Article: Q161730
Product(s): Microsoft FoxPro
Version(s): WINDOWS:5.0,5.0a,6.0
Operating System(s):
Keyword(s): kbprogramming kbvfp kbvfp500aBUG kbvfp500bugkbbuglist
Last Modified: 04-FEB-2000
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The information in this article applies to:
- Microsoft Visual FoxPro for Windows, versions 5.0, 5.0a, 6.0
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SYMPTOMS
========
In Visual FoxPro 3.0 the selected item in a list box can be initialized by
placing the following code in the Form Init or the List box Init:
THISFORM.List1.Selected(1) = .T.
This does not work in Visual FoxPro version 5.0.
RESOLUTION
==========
Place the following code in the Activate method of the form. The variable acts
like a switch so that the code is executed only when the form is first
instantiated. The variable switch should be added to the form as a property.
Switch is initialized to false.
IF !THISFORM.switch
THISFORM.List1.Selected(1)= .T.
THISFORM.switch = .T.
ENDIF
STATUS
======
Microsoft has confirmed this to be a problem in the Microsoft products listed at
the beginning of this article.
MORE INFORMATION
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Steps to Reproduce Behavior
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1. Create a form and add the employee table to the DataEnvironment.
2. Add a list box to the form and set its RowSourceType to 6-Fields. Set its
RowSource to employee.emp_id.
3. Place the following code in the Init of the form or the Init of the list
box.
THISFORM.List1.Selected(1)= .T.
Note that when the form is run, the first item is not selected.
Additional query words: kbvfp500 kbvfp500a kbvfp600
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Keywords : kbprogramming kbvfp kbvfp500aBUG kbvfp500bug kbbuglist
Technology : kbVFPsearch kbAudDeveloper kbVFP500 kbVFP600 kbVFP500a
Version : WINDOWS:5.0,5.0a,6.0
Issue type : kbbug
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