Want to protect your cyber security and still get fast solutions? Ask a secure question today.Go Premium



Posted on 2004-10-31
Medium Priority
Last Modified: 2008-02-26
I hope this is an easy question for someone.  I am using ADO to access some data.  I need to write a SQL statement to narrow that data down.  I can use the LIKE statement successfully for one option, but not multiple.  Here is some code.

'This string will work
strSQL = "Field1 LIKE '%" & txtOption1 & "%'"

'I need something like this to work
strSQL = "Field1 LIKE '%" & txtOption1 & "&' AND Field2 LIKE '%" & txtOption2 & "%' AND Field3 LIKE '%" & txtOption3 & "%'"

Set rsRecordSet = New ADODB.Recordset
With rsRecordSet
    .CursorType = adOpenStatic
    .CursorLocation = adUseClient
    .LockType = adLockPessimistic
    .Source = "SELECT * FROM Table1 WHERE " & strSQL
    .ActiveConnection = connConnection
End With

I get the follwoing error when I run it with the SQL statement that has multiple LIKE statements in it.  ( I put "test" in txtOption2)

**  Syntex error (missing operator) in query expression 'AND Field2 LIKE '%test%"  **

Thanks for the help!
Question by:APlusComp247
  • 3
  • 2
  • 2
  • +2
LVL 55

Expert Comment

by:Ryan Chong
ID: 12457577

strSQL = "Field1 LIKE '%" & replace(txtOption1,"'","''") & "' AND Field2 LIKE '%" & replace(txtOption2,"'","''") & "%' AND Field3 LIKE '%" & replace(txtOption3,"'","''") & "%'"


strSQL = "Field1 LIKE '%" & replace(txtOption1,"'","''") & "%' AND Field2 LIKE '%" & replace(txtOption2,"'","''") & "%' AND Field3 LIKE '%" & replace(txtOption3,"'","''") & "%'"

LVL 14

Expert Comment

by:Shiju Sasidharan
ID: 12457875
Try this
strSQL = "Field1 LIKE '%" & Replace(txtOption1,"'","''") & "%' AND Field2 LIKE '%" & Replace(txtOption2,"'","''") & "%' AND Field3 LIKE '%" & Replace(txtOption3,"'","''")  & "%'"

u put an extra   &   instead of   %


LVL 14

Expert Comment

by:Shiju Sasidharan
ID: 12457891

sorry i am late
ryancys  deserves it

Concerto Cloud for Software Providers & ISVs

Can Concerto Cloud Services help you focus on evolving your application offerings, while delivering the best cloud experience to your customers? From DevOps to revenue models and customer support, the answer is yes!

Learn how Concerto can help you.


Author Comment

ID: 12457915
I understand what you are getting at, but I haven't used any apostraphes in the textboxes so the Replace statement shouldn't change anything.... right?

I tried that and it didn't work.

Any more ideas??
LVL 18

Accepted Solution

JR2003 earned 2000 total points
ID: 12458184
What is in the value of strSQL when you execute the command?  Can you paste it in here so we can have a look?
Just add the line :
Debug.Print "SELECT * FROM Table1 WHERE " & strSQL
when you set the '.Source' to make it appear in the immediate window.

Which database are you using?
LVL 18

Expert Comment

ID: 12459051
There is a very simple way to do this using the "Microsoft Script Control 1.0". All you need to do is create a string which is the expression you want to evaluate and just let the "Microsoft Script Control 1.0" evaluate it.

You can set the 'strExpression' string to any vb expression e.g.: "3 / 4" or "5 * (6 - 1)" etc.

Private Sub Command1_Click()
    Dim dblResult As Double
    Dim strExpression As String
    Dim MyScripControl As Object
    Set MyScripControl = CreateObject("ScriptControl")

    strExpression = "3 / 4"  '<---  You can put any expression you like here

    MyScripControl.Language = "VBScript"
    dblResult = MyScripControl.Eval(strExpression )
    MsgBox dblResult

End Sub
LVL 18

Expert Comment

ID: 12459055
Please ignore the last comment, wrong question answered!
LVL 28

Expert Comment

ID: 12459597
Try to put fields name into a brackets :
[Field1] Like ...

Author Comment

ID: 12460338
Ok, I made a stupid mistake.  I was using a few IF  statements to create the SQL statement depending on whether data was entered into the textbox or not.  I used the following

strSQL = " AND Field1 LIKE '%" & txtOption1 & "%'"

I quickly figured out after using the debug.print than I needed the following

strSQL = strSQL & " AND Field1 LIKE '%" & txtOption1 & "%'"

I was so unsure of my SQL syntax that I completely forgot a little simple debuging.

Thanks for all the help!!

Featured Post


Modern healthcare requires a modern cloud. View this brief video to understand how the Concerto Cloud for Healthcare can help your organization.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Most everyone who has done any programming in VB6 knows that you can do something in code like Debug.Print MyVar and that when the program runs from the IDE, the value of MyVar will be displayed in the Immediate Window. Less well known is Debug.Asse…
Background What I'm presenting in this article is the result of 2 conditions in my work area: We have a SQL Server production environment but no development or test environment; andWe have an MS Access front end using tables in SQL Server but we a…
Get people started with the utilization of class modules. Class modules can be a powerful tool in Microsoft Access. They allow you to create self-contained objects that encapsulate functionality. They can easily hide the complexity of a process from…
Show developers how to use a criteria form to limit the data that appears on an Access report. It is a common requirement that users can specify the criteria for a report at runtime. The easiest way to accomplish this is using a criteria form that a…

578 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question