RE: [PHP] passing variables to a sql statement
| From: | John W. Holmes | Date: | Sat, 18 Jan 2003 23:09:40 +0000 |
| Subject: | RE: [PHP] passing variables to a sql statement | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-132131@lists.php.net to get a copy of this message | ||
> hi, i'm hoping someone can help me out here - im trying to pass a
variable
> from a string in the href tag, however my sql breaks when i put a var
in
> the statment.
> when i replace $foo with a number, the page works fine. can someone
please
> tell me what im doing wrong?
>
> here is the error i get:
>
> Warning: Supplied argument is not a valid MySQL result resource in
>
/home/villany2k1/www.villany2k1.com/htdocs/projects/projects_descrip.php
> on line 11
That means your query failed. Use mysql_error() to find out why.
> here is my code:
>
> <?php
> $hostName="localhost";
> $userName="xxxx";
> $password="xxxx";
> $database= "projects";
> $tableName="project_descrip";
>
> $descripQuery = "SELECT descrip from project_descrip where
rowid=$foo";
> mysql_connect($hostName, $userName, $password);
> $descripResult = mysql_db_query($database,$descripQuery);
> while($row = mysql_fetch_object($descripResult)){
You shouldn't be using mysql_db_query() for one, it's depreciated. Use
mysql_select_db() along with mysql_query().
If rowid is an integer, then you don't need quotes around $foo,
otherwise, if it's a string, then you do.
where rowid='$foo'
---John W. Holmes...
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