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Home/ Questions/Q 8964501
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T16:36:18+00:00 2026-06-15T16:36:18+00:00

Possible Duplicate: PHP: Notice: Undefined variable and Notice: Undefined index if(isset($_POST[‘bul’])) { $marka1=$_POST[‘marka’]; $model1=$_POST[‘model’];

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Possible Duplicate:
PHP: “Notice: Undefined variable” and “Notice: Undefined index”

if(isset($_POST['bul']))
{
    $marka1=$_POST['marka'];
    $model1=$_POST['model'];

   $que="SELECT * from  otomobil_tablosu WHERE markasi='$marka1' AND modeli='$model1'";
   $res=mysql_query($que);

  while($row = mysql_fetch_row($res)){

  echo "<tr>";
  echo "<td>" . $row['otomobilID'] . "</td>";
  echo "</tr>";

  }

when I want to print the row of table there is undefined index warning… I could not find the problem

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  1. Editorial Team
    Editorial Team
    2026-06-15T16:36:19+00:00Added an answer on June 15, 2026 at 4:36 pm

    The specific problem is that mysql_fetch_row() returns an indexed array and not an associative. You want mysql_fetch_assoc().

    The more general problem is that you’re using mysql_*() functions, and are highly vulnerable to SQL injection. Please, don’t use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi – this article will help you decide which. If you choose PDO, here is a good tutorial.

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