using Wild Card for ADO recordset Find in Visual Basic to a SQL 2000

[mgardiner]

I am trying to run a Find on a ADO recordset using
an SQL syntax for the Find.
I need to run for a WildCard Character.
here is my code.
----------------------------------------------------------------

Dim adoQuoteRS As Recordset
'create a recordset to search on
Dim strsql As String
Set adoQuoteRS = New Recordset
strsql = "SELECT tblQuotes.*, tblQuoteLog.*, tblProjects.* " & _
"FROM tblProjects " & _
"INNER JOIN tblQuotes ON tblProjects.ProjectsID = tblQuotes.ProjectID " & _
"INNER JOIN tblQuoteLog ON tblQuotes.QuoteID = tblQuoteLog.QuoteID"

adoQuoteRS.Open strsql, g_objConn, adOpenStatic

If FindLookIn = "QuotationNo" Then
'If FindLookIn = "Entire Database" Then
With adoQuoteRS
.MoveFirst
'.Find (FindLookIn & "= '" & findvalue & "'&quot
.Find ("QuotationNo = '%10881'&quot
'WHERE (QuotationNo LIKE '%10881')
If .BOF Then
MsgBox "Matching Record not Found"
End If
End With
End If

adoQuoteRS.Close
---------------------------------------------------------------

I cannot get VB to accept the LIKE as valid If I just look for the
entire value then I can find it with out using a wildcard character.
any help would be appreciated.

Also when I open my recordset in dynamic mode I get a recordcount of -1 ????

 

[rajeevnandanmishra]

Hi there,
I don't know much about VB but try to use
.Find ("INSTR(QuotationNo,'10881',1)&quot

I don't know if it helps you

 

[mgardiner]

Thanks for your quick response.
This uses a VBA style string search.
I tried it with my recordset .find and it didn't work.
Thanks for all your help though.
Morgan

 

[RJ5]

Hi,

Try replacing the "=" sign with a LIKE.

.Find ("QuotationNo LIKE '%10881'&quot

This might do the trick

Randy : )

 

[mgardiner]

Thanks RJ5
I tried the % in different places and came up with mixed results.
I don't know why yet but I think I can work around it.

.find ("QuotationNO LIKE '%10881%'&quot
Works fine

.find ("QuotationNO LIKE '10881%'&quot
Works fine to find the number and anything behind the text

.find ("QuotationNO LIKE '%10881'&quot
gives me a Error 3001 arguments are of the wrong type.

I don't know why this would be.
but %10881% will have to work for me for right now until someone can
explain it to me.
Thanks Morgan