mysql_fetch_array problem code not processing

Question

My page stops loaded every time I turn on the code below... it looks correct and the tables and fields are correct.

<select name="common" style="width: 136px;">    
<?php
    $group1 = mysql_fetch_array(mysql_query("SELECT country FROM lang_list WHERE grouping = '1' ORDER BY p_order"));
    while($row = $group1){
        echo "<option value=\"$group1\">$group1</option>\n";
    }
?>
</select>

Sabeen Malik · Accepted Answer · 2010-10-14 09:38:17Z

2

<?php
    $group1 = mysql_query("SELECT country FROM lang_list WHERE grouping = '1' ORDER BY p_order");
    while($row = mysql_fetch_array($group1)){
        echo "<option value=\"$row[country]\">$row[country]</option>\n";
    }
?>

answered Oct 14, 2010 at 9:38

Sabeen Malik

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PHPology · Accepted Answer · 2010-10-14 09:38:57Z

1

Try this:

<select name="common" style="width: 136px;">     
<?php 
        $recordset = mysql_query("SELECT country FROM lang_list WHERE grouping = '1' ORDER BY p_order") or die("Error found: " . mysql_error());
    while($row = mysql_fetch_array($recordset)){ 
        echo "<option value=\"".$row['country']."\">".$row['country']."</option>\n"; 
    } 
?> 
</select>

answered Oct 14, 2010 at 9:38

PHPology

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