, 15,16,17,18, 20, 22,23,24,25, 28,29,30

Missing a


5,6,7, 11,12,13,14, 19, 21, 26,27 and 31 on up.

This is the dataset used for all of the following examples.


Finding the next new number at the end of a series with SQL is trivial;
here
is a simplified version of a statement that I found somewhere in the
discussion groups a few years ago:

SELECT Max(MyTable.MySeqFld)+1 FROM MyTable;

This will return a one-record, one-field recordset containing exactly one
value: 31, which is one greater than the largest value so far used in
that
field. This is what you would want to use instead of Access's autonumber,
if
the field is to contain meaningful sequence numbering, rather than just a
unique identifier.

Locating gaps is a little more complicated: it involves a self-join from
N
to N+1 and finding where N+1 doesn't exist, indicating a gap at that
point.

SELECT MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL;

This generates a recordset of 5, 11, 19, 21, 26, 31, where each value is
the
first missing value in a gap, including the "open gap" at the end, and
that's where the trick to a general solution begins. Since these
situations
normally call for either the first (lowest number) gap or last (end of
recordset) gap, you need either the first or last record returned by this
query. Sorting and using the TOP predicate gives you exactly that.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld;

This will again return a one-record, one-field recordset containing
exactly
one value: 5, the first missing number in the first gap in the sequence.
Ascending sort order is the default, so the smallest number is the first
returned.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld DESC;

This will return a one-record, one-field recordset containing exactly one
value: 31, the same "one greater than the highest value so far used in
that
field" that is returned by the first simple example. Specifying the
descending order here is necessary, since we want the last (greatest)
record
from the set and Access SQL does not have a BOTTOM predicate.

Finally, an even more general statement can be used:

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY ((INSTR("LF",[First or Last (F or L)]) *2)-3)*MT1.MySeqFld;

This expects one parameter, F or L and will return either the first
missing
number or the next number at the end of the line. The INSTR expression
evaluates to either -1 or 1 (or -3 if the parameter supplied is neither F
nor L, but that has the same effect as -1 in this instance), that is then
used as a multiplier for the sort field, so the sort is either by the
field
or by the negative of the field (or 3 times the negative of the field),
giving either ascending or descending order and with the TOP 1 predicate
again returns exactly the one value of interest.


--

Pete

This e-mail address is fake to keep spammers and their auto-harvesters
out
of my hair. If you need to get in touch personally, I am 'pdanes' and I
use
Yahoo mail. But please use the newsgroups whenever possible, so that all
may
benefit from the exchange of ideas.

Such situations are common, for a variety of reasons. Depends on the
database and the user and what the data is for. The particular example that
inspired this outburst is a mycological database, where the numbers are used
to sequentially number the scientist's samples. She told me that numbering
is important for others in the field to know roughly how many samples a
particular researcher has, and for internal inventory purposes, that they
don't expect to have holes in the numbering sequence.

If someone who has 1,000 samples in their collection publishes something
about their sample number 10,000 and it is known that the person does not
have anywhere near 10,000 samples, it would be viewed as odd at the very
least, possibly unethical and such a person would find himself not taken
seriously by other researchers. One or two numbers amiss in this situation
is obviously not a major concern.

And for the internal inventory controls, if someone sees sample 152 next to
150, they are going to wonder where is number 151. The inventory methods
used expect sequential numbering and a missing number is an indication of
something wrong. For inventory numbers in the original database, she used
the record number that appears in the text box of Access's navigation
control in conjunction with an autonumber field. You may guess what sort of
hash resulted from that. I started out trying to fix a few things for her
and wound up doing almost a complete re-write of the entire thing and this
numbering issue is one of the things that surfaced. She wants to be able to
fill in all the gaps as well as add new numbers to the end as she collects
new samples.

Pete


"David C. Holley" píse v diskusním príspevku
...
What is the specific *NEED* to find the missing numbers?

Peter Danes wrote:
I occasionally need to determine a number that I don't have in a
sequence, either the first missing one in a gap in a set of sequential
numbers or the next one in line at the end of a numbered series. Always
it meant some fumbling around, with either VBA at first or later with SQL
when I got good enough at it, establishing the proper join parameters and
such. For SQL experts, this is probably routine and trivial, but for me
it was always a bit of a chore. The last straw came with a database which
I recently wrote, where the converted data had such a numbered series,
and the owner wanted to be able to do both, fill in missing numbers in
the gaps AND add new numbers at the end.

Walking home from a bar last night, I got to thinking about it and
realized that both problems are actually fairly similar and that a simple
and general solution is possible.
I put together a simple table containing one field with the following
entries:

1,2,3,4, 8,9,10, 15,16,17,18, 20, 22,23,24,25, 28,29,30

Missing a


5,6,7, 11,12,13,14, 19, 21, 26,27 and 31 on up.

This is the dataset used for all of the following examples.


Finding the next new number at the end of a series with SQL is trivial;
here is a simplified version of a statement that I found somewhere in the
discussion groups a few years ago:

SELECT Max(MyTable.MySeqFld)+1 FROM MyTable;

This will return a one-record, one-field recordset containing exactly one
value: 31, which is one greater than the largest value so far used in
that field. This is what you would want to use instead of Access's
autonumber, if the field is to contain meaningful sequence numbering,
rather than just a unique identifier.

Locating gaps is a little more complicated: it involves a self-join from
N to N+1 and finding where N+1 doesn't exist, indicating a gap at that
point.

SELECT MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL;

This generates a recordset of 5, 11, 19, 21, 26, 31, where each value is
the first missing value in a gap, including the "open gap" at the end,
and that's where the trick to a general solution begins. Since these
situations normally call for either the first (lowest number) gap or last
(end of recordset) gap, you need either the first or last record returned
by this query. Sorting and using the TOP predicate gives you exactly
that.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld;

This will again return a one-record, one-field recordset containing
exactly one value: 5, the first missing number in the first gap in the
sequence. Ascending sort order is the default, so the smallest number is
the first returned.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld DESC;

This will return a one-record, one-field recordset containing exactly one
value: 31, the same "one greater than the highest value so far used in
that field" that is returned by the first simple example. Specifying the
descending order here is necessary, since we want the last (greatest)
record from the set and Access SQL does not have a BOTTOM predicate.

Finally, an even more general statement can be used:

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY ((INSTR("LF",[First or Last (F or L)]) *2)-3)*MT1.MySeqFld;

This expects one parameter, F or L and will return either the first
missing number or the next number at the end of the line. The INSTR
expression evaluates to either -1 or 1 (or -3 if the parameter supplied
is neither F nor L, but that has the same effect as -1 in this instance),
that is then used as a multiplier for the sort field, so the sort is
either by the field or by the negative of the field (or 3 times the
negative of the field), giving either ascending or descending order and
with the TOP 1 predicate again returns exactly the one value of interest.

I'm sorry, I don't understand what you mean by "invoice 11"

--
Doug Steele, Microsoft Access MVP
http://I.Am/DougSteele
(no e-mails, please!)


"Peter Danes" wrote in message
...
Thank you Doug, interesting article. I like your addition of the range, I
think I'll be able to use that in something I'm working on now. And many
of
the other titles look intriguing as well - time to do some reading.

(BTW, the description for invoice 11 says how about sending me an e-mail,
but your signature says no e-mails, please. I'm feeling schizophrenic.
Maybe
if I write you one but don't send it...?)

Pete


"Douglas J Steele" pí¹e v diskusním
pøíspìvku ...
You might be interested in the analysis I had in my April, 2004 "Access
Answers" column in Pinnacle Publication's "Smart Access". You can
download
the column (and sample database) for free from
http://www.accessmvp.com/djsteele/SmartAccess.html

--
Doug Steele, Microsoft Access MVP
http://I.Am/DougSteele
(no e-mails, please!)


"Peter Danes" wrote in message
...
I occasionally need to determine a number that I don't have in a
sequence,
either the first missing one in a gap in a set of sequential numbers or
the
next one in line at the end of a numbered series. Always it meant some
fumbling around, with either VBA at first or later with SQL when I got
good
enough at it, establishing the proper join parameters and such. For SQL
experts, this is probably routine and trivial, but for me it was always
a
bit of a chore. The last straw came with a database which I recently
wrote,
where the converted data had such a numbered series, and the owner
wanted
to
be able to do both, fill in missing numbers in the gaps AND add new
numbers
at the end.

Walking home from a bar last night, I got to thinking about it and
realized
that both problems are actually fairly similar and that a simple and
general
solution is possible.
I put together a simple table containing one field with the following
entries:

1,2,3,4, 8,9,10, 15,16,17,18, 20, 22,23,24,25, 28,29,30

Missing a


5,6,7, 11,12,13,14, 19, 21, 26,27 and 31 on up.

This is the dataset used for all of the following examples.


Finding the next new number at the end of a series with SQL is trivial;
here
is a simplified version of a statement that I found somewhere in the
discussion groups a few years ago:

SELECT Max(MyTable.MySeqFld)+1 FROM MyTable;

This will return a one-record, one-field recordset containing exactly
one
value: 31, which is one greater than the largest value so far used in
that
field. This is what you would want to use instead of Access's
autonumber,
if
the field is to contain meaningful sequence numbering, rather than just
a
unique identifier.

Locating gaps is a little more complicated: it involves a self-join
from
N
to N+1 and finding where N+1 doesn't exist, indicating a gap at that
point.

SELECT MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL;

This generates a recordset of 5, 11, 19, 21, 26, 31, where each value
is
the
first missing value in a gap, including the "open gap" at the end, and
that's where the trick to a general solution begins. Since these
situations
normally call for either the first (lowest number) gap or last (end of
recordset) gap, you need either the first or last record returned by
this
query. Sorting and using the TOP predicate gives you exactly that.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld;

This will again return a one-record, one-field recordset containing
exactly
one value: 5, the first missing number in the first gap in the
sequence.
Ascending sort order is the default, so the smallest number is the
first
returned.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld DESC;

This will return a one-record, one-field recordset containing exactly
one
value: 31, the same "one greater than the highest value so far used in
that
field" that is returned by the first simple example. Specifying the
descending order here is necessary, since we want the last (greatest)
record
from the set and Access SQL does not have a BOTTOM predicate.

Finally, an even more general statement can be used:

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY ((INSTR("LF",[First or Last (F or L)]) *2)-3)*MT1.MySeqFld;

This expects one parameter, F or L and will return either the first
missing
number or the next number at the end of the line. The INSTR expression
evaluates to either -1 or 1 (or -3 if the parameter supplied is neither
F
nor L, but that has the same effect as -1 in this instance), that is
then
used as a multiplier for the sort field, so the sort is either by the
field
or by the negative of the field (or 3 times the negative of the field),
giving either ascending or descending order and with the TOP 1
predicate
again returns exactly the one value of interest.


--

Pete

This e-mail address is fake to keep spammers and their auto-harvesters
out
of my hair. If you need to get in touch personally, I am 'pdanes' and I
use
Yahoo mail. But please use the newsgroups whenever possible, so that
all
may
benefit from the exchange of ideas.

Have you thought about using a DAO approach where you loop through the
records one by one and compare the current value to the previous?

Peter Danes wrote:
There are three differences:

1. Your example is the same as my first example which returns only the
"greatest +1", except that you additionally include an alias to the table,
the "AS MT1" at the end of the statement. It doesn't hurt anything, but
isn't really necessary.

2. Youe example doesn't call for a parameter, mine does, to determine the
sort order and so whether you get the first missing number or the next in
line greater than all numbers used so far.

3. Obviously, the example you posted is considerably simpler, and if you
only need what it returns, simpler is preferable. The point of my 'lecture'
was simply that a general solution to these related problems is possible
with a single SQL statement. I do not claim that it is preferable in all
situations, or even any particular situation.

Pete


"Chris2" pí¹e v diskusním
pøíspìvku ...

"Peter Danes" wrote in message
.. .

snip

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON

MT1.MySeqFld+1=MT2.MySeqFld

WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld DESC;

This will return a one-record, one-field recordset containing

exactly one

value: 31, the same "one greater than the highest value so far

used in that

field" that is returned by the first simple example. Specifying

the

descending order here is necessary, since we want the last

(greatest) record

from the set and Access SQL does not have a BOTTOM predicate.

Peter Danes,

I am not sure what the difference is between the above and the
below.

SELECT MAX(MT1.MySeqFld) + 1
FROM MyTable AS MT1;


Sincerely,

Chris O.

Small joke. (Very small) In the sample database for the article to which you
referred me, you have two tables, PossibleInvoices and Invoices, one with
just numbers and one with numbers and a text field. In the text field next
to invoice number 11, you had this comment "If so, how about sending me an
e-mail?"

Sorry for the confusion.

Pete


"Douglas J Steele" pí¹e v diskusním
pøíspìvku ...
I'm sorry, I don't understand what you mean by "invoice 11"

--
Doug Steele, Microsoft Access MVP
http://I.Am/DougSteele
(no e-mails, please!)


"Peter Danes" wrote in message
...
Thank you Doug, interesting article. I like your addition of the range, I
think I'll be able to use that in something I'm working on now. And many
of
the other titles look intriguing as well - time to do some reading.

(BTW, the description for invoice 11 says how about sending me an e-mail,
but your signature says no e-mails, please. I'm feeling schizophrenic.
Maybe
if I write you one but don't send it...?)

Pete


"Douglas J Steele" pí¹e v diskusním
pøíspìvku ...
You might be interested in the analysis I had in my April, 2004 "Access
Answers" column in Pinnacle Publication's "Smart Access". You can
download
the column (and sample database) for free from
http://www.accessmvp.com/djsteele/SmartAccess.html

--
Doug Steele, Microsoft Access MVP
http://I.Am/DougSteele
(no e-mails, please!)


"Peter Danes" wrote in message
...
I occasionally need to determine a number that I don't have in a
sequence,
either the first missing one in a gap in a set of sequential numbers
or
the
next one in line at the end of a numbered series. Always it meant some
fumbling around, with either VBA at first or later with SQL when I got
good
enough at it, establishing the proper join parameters and such. For
SQL
experts, this is probably routine and trivial, but for me it was
always
a
bit of a chore. The last straw came with a database which I recently
wrote,
where the converted data had such a numbered series, and the owner
wanted
to
be able to do both, fill in missing numbers in the gaps AND add new
numbers
at the end.

Walking home from a bar last night, I got to thinking about it and
realized
that both problems are actually fairly similar and that a simple and
general
solution is possible.
I put together a simple table containing one field with the following
entries:

1,2,3,4, 8,9,10, 15,16,17,18, 20, 22,23,24,25, 28,29,30

Missing a


5,6,7, 11,12,13,14, 19, 21, 26,27 and 31 on up.

This is the dataset used for all of the following examples.


Finding the next new number at the end of a series with SQL is
trivial;
here
is a simplified version of a statement that I found somewhere in the
discussion groups a few years ago:

SELECT Max(MyTable.MySeqFld)+1 FROM MyTable;

This will return a one-record, one-field recordset containing exactly
one
value: 31, which is one greater than the largest value so far used in
that
field. This is what you would want to use instead of Access's
autonumber,
if
the field is to contain meaningful sequence numbering, rather than
just
a
unique identifier.

Locating gaps is a little more complicated: it involves a self-join
from
N
to N+1 and finding where N+1 doesn't exist, indicating a gap at that
point.

SELECT MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL;

This generates a recordset of 5, 11, 19, 21, 26, 31, where each value
is
the
first missing value in a gap, including the "open gap" at the end, and
that's where the trick to a general solution begins. Since these
situations
normally call for either the first (lowest number) gap or last (end of
recordset) gap, you need either the first or last record returned by
this
query. Sorting and using the TOP predicate gives you exactly that.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld;

This will again return a one-record, one-field recordset containing
exactly
one value: 5, the first missing number in the first gap in the
sequence.
Ascending sort order is the default, so the smallest number is the
first
returned.

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY MT1.MySeqFld DESC;

This will return a one-record, one-field recordset containing exactly
one
value: 31, the same "one greater than the highest value so far used in
that
field" that is returned by the first simple example. Specifying the
descending order here is necessary, since we want the last (greatest)
record
from the set and Access SQL does not have a BOTTOM predicate.

Finally, an even more general statement can be used:

SELECT TOP 1 MT1.MySeqFld+1
FROM MyTable AS MT1 LEFT JOIN MyTable AS MT2 ON
MT1.MySeqFld+1=MT2.MySeqFld
WHERE MT2.MySeqFld IS NULL
ORDER BY ((INSTR("LF",[First or Last (F or L)]) *2)-3)*MT1.MySeqFld;

This expects one parameter, F or L and will return either the first
missing
number or the next number at the end of the line. The INSTR expression
evaluates to either -1 or 1 (or -3 if the parameter supplied is
neither
F
nor L, but that has the same effect as -1 in this instance), that is
then
used as a multiplier for the sort field, so the sort is either by the
field
or by the negative of the field (or 3 times the negative of the
field),
giving either ascending or descending order and with the TOP 1
predicate
again returns exactly the one value of interest.


--

Pete

This e-mail address is fake to keep spammers and their
auto-harvesters
out
of my hair. If you need to get in touch personally, I am 'pdanes' and
I
use
Yahoo mail. But please use the newsgroups whenever possible, so that
all
may
benefit from the exchange of ideas.