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Type ahead function for listbox in vb6

Posted on 2004-08-20
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Last Modified: 2013-12-25
Anyone know how to do a type ahead for a listbox in vb6?

ex. The listbox contains:

Acorn
Apple
Banana

When you type A then P it should change the focus to Apple.
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Question by:n0ell
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5 Comments
 
LVL 4

Expert Comment

by:llcooljayce
ID: 11856058
0
 
LVL 5

Accepted Solution

by:
gary_j earned 1000 total points
ID: 11856210
I add a "navigator" text box, then do this in the textbox change event:

dim i as integer

if list1.listcount > 0 then
     for i = 0 to list1.listcount - 1
           if left(list1.list(i),len(trim(text1.text))) = trim(text1.text) then
                 list1.listindex = i
                 exit for
           endif
     next 'i
endif
0
 

Author Comment

by:n0ell
ID: 11869938
Hi gary j,

Thanks for responding to my question. I tried to implement this code and it worked when i type on the navigator text box however i want to implement this using only the Listbox so that when i type on it, it searches for a match in my list. What i did was to hide the navigator text box and then passed the search criteria using the keypressed event of my listbox. For some reason that i still have not figured out yet, it jumps an extra line after it finds a match in my list. e.g.

abc
abcappend
acc

When I use the text "abca" as my search criteria, the application is able to find abcappend but then jumps to the next one (acc)

here are the code that i used :
'code for my listbox
Private Sub lstMed_KeyPress(KeyAscii As Integer)
   If KeyAscii = 13 Then
        Text1.Text = ""
    ElseIf KeyAscii = 8 Then
        If Text1.Text <> "" Then
            Text1.Text = Left(Text1.Text, Len(Text1.Text) - 1)
        End If
    Else
         Text1.Text = Text1.Text & Chr(KeyAscii)
    End If
End Sub
 
'code for hidden navigator textbox
Private Sub Text1_Change()
If lstMed.ListCount > 0 Then
     For i = 0 To lstMed.ListCount - 1
           If Left(lstMed.List(i), Len(Trim(Text1.Text))) = Trim(Text1.Text) Then
                 lstMed.ListIndex = i
                 Exit For
           End If
     Next 'i
End If
End Sub
 
Thanks.
0
 
LVL 5

Expert Comment

by:gary_j
ID: 11870213
I tested this and I think it's because your lstMed_KeyPress event is still in "control" -- in other words, with the keystrokes you're making, the normal operation of that event will move focus to index 2.  That normal operation as you leave the keypress event is over-riding what you're doing in the text1 change event.  
0
 

Author Comment

by:n0ell
ID: 11870585
Gary J,

Thanks for the speedy reply. What I did was to put the text1_change code and put them in the lstMed_KeyPress function and just set the keyascii = 0 so that it will not override the the setting of the list index. It's now working.

Here's the update code:
Public Searchkey as string

Private Sub lstMed_KeyPress(KeyAscii As Integer)
     If KeyAscii = 13 Then
        searchkey = ""
    ElseIf KeyAscii = 8 Then
        If searchkey <> "" Then
            searchkey = Left(searchkey, Len(searchkey) - 1)
        End If
    Else
         searchkey = searchkey & Chr(KeyAscii)
    End If


    Dim i As Integer
   
    If lstMed.ListCount > 0 Then
         For i = 0 To lstMed.ListCount - 1
               If Left(lstMed.List(i), Len(Trim(searchkey))) = Trim(searchkey) Then
                     lstMed.ListIndex = i
                     Exit For
               End If
         Next 'i
    End If
    KeyAscii = 0
End Sub

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