var undefined getting value $ _GET ['page']

Question

I want paging is a content of a news blog. Everything works correctly, the page content is successful. But I get an error screen displays PHP:

Notice: Undefined index: page in C:\wamp\www\index.php on line 146

code with the line that gives the error:

$maxreg = 1;

    $pag = $_GET['page'];

if (!isset($pag) || empty($pag)){

      $min = 0;
      $pag = 1;  

}else{

      if($pag == 1){

            $min = 0;

      }else{

            $min = $maxreg * $pag;
            $min = $min - $maxreg;

      }
}

include("js/class.AutoPagination.php");
$obj = new AutoPagination(contar_contenido(), $pag);

mostrar_contenido($min,$maxreg);

echo $obj->_paginateDetails();

The line gives the error is this:

$ page = $ _GET ['page'];

The first page by default index.php and contains no var in the url. I do not understand why if fails below by a conditional var I determine if that has content or not.

Should not show any error php, could someone give me a solution? Thanks

make sure `?page=123` is in your URL (in your browser URL where you type) — pavlovich, Nov 28 '15 at 08:19
It works perfectly, I went to page three and takes me to page 3. On that page does not fail. The url: http://localhost/index.php?&page=3 Everything is perfect, only fails when you first open the page and there is no var on url. — PlayerWet, Nov 28 '15 at 08:23

score 2 · Answer 1 · answered Nov 28 '15 at 08:30

2

You don't need to do

$pag = $_GET['page'];

Check if $_GET['page'] is set first:

if(!isset($_GET['page']) || empty($_GET['page'])) {

    $min = 0;
    $pag = 1;

}else{

    if($pag == 1){

        $min = 0;

    }else{

        $min = $maxreg * $pag;
        $min = $min - $maxreg;

    }
}

answered Nov 28 '15 at 08:30

pavlovich

1,935
15
20

Thus it gives me error when moving to page 2. It drives me to report: Undefined variable: p in C: \ wamp \ www \ index.php on line 146 :( – PlayerWet Nov 28 '15 at 08:37
It happens because the value of the url var page to be applied. Otherwise not work, you know? – PlayerWet Nov 28 '15 at 08:40
To find out which page these need to: $pag = $_GET['page']; – PlayerWet Nov 28 '15 at 08:42

score 1 · Accepted Answer · edited May 23 '17 at 12:23

1

I quote;

Declare your variables. Or use isset() to check if they are declared before referencing them;

PHP: "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset"

edited May 23 '17 at 12:23

Community

1
1

answered Nov 28 '15 at 08:23

Dave LB

26
4

Is that the problem, but how could build conditional to not show any error? – PlayerWet Nov 28 '15 at 08:25
But it is that it is, then the variable has no value. Should not display the error because the variable does not exist at that instant. – PlayerWet Nov 28 '15 at 08:28
1

The answer is described there.. `$pag = isset($_GET['page']) ? $_GET['page'] : '';` – Dave LB Nov 30 '15 at 02:11
True, he had not noticed. Thank you !! – PlayerWet Nov 30 '15 at 05:30

score 0 · Answer 3 · answered Nov 28 '15 at 08:59

I found the solution, I've built so, assigning a value to the variable empty get in if it is indefinite. Maybe it's a very elegant solution, but it works perfect paging.

 if(!isset($_GET['page']) || empty($_GET['page'])) {

        $_GET['page']="";

        }


        $maxreg = 2;
        $min = 0;
        $pag = $_GET['page'];

I think the second conditional left over.

I leave it here in case anyone has the same problem. This way you can fix it.

var undefined getting value $ _GET ['page']

3 Answers3