cornelombaard Posted January 7, 2009 Share Posted January 7, 2009 The following code gives me the error at the bottom of the page. What's wrong and how do I fix it? The book does not explain properly. Does anyone know what is the best book for learning PHP MySQL and Javascript for the web? The code: <html> <head> <title> </title> </head> <body> <pre> <?php $connection = mysql_connect("localhost", "corne", "c"); mysql_select_db("winestore", $connection); $result = mysql_query("select winery_name, phone, fax from winery limit 20", $connection); while($row = mysql_fetch_array($result)) { print "The {$row["Winery Name"]} Winery's fax is {$row["fax"]} Their phone is {$row["phone"]}. \n"; } ?> </pre> </body> </html> The Error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/html/testphp/database.php on line 15 Also what is MYSQ_NUM? Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted January 7, 2009 Share Posted January 7, 2009 The error means that your mysql_query() failed and it returned a FALSE value instead of a result resource. See the section in the php manual for a complete description of mysql_query mysql_query can fail either due to a problem with your php/mysql configuration, a problem with the mysql server, a problem with the selected database, or due to a syntax error in your query. To get php/mysql to tell you why it failed use some error checking logic. For the first possibility (php/mysql configuration) add the following two lines of code immediately after your first opening <?php tag - ini_set ("display_errors", "1"); error_reporting(E_ALL); For the rest of the possible reasons add an or die(mysql_error()); statement to the end of your mysql_connect, mysql_select_db, and mysql_query() statements - $connection = mysql_connect("localhost", "corne", "c") or die('Could not connect to database server ' . mysql_error()); mysql_select_db("winestore", $connection) or die('Could not select database ' . mysql_error()); $result = mysql_query("select winery_name, phone, fax from winery limit 20", $connection) or die('Query failed ' . mysql_error()); Quote Link to comment Share on other sites More sharing options...
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