Re: Passing on an "ID"
| From: | David Robley | Date: | Mon, 22 Apr 2002 11:35:34 +0000 |
| Subject: | Re: Passing on an "ID" | ||
| References: | 1 2 3 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-18722@lists.php.net to get a copy of this message | ||
In article <20020422105225.76797.qmail@pb1.pair.com>,
afrancis@camerondesign.co.uk says...
>
> The button "Email Comments" links to the following page:
>
> <?include ("../config.inc.php"); ?>
>
> <?php
>
> // setup SQL statement to retrieve link that we want to edit
> $SQL = " SELECT * FROM $tablename1 ";
> $SQL = $SQL . " WHERE id = $id ";
>
> // execute SQL statement
> $ret = mysql_db_query($dbname,$SQL,$link);
>
> // retrieve values
> $row = mysql_fetch_array($ret);
> $email = $row["email"];
> $title = $row["title"];
> $comments = $row["comments"];
> $title = $row["title"];
>
> else {echo(mysql_error());
>
> I get an error at $row = mysql_fetch_array($ret);
Wiyhout seeing the message this is a guess, but have you considered using
mysql_error to check the status of your query?
--
David Robley
Temporary Kiwi!
Quod subigo farinam