Javascript 使用PHP和js在Chart.js中显示获取动态数据

我想在我的页面上使用Chart.js动态显示来自不同用户的统计信息

我已经可以通过一个用户的清晰数据查询来显示数据，但是应该显示几个相同的具有不同数据的引导卡。如何将用户变量动态传递到playerne.php中的mysqli_查询

playerne.php

header('Content-Type: application/json');

include "../../../includes/db.php";

$query = "SELECT SUM(game_stats.match_stats_kills) AS Kills, SUM(game_stats.match_stats_deaths) AS Deaths FROM game_stats WHERE game_stats.user_id = 1";
$select_kd = mysqli_query($connection, $query);

$data = array();
foreach($select_kd as $row) {
    $data[] = $row;
}

mysqli_close($connection);

echo json_encode($data);

stats.js

$(document).ready(function() {
    showData();
});

function showData() {
{
    ($.post("includes/stats/playerOne.php",
    function(data) {

        var kills = [];
        var deaths = [];

        for(var i in data) {
            kills.push(data[i].Kills)
            deaths.push(data[i].Deaths);
        }

        var pieChartData = {
            labels: [
                'Kills', 'Deaths'
            ],
            datasets: [
                {
                    backgroundColor: ['#f56954', '#00c0ef'],
                    data: [kills, deaths]
                }
            ]
        };

        var pieChartTarget = $('#playerKD').get(0).getContext('2d');

        var pieChart = new Chart(pieChartTarget, {
            type: 'pie',
            data: pieChartData
        });
    }));
}
}

---

您可以在url上发送变量，此处

($.post("includes/stats/playerOne.php?user=1",  // <-- add variable here -- ?user=1

e、 g

$query = "SELECT SUM(game_stats.match_stats_kills) AS Kills, SUM(game_stats.match_stats_deaths) AS Deaths FROM game_stats WHERE game_stats.user_id = " + $_GET['user'];

---

您可以在url上发送变量，此处

($.post("includes/stats/playerOne.php?user=1",  // <-- add variable here -- ?user=1

e、 g

$query = "SELECT SUM(game_stats.match_stats_kills) AS Kills, SUM(game_stats.match_stats_deaths) AS Deaths FROM game_stats WHERE game_stats.user_id = " + $_GET['user'];