mysql_fetch_array code de problème ne pas traiter

Question

Ma page s'arrête chargé chaque fois que j'allume le code ci-dessous ... il semble correct et les tables et les champs sont corrects.

<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> 

Answer 1

<?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"; 
    } 
?> 

Answer 2

Essayez ceci:

<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>