Jackthumper Posted November 22, 2010 Share Posted November 22, 2010 @$pet = mysql_fetch_array(mysql_query("SELECT * FROM pets)); Quote Link to comment Share on other sites More sharing options...
jim_keller Posted November 22, 2010 Share Posted November 22, 2010 It's probably best to describe your problem in more detail, but my first guess is that the @ sign shouldn't go before the variable name. if you really need to use it (you probably don't), put it before @mysql_fetch_array(). Quote Link to comment Share on other sites More sharing options...
Jackthumper Posted November 22, 2010 Author Share Posted November 22, 2010 it show my "Array". Quote Link to comment Share on other sites More sharing options...
Pikachu2000 Posted November 22, 2010 Share Posted November 22, 2010 Some context would be helpful. Quote Link to comment Share on other sites More sharing options...
Jackthumper Posted November 22, 2010 Author Share Posted November 22, 2010 heres the page http://toxicpets.co.uk/search.php it just dosent get the stuff from the database Quote Link to comment Share on other sites More sharing options...
litebearer Posted November 22, 2010 Share Posted November 22, 2010 As a simple test, try changing this @$pet = mysql_fetch_array(mysql_query("SELECT * FROM pets)); to this $query ="SELECT * FROM pets"; $result = mysql_query($query); while($row = mysql_fetch_array($result)) { echo $row[0] . " - " . $row[1] . "<br>"; } Quote Link to comment Share on other sites More sharing options...
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