Styjun

I want to use the value of limit and offset into my PHP code but I can't.

Here is my code:

var maxData = 0;
var limitt=6;
var offsett=1;

$.ajax(
 url: "../model/conn.php",
 type: 'POST',
 data: 'getData='+'&limit='+limitt+'&offset='+offsett,
 ).done(function( data ) 
 $("#d1 ").html(data);
 while (limitt<maxData)
limitt= limitt+6;
offsett=offsett+6;


); 

<?php
if(isset($_POST['getData'])) {
$serv = "localhost";
$user = "root";
$psrd = "";
$db = "nonc";

$conn = mysqli_connect($serv, $user, $psrd, $db);

$limit=$_POST['&limit'];

$offs=$_POST['&offset'];

$sql = "SELECT * FROM non_confor limit $offs, $limit;";
$resltt = mysqli_query($conn, $sql);
$checkk = mysqli_num_rows($resltt);
?>

When I run my PHP page they show me that I have errors on $limt and $offs, because they don't receuve the data from AJAX.

This is a syntax problem , here is the correction :

$.ajax(
 type: 'POST',
 url: '../model/conn.php',
 data: 
 'getData':limit,
 'offset':offsett
 ,
 success: function(msg)
 // rest of your code 
 
);

Data attribute should have as value a json format

First thing, you are using POST method for form submission and passing data as a query string which is making an error. Correct that as below:

$.ajax(
 url: "../model/conn.php",
 type: 'POST',
 data: 
 'getData': 1, // passing 1 as you are using getData in php.
 'offset': offsett,
 'limit': limitt
 ,
).done(function(data) 
 $("#d1 ").html(data);
 while (limitt<maxData)
 limitt= limitt+6;
 offsett=offsett+6; 
 
);

After this, you need to make modification in your PHP code as below:

$limit=$_POST['limitt'];
$offs=$_POST['offsett'];

After this, your code should work fine. Hope it helps you.