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mysql_fetch_array问题代码未处理
原标题:mysql_fetch_array problem code not processing
  • 时间:2010-10-14 09:36:13
  •  标签:
  • php
  • mysql

每当我打开下面的代码时,我的页面就会停止加载。。。它看起来是正确的,表和字段也是正确的。

<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>
";
    }
?>
</select>
最佳回答
<?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>
";
    }
?>
问题回答

试试这个:

<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>
"; 
    } 
?> 
</select> 




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