using the keyword "LIKE" with SELECT not working

  • Thread starter Thread starter Guest
  • Start date Start date
G

Guest

Hi,
I have created a db with the following contents. Table Name: "EMPLOYEE"

LASTNAME FIRSTNAME AREACODE PHONE ST ZIP
--------------- ---------------- -------- -------- -- ------
BUNDY AL 100 555-1111 IL 22333
MEZA AL 200 555-2222 UK 234324

when i give the query SELECT * FROM EMPLOYEE WHERE LASTNAME LIKE 'M%'
it doesnt return the MEZA record.

but the query SELECT * FROM EMPLOYEE WHERE LASTNAME LIKE 'MEZA'
runs fine..

I dont see any problem with the first query. Someone please clarify.

Thanks,
Vijender Reddy


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pacman wrote in message
Hi,
I have created a db with the following contents. Table Name: "EMPLOYEE"

LASTNAME FIRSTNAME AREACODE PHONE ST ZIP
--------------- ---------------- -------- -------- -- ------
BUNDY AL 100 555-1111 IL 22333
MEZA AL 200 555-2222 UK 234324

when i give the query SELECT * FROM EMPLOYEE WHERE LASTNAME LIKE 'M%'
it doesnt return the MEZA record.

but the query SELECT * FROM EMPLOYEE WHERE LASTNAME LIKE 'MEZA'
runs fine..

I dont see any problem with the first query. Someone please clarify.

Thanks,
Vijender Reddy


----------------
This post is a suggestion for Microsoft, and Microsoft responds to the
suggestions with the most votes. To vote for this suggestion, click the "I
Agree" button in the message pane. If you do not see the button, follow this
link to open the suggestion in the Microsoft Web-based Newsreader and then
click "I Agree" in the message pane.

http://www.microsoft.com/office/com...3b3e8502d6&dg=microsoft.public.access.queries
Click to expand...

The usual wildcard in Access is * not %. % is used for instance when
working with ADO.
 
yes thats right.. it works. thanks!

RoyVidar said:
pacman wrote in message


The usual wildcard in Access is * not %. % is used for instance when
working with ADO.
Click to expand...
 
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