CODEWITHSUNDEEP
phpmysql
Mark Gerryl Mirandilla
Mark Gerryl Mirandilla

Reputation: 823

php pagination, something is wrong

I found a tutorial according pagination. http://papermashup.com/easy-php-pagination/

I tried the code and it work. but every time the first index is reload(first result in pagination).

I got this message.

Notice: Undefined index: page in C:\xampp\htdocs\sim\registrar\index.php on line 207

how can I remove this message?

here's the code..

$targetpage = "index.php";
$limit = 3;

$query = "SELECT COUNT(*) as num FROM advisoryclass";

$total_pages = mysql_fetch_array(mysql_query($query));

$total_pages = $total_pages['num'];

$stages = 3;

    $page = mysql_escape_string($_GET['page']);//this part

if($page){
$start = ($page - 1) * $limit;

}else{

$start = 0;
}

thanks

Upvotes: 0

Views: 50

Answers (3)

MaggsWeb
MaggsWeb

Reputation: 3027

You need to handle if $page does not exist.

$page = mysql_escape_string(isset($_GET['page']) ? $_GET['page'] : 0);

Upvotes: 0

taxicala
taxicala

Reputation: 21759

add a check in order to see if the parameter is in the get array:

$page = mysql_escape_string(isset($_GET['page']) ? $_GET['page'] : 0);

this is a ternary comparison in order to make it shorter than an if statement.

Upvotes: 1

Nathan Dawson
Nathan Dawson

Reputation: 19308

You're assuming page is set when that's not always going to be the case; hence the error.

Change this:

$page = mysql_escape_string($_GET['page']);//this part

To:

$page = ( isset( $_GET['page'] ) ) ? mysql_escape_string( $_GET['page'] ) : 1;

In the code above we're checking the page $_GET variable has been set and using 1 as a fallback.

You may want to look at the functions you're using with the database too.

Upvotes: 1

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