PHP: Undefined index in $_POST

Question

I am using forms select. I just want to check what user selects by echo-ing the result on the same page so I kept the action="". But its showing error undefined index slct. Can any one please help me

<form action="" method="post">
<select name="slct">
<option value="yes" selected="selected"> yes </option>
<option value="no"> no </option>
</select>
<input type="button" value="Submit" />
</form>


<?php 
$tofd = $_POST["slct"];
echo $tofd; 
?>

Why its showing the error

Notice: Undefined index: slct in C:\wamp\www\Univ Assignment\Untitled-4.php on line 21

That will display value if form is submitted. btw good luck for your assignment. ;) — Riz
– Riz, Commented Sep 21, 2012 at 11:40
possible duplicate of PHP: "Notice: Undefined variable" and "Notice: Undefined index" — Jocelyn
– Jocelyn, Commented Mar 29, 2013 at 11:56

Alaa Gamal · Accepted Answer · 2012-09-21 11:41:30Z

1

You should use button type submit NOT button

<input type="submit" value="submit" />

And then test IT like

echo (isset($_POST['slct']))? $_POST['slct'] : 'Variable undefined..';

answered Sep 21, 2012 at 11:41

Alaa Gamal

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Comments

Baba · Accepted Answer · 2012-09-21 11:39:41Z

0

Use PHP isset to check if its exist first

Example :

$tofd = isset($_POST["slct"]) ? $_POST["slct"] : null ;

Example 2 Using a function

function __POST($var)
{
    return  isset($_POST[$var]) ? $_POST[$var] : null ;
}

$tofd = __POST("slct");

answered Sep 21, 2012 at 11:39

Baba

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Comments

codingbiz · Accepted Answer · 2012-09-21 11:40:44Z

0

If they are on the same page, initaially, $_POST would be empty because the user has not posted anything. So you have to handle that.

if(isset($_POST["slct"]))
    $tofd = $_POST["slct"];

answered Sep 21, 2012 at 11:40

codingbiz

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Comments

Afshin · Accepted Answer · 2012-09-21 13:58:16Z

0

<?php
  if (isset($_POST["slct"])){
  $tofd = $_POST["slct"];
  echo $tofd; }
?>

answered Sep 21, 2012 at 13:58

Afshin

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