Re: PHP - MySQL Query...

From: Date: Thu, 14 Aug 2003 11:26:43 +0000
Subject: Re: PHP - MySQL Query...
References: 1  Groups: php.general 
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Try this, $result = mysql_query($query,$dbh); -Murugesan ----- Original Message ----- From: "Jay Blanchard" <jay.blanchard@niicommunications.com> To: "Steven Kallstrom" <stevenkallstrom@insightbb.com>; "PHP List" <php-general@lists.php.net> Sent: Thursday, August 14, 2003 4:50 PM Subject: RE: [PHP] PHP - MySQL Query... [snip] $dbh = mysql_connect("localhost", "login", "password") or die('cannot connect to the database because: ' . mysql_error()); mysql_select_db("database"); $query = 'SELECT * FROM cities'; $result = mysql_query($query); while ($row = mysql_fetch_row($result)) { echo ('<tr> <td class="city">' . $row[0] . ', ' . $row[1] . '</td> <td>' . $row[2] . '</td> <td>' . $row[3] . "</td> </tr>\n"); } getting this error: *Warning*: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in... so... is the problem with the $query? I don't see anything wrong (assuming my login and database selection is correct) what are the common errors here? [/snip] mysql_query needs both the query variable and the connection variable.... mysql_query($query, $dbh); HTH! -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php

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