Ok I am looking at the code an I'm thinking it has to be in the code because
I tried it on my box at home and it did the same thing.

So while looking hard at the code I came up with this:
if($searchword == $iname) this is the problem

if $searchword == $iname then it will echo everything I want it to.
What I needed to do is have it where it isn't == to anything in otherwords a
null value.

So if $searchword = $iname then it will work as it is suppose to.

Thanks everyone for your time, effort and help.
Jennifer



"Jennifer Downey" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Hi all,
>
> I have been trying to build a search script for my site that deals with
only
> one table in my db.
> As Julie Meloni pointed out look in the MySQL manual for LIKE clauses I
> can't seem to locate that clause in ether
> manual.
> Dan Brunner gave me this to go on:
>
> $query = "SELECT uid, id, image, iname, quantity, type FROM
> {$config["prefix"]}_shop WHERE  iname  LIKE = '%$shopsearch% ORDER BY
> iname'";
> $ret = mysql_query($query);
> while(list($quantity)=mysql_fetch_row($ret))
>
> But I can't seem to get this to work. I have never worked with LIKE before
> and would appreciate any help on this.
>
> Thanks Julie and Dan for your time and effort.
>
> Thanks all for your time and help
> Jennifer
>
>
> --
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>
>
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