Convert variable #'s in a gen. format to a # that can be used w/ma

I have this data that is pasted to my worksheet:

Period: 01/23/2010 to 01/29/2010 - Days: 6
It is in general format. The day numbers will always be in the last, last 2
, or last 3 positions.
Depending on the period, the days may be any # up to the 365.
I need to use that day #, be it 6, or 25, or 147, whatever, as a divisor of
another number. E.g. if 6, then 1234 / 6 = 206
if 25, then 5678 / 25 = 227, etc/
The variable #'s will be in the same cell, so I basically want: E29/6 or
E29/25, or E29/147, etc.
Basically, those general formatted #'s converted to a usable/divisable #.

Thanks,

Steve

Steve wrote:
I have this data that is pasted to my worksheet:

Period: 01/23/2010 to 01/29/2010 - Days: 6
It is in general format. The day numbers will always be in the last, last 2
, or last 3 positions.
Depending on the period, the days may be any # up to the 365.
I need to use that day #, be it 6, or 25, or 147, whatever, as a divisor of
another number. E.g. if 6, then 1234 / 6 = 206
if 25, then 5678 / 25 = 227, etc/
The variable #'s will be in the same cell, so I basically want: E29/6 or
E29/25, or E29/147, etc.
Basically, those general formatted #'s converted to a usable/divisable #.

Thanks,

Steve




One way:

=--TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",10)),10))

One way...

Assuming the word "Days:" is *always* present.

=--MID(A1,SEARCH("Days:",A1)+5,5)

--
Biff
Microsoft Excel MVP


"Steve" wrote in message
...
I have this data that is pasted to my worksheet:

Period: 01/23/2010 to 01/29/2010 - Days: 6
It is in general format. The day numbers will always be in the last, last
2
, or last 3 positions.
Depending on the period, the days may be any # up to the 365.
I need to use that day #, be it 6, or 25, or 147, whatever, as a divisor
of
another number. E.g. if 6, then 1234 / 6 = 206
if 25, then 5678 / 25 = 227, etc/
The variable #'s will be in the same cell, so I basically want: E29/6 or
E29/25, or E29/147, etc.
Basically, those general formatted #'s converted to a usable/divisable #.

Thanks,

Steve

I'm getting a #value!
A1 is where my general format data is, correct ?

"Glenn" wrote:

Steve wrote:
I have this data that is pasted to my worksheet:

Period: 01/23/2010 to 01/29/2010 - Days: 6
It is in general format. The day numbers will always be in the last, last 2
, or last 3 positions.
Depending on the period, the days may be any # up to the 365.
I need to use that day #, be it 6, or 25, or 147, whatever, as a divisor of
another number. E.g. if 6, then 1234 / 6 = 206
if 25, then 5678 / 25 = 227, etc/
The variable #'s will be in the same cell, so I basically want: E29/6 or
E29/25, or E29/147, etc.
Basically, those general formatted #'s converted to a usable/divisable #.

Thanks,

Steve




One way:

=--TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",10)),10))
.

I do believe the "Days" will always be present, and your formula works to
produce either the single #, tens, or hundreds, but when I use that result as
a divisor, it still produces a #value!
E.g. Formula result is 6(in J3).... e19/j3 = #value!

"T. Valko" wrote:

One way...

Assuming the word "Days:" is *always* present.

=--MID(A1,SEARCH("Days:",A1)+5,5)

--
Biff
Microsoft Excel MVP


"Steve" wrote in message
...
I have this data that is pasted to my worksheet:

Period: 01/23/2010 to 01/29/2010 - Days: 6
It is in general format. The day numbers will always be in the last, last
2
, or last 3 positions.
Depending on the period, the days may be any # up to the 365.
I need to use that day #, be it 6, or 25, or 147, whatever, as a divisor
of
another number. E.g. if 6, then 1234 / 6 = 206
if 25, then 5678 / 25 = 227, etc/
The variable #'s will be in the same cell, so I basically want: E29/6 or
E29/25, or E29/147, etc.
Basically, those general formatted #'s converted to a usable/divisable #.

Thanks,

Steve




.

E.g. Formula result is 6(in J3).... e19/j3 = #value!

What's in E19?

When a formula like that resturns a #VALUE! error it usually means one (or
possibly both) cells referenced contain TEXT. Since we know that J3 contains
numeric 6 then the problem has to be with cell E19.

--
Biff
Microsoft Excel MVP


"Steve" wrote in message
...
I do believe the "Days" will always be present, and your formula works to
produce either the single #, tens, or hundreds, but when I use that result
as
a divisor, it still produces a #value!
E.g. Formula result is 6(in J3).... e19/j3 = #value!

"T. Valko" wrote:

One way...

Assuming the word "Days:" is *always* present.

=--MID(A1,SEARCH("Days:",A1)+5,5)

--
Biff
Microsoft Excel MVP


"Steve" wrote in message
...
I have this data that is pasted to my worksheet:

Period: 01/23/2010 to 01/29/2010 - Days: 6
It is in general format. The day numbers will always be in the last,
last
2
, or last 3 positions.
Depending on the period, the days may be any # up to the 365.
I need to use that day #, be it 6, or 25, or 147, whatever, as a
divisor
of
another number. E.g. if 6, then 1234 / 6 = 206
if 25, then 5678 / 25 = 227, etc/
The variable #'s will be in the same cell, so I basically want: E29/6
or
E29/25, or E29/147, etc.
Basically, those general formatted #'s converted to a usable/divisable
#.

Thanks,

Steve




.

Yes, A1 should have "Period: 01/23/2010 to 01/29/2010 - Days: 6". You
probably have trailing spaces in your data, which could cause the #Value! error.
Try this instead:

B1 =--RIGHT(SUBSTITUTE(TRIM(A1)," ",REPT(" ",3)),3)

Steve wrote:
I'm getting a #value!
A1 is where my general format data is, correct ?

"Glenn" wrote:

Steve wrote:
I have this data that is pasted to my worksheet:

Period: 01/23/2010 to 01/29/2010 - Days: 6
It is in general format. The day numbers will always be in the last, last 2
, or last 3 positions.
Depending on the period, the days may be any # up to the 365.
I need to use that day #, be it 6, or 25, or 147, whatever, as a divisor of
another number. E.g. if 6, then 1234 / 6 = 206
if 25, then 5678 / 25 = 227, etc/
The variable #'s will be in the same cell, so I basically want: E29/6 or
E29/25, or E29/147, etc.
Basically, those general formatted #'s converted to a usable/divisable #.

Thanks,

Steve



One way:

=--TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",10)),10))
.

This produced :6
I changed the 3's to 2's and got the 6, but I don't think my fix will
account for 2 or 3 digit numbers, will it ? Same as other, taking a number in
cell E19, and dividing by the resulting 6, still produces a #value!


"Glenn" wrote:

Yes, A1 should have "Period: 01/23/2010 to 01/29/2010 - Days: 6". You
probably have trailing spaces in your data, which could cause the #Value! error.
Try this instead:

B1 =--RIGHT(SUBSTITUTE(TRIM(A1)," ",REPT(" ",3)),3)

Steve wrote:
I'm getting a #value!
A1 is where my general format data is, correct ?

"Glenn" wrote:

Steve wrote:
I have this data that is pasted to my worksheet:

Period: 01/23/2010 to 01/29/2010 - Days: 6
It is in general format. The day numbers will always be in the last, last 2
, or last 3 positions.
Depending on the period, the days may be any # up to the 365.
I need to use that day #, be it 6, or 25, or 147, whatever, as a divisor of
another number. E.g. if 6, then 1234 / 6 = 206
if 25, then 5678 / 25 = 227, etc/
The variable #'s will be in the same cell, so I basically want: E29/6 or
E29/25, or E29/147, etc.
Basically, those general formatted #'s converted to a usable/divisable #.

Thanks,

Steve



One way:

=--TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",10)),10))
.

.

E19 has a number in it. Formatted as a number.
Also, in playing around, I may have changed the formatting of J3. It is
currently formatted as a number.

Thanks,

Steve

"T. Valko" wrote:

E.g. Formula result is 6(in J3).... e19/j3 = #value!

What's in E19?

When a formula like that resturns a #VALUE! error it usually means one (or
possibly both) cells referenced contain TEXT. Since we know that J3 contains
numeric 6 then the problem has to be with cell E19.

--
Biff
Microsoft Excel MVP


"Steve" wrote in message
...
I do believe the "Days" will always be present, and your formula works to
produce either the single #, tens, or hundreds, but when I use that result
as
a divisor, it still produces a #value!
E.g. Formula result is 6(in J3).... e19/j3 = #value!

"T. Valko" wrote:

One way...

Assuming the word "Days:" is *always* present.

=--MID(A1,SEARCH("Days:",A1)+5,5)

--
Biff
Microsoft Excel MVP


"Steve" wrote in message
...
I have this data that is pasted to my worksheet:

Period: 01/23/2010 to 01/29/2010 - Days: 6
It is in general format. The day numbers will always be in the last,
last
2
, or last 3 positions.
Depending on the period, the days may be any # up to the 365.
I need to use that day #, be it 6, or 25, or 147, whatever, as a
divisor
of
another number. E.g. if 6, then 1234 / 6 = 206
if 25, then 5678 / 25 = 227, etc/
The variable #'s will be in the same cell, so I basically want: E29/6
or
E29/25, or E29/147, etc.
Basically, those general formatted #'s converted to a usable/divisable
#.

Thanks,

Steve




.



.

What EXACTLY is in A1 (or whatever cell it is that you are looking at)?

Maybe try this, which assumes you are just subtracting the first date from the
second date to get "Days":

=MID(A1,FIND(" to ",A1)+4,11)-MID(A1,FIND(" to ",A1)-10,11)

As for your other problem, put this in a blank cell and tell us what the result is:

=ISNUMBER(E19)


Steve wrote:
This produced :6
I changed the 3's to 2's and got the 6, but I don't think my fix will
account for 2 or 3 digit numbers, will it ? Same as other, taking a number in
cell E19, and dividing by the resulting 6, still produces a #value!


"Glenn" wrote:

Yes, A1 should have "Period: 01/23/2010 to 01/29/2010 - Days: 6". You
probably have trailing spaces in your data, which could cause the #Value! error.
Try this instead:

B1 =--RIGHT(SUBSTITUTE(TRIM(A1)," ",REPT(" ",3)),3)

Steve wrote:
I'm getting a #value!
A1 is where my general format data is, correct ?

"Glenn" wrote:

Steve wrote:
I have this data that is pasted to my worksheet:

Period: 01/23/2010 to 01/29/2010 - Days: 6
It is in general format. The day numbers will always be in the last, last 2
, or last 3 positions.
Depending on the period, the days may be any # up to the 365.
I need to use that day #, be it 6, or 25, or 147, whatever, as a divisor of
another number. E.g. if 6, then 1234 / 6 = 206
if 25, then 5678 / 25 = 227, etc/
The variable #'s will be in the same cell, so I basically want: E29/6 or
E29/25, or E29/147, etc.
Basically, those general formatted #'s converted to a usable/divisable #.

Thanks,

Steve


One way:

=--TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",10)),10))
.

.