Php Passing values

Question

I have a problem with this code:

    <?php   echo "<meta http-equiv='refresh' content='3;search_enroll.php?&id='.$id />"; ?>

I'm using this code to pass a value from this page to this page with $id and it is not empty I echoed $id and it holds the value. And this is the code on the receiving end:

   <?php
            if (isset($_POST['SearchS'])){
                $id = $_POST['searchstudent'];

            }else if(!empty($_GET['id'])){
                $id = $_GET['id'];
            }
            else if(!empty($_GET['student_id'])){
                $id = $_GET['student_id'];
            }

            else {
                $id= $_REQUEST['student_id']; <--- this is line 37
            }
            ?>

currently having this error note and I expect the 2nd else statement should retrieve the code.

Notice: Undefined index: student_id in C:\xampp\htdocs\Thesis\search_enroll.php on line 37

Your string is messed up, count your quotes. Also, be sure to sanitize any variables before doing anything with them. i.e: addslashes, check if is number, etc. — David Houde
– David Houde, Commented Aug 29, 2012 at 12:38

F.P · Accepted Answer · 2012-08-29 12:36:18Z

2

Do your string escaping right or don't do it at all:

<?php
    echo "<meta http-equiv='refresh'
                content='3;search_enroll.php?id=".$id."' />"; ?>

answered Aug 29, 2012 at 12:36

F.P

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1 Comment

F.P Over a year ago

Not a problem, but you should really look into that matter right away. If you learn it wrong, you'll never get the hang of it!

Mr_DeLeTeD · Accepted Answer · 2012-08-29 12:40:39Z

0

The meta is not used like that. You hate to specify the URL on a URL subattribute and not just on the CONTENT attribute like that

<META HTTP-EQUIV="Refresh" CONTENT="n; URL=MyURL">

http://en.wikipedia.org/wiki/Meta_refresh

there, your param is never setted

answered Aug 29, 2012 at 12:40

Mr_DeLeTeD

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Comments

Mike Mackintosh · Accepted Answer · 2012-08-29 14:31:05Z

0

Use the @ suppressor in front of a variable when Undefined index error occur.

This error occur when variable is not declared first and we used in code:

For eg.

 else {
     @$id= @$_REQUEST['student_id']; <--- this is line 37
 }

But in your case there this else condition should not be run .

edited Aug 29, 2012 at 14:31

Mike Mackintosh

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answered Aug 29, 2012 at 12:42

arpan

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Comments

Mike Mackintosh · Accepted Answer · 2012-08-29 14:32:29Z

0

You should do:

else {
    if(isset($_REQUEST['student_id'])){
        $id= $_REQUEST['student_id'];
    }
}

This will make sure the student_id key exists in $_REQUEST before you try to access it.

Without a check, it will throw the Notice

answered Aug 29, 2012 at 14:32

Mike Mackintosh

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Comments

Mike Mackintosh · Accepted Answer · 2012-08-29 14:32:38Z

0

This error appears because of your PHP error reporting settings.

Usually, it appears when your variable is not properly set.

Check if $_POST['action'] is set before using it.

edited Aug 29, 2012 at 14:32

Mike Mackintosh

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answered Aug 29, 2012 at 12:38

Dr James Sprinks

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