PHP mySQL UPDATE script wont execute

Question

I can't understand why this will not be executed into the mySQL database? Everything is set and all it does is to update some values.

    <?php
$con=mysqli_connect(This is filled.);
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

mysqli_query($con,"UPDATE `members` SET `rank`='$_POST[rank]' WHERE `id`='$_POST[id]'");


mysqli_close($con);
header("Location: index.php");
?>

Heres the thing that should fill the $_POSTs

<form action="setrank.php" method="post">
User: 
  <select>
      <?php
  $result = mysqli_query($con,"SELECT * FROM members ORDER BY id");

while($row = mysqli_fetch_array($result))
  {
  echo "<option name='id' id='id' value='" . $row['id'] . "'>" . $row['username'] . " (" . $row['id'] . ")</option>";
  }
?>
  </select><br />
Set rank to: 
    <select><option name="rank" id="rank" value="0" >Guest.</option><option name="rank" id="rank" value="1" >Moderator.</option><option value="2" name="rank" id="rank">Administrator.</option><option value="3" name="rank" id="rank">Owner.</option></select>
<br /><input type="submit">
</form>

Thank you very much for help, i've been using hours. :/

The name attribute belongs on <select name='rank'> not on <option name='rank'>, same for id You should be checking that the form was posted before attempting to do the UPDATE, and most importantly, the code is vulnerable to SQL injection. It is important to start learning how to use mysqli::prepare() to create prepared statements, safe from SQL injection. — Michael Berkowski
– Michael Berkowski, Commented Mar 8, 2014 at 3:02

Girish · Accepted Answer · 2014-03-08 03:02:37Z

1

The form select element define incorrect, add form field name in select box not an options please try this way

<select name="id"> 
 <option value="xyz">Name</option>
</select>

answered Mar 8, 2014 at 3:02

Girish

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1 Comment

Girish Over a year ago

@KinJacob, ok good, if you answer useful then please correct.

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