RE: [PHP-DB] Warning - newby question -- $_GET
| From: | John W. Holmes | Date: | Thu, 07 Nov 2002 02:19:42 +0000 |
| Subject: | RE: [PHP-DB] Warning - newby question -- $_GET | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-23070@lists.php.net to get a copy of this message | ||
> I'm trying to call a script with the "?=" after the php script name in
the
> URL, but can't seem to pick up the variable to use in my script...
> any help would be greatly appreciated :)
>
> I'm calling the script with this url:
> http://localhost/Data/test/dbconnect2c.php?id=TestPage
>
> The section of the script i'm using is:
> $id = $_GET['id'];
> $sql="SELECT * FROM content where title=$id";
> $result=mysql_query($sql,$db);
Well, for your example, $id is going to equal "TestPage", so you are
making an invalid query: where title=TestPage. If you echo
mysql_error(), it'll probably say something about unknown column
TestPage.
At the very least, use quotes:
where title='$id'
> Error message returned:
> Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
> result
> resource in...
This means your query failed. Echo mysql_error() to determine why. Get
used to using mysql_error for good debugging.
---John Holmes...