Re: Search Script
| From: | Jennifer Downey | Date: | Thu, 25 Apr 2002 21:28:23 +0000 |
| Subject: | Re: Search Script | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-18815@lists.php.net to get a copy of this message | ||
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" <jendowney@attbi.com> wrote in message
news:20020425141620.75561.qmail@pb1.pair.com...
> 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
>
>
> --
> The sleeper has awaken
>
>
> ---
> Outgoing mail is certified Virus Free.
> Checked by AVG anti-virus system (http://www.grisoft.com).
> Version: 6.0.351 / Virus Database: 197 - Release Date: 4/19/2002
>
>
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.351 / Virus Database: 197 - Release Date: 4/19/2002