使用ajax发布按钮值

Question

I am trying to post a value to a php script using post method in ajax, however I am getting

Notice: Undefined index: gamekey

The following is my JavaScript code

function adminchat(row) {
    var to_user_id = $('#start_chat_' + row).data('key_' + row);
    var dataString = 'gamekey=' + to_user_id;

    fetch_chat()

    setInterval(function () {
        fetch_chat();
    }, 5000)

    setInterval(function () {
        autoRefresh_div();
    }, 1000);

    function fetch_chat() {
        $.ajax({
            url: 'admin_dispute_handler.php',
            data: dataString,
            type: 'post',,
            success: function () {
                window.location = 'admin_dispute_handler.php';
            },
            error: function (request, status, error) {
                alert(error);
                alert(status);
                alert(request);
            }
        });
    }

    function autoRefresh_div() {
        $("#result").load("load.php").show(); // a function which will load data from other file after x seconds
    }

};

Below is the html:

<button type="button" class="btn btn-info btn-xs start_chat" id="start_chat_1" onclick="adminchat(1)" data-key_1="34354661" data-toggle="modal" data-target="#exampleModal">Start Chat</button>

The section of the php code giving me issue:

<?php
    if ($_SESSION['logged_in']==FALSE) {
        header("Location:http://localhost/the_site/login/login.html");
    } else {
        require_once 'connection.php';

        $conn = new mysqli(host,user,password,db);
        session_start();
        $Game_key   = $_POST['gamekey'];
        $_SESSION['gamekey'] = $_POST['gamekey'];
        $username   = $_SESSION['username'];
        $query = "SELECT gamername,score From ongoing_game_list Where Game_Key = $Game_key";

        $result = $conn->query($query);

        /* numeric array */
        $row = $result->fetch_array(MYSQLI_ASSOC);

        $gamername  = $row["gamername"];
            $score      = $row["score"];
    }
?>

Answer 1

First, you have a syntax error on type: 'post',, , browser should raise an error message on developer console. And use object as param in .ajax() :

$.ajax({
    url: 'admin_dispute_handler.php',
    data: {
        gamekey: to_user_id
    },
    // Original is: type: 'post',,
    type: 'post',
    success: function () {
        window.location = 'admin_dispute_handler.php';
    },
    error: function (request, status, error) {
        alert(error);
        alert(status);
        alert(request);
    }
});

You can learn more about jQuery AJAX usage at official docs example section .

Answer 2

If you facing issue after redirecting then try below solution

$headers = apache_request_headers();
$is_ajax = (isset($headers['X-Requested-With']) && $headers['X-Requested-With'] == 'XMLHttpRequest');

here you got $is_ajax true if send request using AJAX other wise return false(null).

Hope this help to you.