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?
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.
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