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From: Richard on 16 Mar 2010 19:35
I use the below code to filter my form. This filter matches any where in the
phone string. What I am looking for is the exact match. Thanks for any help.

Private Sub cmdPhoneNumber_Click()
' Find the record that matches the control.
Dim stFilter As String

Me.FilterOn = False

stFilter = "[Phone] Like '*" & Me.[txtPhoneNumber] & "*'"


Me.Filter = stFilter
Me.FilterOn = True

End Sub
From: Mr. B on 16 Mar 2010 20:11
You need to remove the Like and "*" from the equation

stFilter = "[Phone] = " & Me.[txtPhoneNumber] & "'"

-----
HTH
Mr. B
http://www.askdoctoraccess.com/
Doctor Access Downloads Page:
http://www.askdoctoraccess.com/DownloadPage.htm


"Richard" wrote:

> I use the below code to filter my form. This filter matches any where in the
> phone string. What I am looking for is the exact match. Thanks for any help.
>
> Private Sub cmdPhoneNumber_Click()
> ' Find the record that matches the control.
> Dim stFilter As String
>
> Me.FilterOn = False
>
> stFilter = "[Phone] Like '*" & Me.[txtPhoneNumber] & "*'"
>
>
> Me.Filter = stFilter
> Me.FilterOn = True
>
> End Sub
From: Richard on 16 Mar 2010 23:39
I changed the code to your suggestion but I now get a run-time error '2448'
"You can't assign a value to this object". Also the code breaks on the
following line:

Me.Filter = stFilter

Richard


"Mr. B" wrote:

> You need to remove the Like and "*" from the equation
>
> stFilter = "[Phone] = " & Me.[txtPhoneNumber] & "'"
>
> -----
> HTH
> Mr. B
> http://www.askdoctoraccess.com/
> Doctor Access Downloads Page:
> http://www.askdoctoraccess.com/DownloadPage.htm
>
>
> "Richard" wrote:
>
> > I use the below code to filter my form. This filter matches any where in the
> > phone string. What I am looking for is the exact match. Thanks for any help.
> >
> > Private Sub cmdPhoneNumber_Click()
> > ' Find the record that matches the control.
> > Dim stFilter As String
> >
> > Me.FilterOn = False
> >
> > stFilter = "[Phone] Like '*" & Me.[txtPhoneNumber] & "*'"
> >
> >
> > Me.Filter = stFilter
> > Me.FilterOn = True
> >
> > End Sub
From: Duane Hookom on 16 Mar 2010 23:52
It looks like there is a missing single quote after Phone =. Try:

stFilter = "[Phone] = '" & Me.[txtPhoneNumber] & "'"

--
Duane Hookom
MS Access MVP

"Richard" <Richard(a)discussions.microsoft.com> wrote in message
news:53923D75-FDB0-4093-85A3-C5C7628CC404(a)microsoft.com...
> I changed the code to your suggestion but I now get a run-time error
> '2448'
> "You can't assign a value to this object". Also the code breaks on the
> following line:
>
> Me.Filter = stFilter
>
> Richard
>
>
> "Mr. B" wrote:
>
>> You need to remove the Like and "*" from the equation
>>
>> stFilter = "[Phone] = " & Me.[txtPhoneNumber] & "'"
>>
>> -----
>> HTH
>> Mr. B
>> http://www.askdoctoraccess.com/
>> Doctor Access Downloads Page:
>> http://www.askdoctoraccess.com/DownloadPage.htm
>>
>>
>> "Richard" wrote:
>>
>> > I use the below code to filter my form. This filter matches any where
>> > in the
>> > phone string. What I am looking for is the exact match. Thanks for any
>> > help.
>> >
>> > Private Sub cmdPhoneNumber_Click()
>> > ' Find the record that matches the control.
>> > Dim stFilter As String
>> >
>> > Me.FilterOn = False
>> >
>> > stFilter = "[Phone] Like '*" & Me.[txtPhoneNumber] & "*'"
>> >
>> >
>> > Me.Filter = stFilter
>> > Me.FilterOn = True
>> >
>> > End Sub

From: John W. Vinson on 17 Mar 2010 00:05
On Tue, 16 Mar 2010 20:39:01 -0700, Richard
<Richard(a)discussions.microsoft.com> wrote:

>I changed the code to your suggestion but I now get a run-time error '2448'
>"You can't assign a value to this object". Also the code breaks on the
>following line:
>
>Me.Filter = stFilter
>
>Richard
>
>
>"Mr. B" wrote:
>
>> You need to remove the Like and "*" from the equation
>>
>> stFilter = "[Phone] = " & Me.[txtPhoneNumber] & "'"

Possibly Mr. B had a tiny typo, which might be causing this problem: there's a
missing quotemark. Try

stFilter = "[Phone] = '" & Me.[txtPhoneNumber] & "'"

This will evaluate to

[Phone] = '(222) 222-2222'

which should work correctly as a filter string.

I'm thinking this may be the case because your original LIKE filter string was
(sort of) working....
--

John W. Vinson [MVP]
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