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PHP- why does this not select the posted value?

Posted on 2006-06-26
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Last Modified: 2010-04-17


<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>

<body>

<form action="testy.php" method="post">
<select name="listname">
<option value="php3"<? if(isset($_POST['listname']) && $_POST['listname'] == 'php3'){echo  ' selected = "selected" '; }?>> PHP 3</option>

<option value="php4"<? if(isset($_POST['listname']) && $_POST['listname'] == 'php3'){echo  ' selected = "selected"'; }?>> PHP 4</option>

<option value="php5"<? if(isset($_POST['listname']) && $_POST['listname'] == 'php3'){echo  ' selected = "selected"'; }?>>PHP 5</option>

</select>

<input type="submit" name="Submit" value="Submit" />
</form>
</body>
</html>
0
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Question by:roscoeh23
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2 Comments
 
LVL 3

Accepted Solution

by:
AndyWHV earned 500 total points
ID: 16982885
The Problem is the login in the if structure...

look at
if ( (isset($_POST['listname'])) && ($_POST['listname'] == 'php3') )

you check for php3 every time.

Greetz
Andy ;-)
0
 
LVL 2

Expert Comment

by:T3Logic
ID: 16984016
<option value="php3"<? if(isset($_POST['listname']) == 'php3'){echo  ' selected = "selected" '; }?>> PHP 3</option>

<option value="php4"<? if(isset($_POST['listname']) == 'php4'){echo  ' selected = "selected"'; }?>> PHP 4</option>

<option value="php5"<? if(isset($_POST['listname']) == 'php5'){echo  ' selected = "selected"'; }?>>PHP 5</option>
0

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