Simple SQL Problem

rpupkin77

Can someone tell me why my database is giving me an error on this sqript..

select * from ads where client LIKE '$client'

This translates into:

select * from ads where client LIKE 'Millbrook'

I haven't got a clue what is wrong.

Thanks.

😕

riffy

what are you exactly trying to do and what is the error it's given you???

can you elaborate a little more??

rpupkin77

The actual PHP page generates this error:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/nemp3/allenrochegroup-www/admanager/searchresults.php on line 100

The MySql interface on my webserver simply tells me I have an error in my code near 'LIKE 'client'.

Not much help, I have done queries like this in the past (http://www.greenbrookauto.com/searchnew.htm) and I am fairly adept at PHP and MYSQL (for someone whose been using it for 3 months), I can't figure the issue, Sometimes I think my webhost has MySql issues.

steved

Normally when you use LIKE, you would also use wildcards. Try this:

select * from ads where client LIKE '%$client%'

ratass2002

What have you tried to test using that SELECT statement.

You syntax is correct, based on the documentation at www.mysql.com.

Does it work using a literal string in place of your variable?
if not then there is something wrong with the syntax.

Otherwise print your variable before it's used in the SELECT statement, see if there is anything wrong with it.

Does any SELECT with LIKE works?

rpupkin77

Yes they do...sometimes... I am using the same host for various sites, On the mysql interface they provide, it does not work EVER, I do not know why, however when i place them in a page and use PHP to generate the query it usually does. It is odd and confusing and nobody at the webhost really knows a lot about MySql. I will try the Wild Cards.