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Command Button to Open Form and Filter using Wildcard

Posted on 2013-12-11
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Last Modified: 2013-12-11
We have Office 2010.  In Access 2010 I have a form named "frmDocket".  On this form I have a Text Box named "txtInitials".  I also have a Command Button which opens the form "frmDktAtty".  On the form "frmDktAtty" I have a field named "RoutedTo".

When you click the Command Button on the form "frmDocket" it currently filters with the logic (txtInitials = RoutedTo).

Here is my code:
    DoCmd.OpenForm "frmDktAtty", acNormal, "", _
    "[RoutedTo]LIKE " & "'" & "*" & Me![txtInitials] & "*" & "'", , acNormal

What I really need is the following logic, but I cannot figure out how to alter my code (above) without getting an error.  I've tried so many variations so please don't ask me what I've already tried.

txtInitials = (Like "* [RoutedTo]*")

Can anyone assist me with this.

Thanks in advance!
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Question by:Senniger1
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Accepted Solution

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PatHartman earned 2000 total points
ID: 39711995
Are you really supplying only a partial string?  If you are supplying the actual initials then you should not be using like:

 DoCmd.OpenForm "frmDktAtty", acNormal, "", _
    "[RoutedTo] = '" & Me![txtInitials] & "'", , acNormal

If you use like, then

DoCmd.OpenForm "frmDktAtty", acNormal, "", _
    "[RoutedTo] Like  '*" & Me![txtInitials] & "*'", , acNormal
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Author Comment

by:Senniger1
ID: 39712036
The "txtInitials" field contains the User's initials like LDM, for example.

The "RoutedTo" field contains entries like the following:
   JDP, AXT, LDM
   MEN, LDM, MCP, PCV
   PXT, JDP, JJK
   LDM, JJK

I'm tring to filter so that when LDM is in the "txtInitials" field, then only the records which contain LDM in the RoutedTo field appear.
.
0
 

Author Closing Comment

by:Senniger1
ID: 39712259
I used the following and it worked.

DoCmd.OpenForm "frmDktAtty", acNormal, "", _
    "[RoutedTo] Like  '*" & [txtInitials] & "*'", , acNormal

I realized I had another filter on my form which was causing some of my attempts to fail.

Thanks so much!
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