在图表中显示获取动态数据.js使用 PHP 和 JS

Question

我想在我的页面上动态显示来自不同用户的图表.js统计数据。

我已经可以通过一个用户的清晰数据查询显示数据，但应该显示几个具有不同数据的相同引导卡。如何将用户变量动态传递给playerOne中的mysqli_query.php？

玩家一.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);

统计.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
});
}));
}
}

Answer 1

您可以在URL上发送变量，在这里...

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

然后在您的 PHP 中，使用...

$_GET['user']

例如

$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'];