Question

i wrote in file showList.php the following form, which select items from the database and show them in drop-down list:

<form id="selForm" name="selForm" action="index.php" method="post">
<select name="selection" id="selection">
<option id="nothingSelected" >--Choose form---></option>
<?php

$con=mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("myDatabase",$con);
$result = mysql_query("SELECT * FROM formsTable");

while($row = mysql_fetch_array($result))
  {
  $selection_id=$row['id'];
if($_POST['selection']==$selection_id)$selElement="selected";
  echo "<option  id='$selection_id' name=\"sectionid\"  value='$selection_id' >";
  echo $row['nummer'] . " " . $row['titel']. " ";
  echo "</option>";
  }
?>

</select>
<input type="button" value="load form" onClick="validateForm(document.selForm)">
<input type="button" value="delete form" onClick="deleteForm(document.selForm);">
</form>

I include this file in index.php as follows:

<?php include('showList.php');?>

Now when I call index.php, a list of found forms will be displayed in a drop-down list.

This works fine in firefox, my Problem is when I call index.php in internetexplorer, I get the following error:

Notice: Undefined index: selection in C:\path\showList.php on line 43

Line 43 is:

if($_POST['selection']==$selection_id)$selElement="selected";

as you can see in the form above.
Any idea?

Answer 1

You need to change the problem line from:

if($_POST['selection']==$selection_id)$selElement="selected";

to:

if(isset($_POST['selection']) && ($_POST['selection']==$selection_id))
    $selElement="selected";

to check that a value for (as @b1onic suggested).

Obviously nothing will be POSTed the first time the form is shown in the browser – whichever browser you are using – so you will get that error.