>Hi
>I'm a junior programmer looking for some help. I have a form, the control source for which is a table(customerinfo) in my project. On this form I need a control which when clicked launches an image file. The path to the image file is located in the "image" field of customerinfo table, I want to access the path via an event
>I've been trying to come up with something based on the information from the Programmer's Guide, but the solution eludes me.
>Windows 98
>VFP 6.0
>Any help you can give is appreciated.
>Thanks,
>DG

We do precisely that here, we have a table that references some technical drawings. When the user wishes to view the drawing then they just click a button on the form to call their image viewer with the drawing they requested. The button click method code looks as follows:

* We have a Drawing and the user has a tif viewer so lets call riscrs to view the drawing
THISFORM.RiscrsUtilities.ViewRevision(docacmds.docnumber, docacmds.docrevision)

This above method calls upon a class that is on the form called RiscrsUtilities it is calling a method called ViewRevision which needs as properties the document number and revision.

The viewrevision method basically opens the table that holds where the drawing is physically located. If finds the drawing number and revision in the library table. Once the record is found then the path is extracted to the TIF image. The full path is passed to another class method that calls up the users image viewing program and displays the drawing to the user.

Hope this helps