> <?php > //.../... first part of the code is to connect to the right DB on a MySQL > server. It works fine > //This code to show the query. It runs well under MySQL and gives 1 result : > $query="SELECT Login, Password FROM `user` WHERE user.login=''$login'' and > user.password=''$password''"; > //This var should contain a query result ressource: > $result_query=mysql_query($query); > //Here's my error message after executing: > // Supplied argument is not a valid MySQL result resource in this line: > $result_table=mysql_fetch_array($result_query); > // I thought that this var ($result_table) should contain the row of 2 cells > //And I expected to read $result_table['login'] or $result_table[0] as the > first cell of the row... > // I any of you can help me, poor php newby... :o) Great thanks !! > ?>
=Please check the archives for advice about checking the result of every mysql_...() call. It's much better to know/be told by PHP/MySQL than to say "I thought that...I expected to ..." =the $query assignment statement contains multiple double-quotes. These cannot be 'nested'. Use a mixture of single and double-quotes or 'escape' the inner set(s). =also (and this may be a function of our email packages not a PHP thing) are those single quotes around user? =dn -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
