November 13th, 2006, 02:12 AM
posted to microsoft.public.access.forms,microsoft.public.access.formscoding,microsoft.public.access.modulesdaovba,microsoft.public.access.reports
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conditional macro or code
Thanks for the solution. If I want to get the filename in testit() , as an
example, how can I get? Thanks.
"Allen Browne" wrote in message
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You cannot record macros in Access the way you can in Excel or Word, so if
you have no experience with writing VBA code, this task will be beyond
you.
If you want the user to select the file, the code to open the FileOpen
dialog is he
http://www.mvps.org/access/api/api0001.htm
To perform the import, you need TransferText, TransferSpreadsheet, or
TransferDatabase, depending on the kind of file you want to import.
To open a form so the user can edit it, use OpenForm.
To export the data out again, use the appropriate Transferxxx method from
the 3 above.
--
Allen Browne - Microsoft MVP. Perth, Western Australia
Tips for Access users - http://allenbrowne.com/tips.html
Reply to group, rather than allenbrowne at mvps dot org.
"00KobeBrian" wrote in message
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I want to use a macro or code to do the actions as: Open a open
file dialogue, select the file, then transfer that file to a table, open
an access form based on that table and change some data in the table,
finally transfer that table to another file. Can anyone please advise me
how to accomplish this with conditional macro or module code? Thanks.
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