Re: A very simple MySQL + PHP error
| From: | Hoth | Date: | Sun, 08 Jul 2001 19:20:55 +0000 |
| Subject: | Re: A very simple MySQL + PHP error | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-10151@lists.php.net to get a copy of this message | ||
In my oppinion the error is in the Line:
WHERE title LIKE "%$title%")";
because you have a ")" too much and write the search String a follow:
WHERE title LIKE '%$title%';
important is the symbol ' not " .
I hope this helps you.
bye
Hoth
"Brad Lipovsky" <syzme@home.com> schrieb im Newsbeitrag
news:20010708182828.84349.qmail@pb1.pair.com...
> I get the following error:
>
> //error
> Warning: Supplied argument is not a valid MySQL result resource in
> /web/sites/184/syzme/www.syzme.f2s.com/Commerce_System/search.php on line
22
> //error
>
> when I try to use this code:
>
> <?php
> $Link = mysql_connect ($Host, $User, $Password);
>
> //code
> if ($category) {
> $Query = "SELECT * from $TableName WHERE ((category)=$category)";
> } else { if ($title) {
> $Query = "SELECT * from $TableName WHERE title LIKE "%$title%")";
> }
> }
>
> $Result = mysql_db_query ($DBName, $Query, $Link);
>
> print "Your search query of <b> $category </b> returned the following
> results:<br><br>";
>
> while ($Row = mysql_fetch_array ($Result)) {
> print "<a href=item.php?p=$Row[id]>";
> echo ( $Row[title] );
> print "</a><br>";
> }
>
>
> mysql_close ($Link);
> ?>
> //code
>
> on my site. If i pass in a value for $category, it works fine... the
error
> arises when I try to pass in a value for $title. I want it to search a
> MySQL DB for anything with the string $title in it, and then print it out
in
> the while structure. Thanks
>
> Brad
>
>