duangan4406 2014-04-09 18:17
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PHP注意:未定义的偏移量:1

i fetch MySql result using PHP array using this Code:

$sql = 'SELECT * FROM ' . LOCATION . ' ORDER BY category ';
$r = $db->query ( $sql ) or error ('Critical Error', mysql_error () );

while ($ROW = $db->fetcharray($r))
{
     if ($ROW[1] == '') //line 15
          $ROW[1] = $ROW['subcategory'];
} 

fetcharray function:

function fetcharray ($query_id)

    {
        if(!$query_id)
        {
            $query_id = $this->query_res;
        }

        if($query_id)
    {
        $id = (int) $query_id;

        $this->q_array[$id] = @mysql_fetch_array($query_id,MYSQL_ASSOC); // LINE 124
        return $this->q_array[$id]; 
    }

        else

        {
            return false;
        }
    }

I see this error:

Notice: Undefined offset: 1 in C:\xampp\htdocs\script\state.php on line 15

I change if ($ROW[1] == '') to if (isset($ROW[1])) but I see again error.

NOTE: i remove **MYSQL_ASSOC** from fetcharray function and fix error. i think my problem with fetcharray function how to fix this?

How I can Fix this error?

  • 写回答

2条回答 默认 最新

  • duanchui1279 2014-04-09 18:34
    关注

    This is not directly answer to your question BUT you implemented some kind of wrapper for old and deprecated functions.

    You should use MySQLi or PDO.

    In addition, you have inconsistent coding style. Once you use small letters for variables, another time you use capital letters.

    Instead of this:

    if ($ROW[1] == '') //line 15
              $ROW[1] = $ROW['subcategory'];
    

    you should use:

    if (!isset($ROW[1])) //line 15
              $ROW[1] = $ROW['subcategory'];
    

    or this:

    if (empty($ROW[1])) //line 15
              $ROW[1] = $ROW['subcategory'];
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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