将嵌套的json数据解析为ChartJS中使用的JQUERY变量

Question

我通过PHP的平面JSON输出获得了一个MySQL数据ChartJS图。ChartJS发出以下代码：

function showGraph()
{
    {
        $.post("data.php?id=0",
        function (data)
        {
            console.log(data);
            const date = [];
            const time = [];
            const temperature = [];
            const humidity = [];
            //const name = [];

            for (var i in data) {
                date.push(data[i].date);
                time.push(data[i].time);
                temperature.push(data[i].temperature);
                humidity.push(data[i].humidity);
                name.push(data[i].name);
            }

PHP脚本提供了平面JSON数据，例如：

{
    "timestamp": "1582877101",
    "time": "09:05",
    "temperature": "18.90",
    "humidity": "50.70"
},
{
    "timestamp": "1582877161",
    "time": "09:06",
    "temperature": "18.90",
    "humidity": "50.70"
},

由于JSON更改为嵌套数据，因此我的图形不再起作用。新的嵌套JSON看起来像：

[
    {
        "id": "0",
        "name": "Logger 0",
        "data": [
            {
                "date": "2020-02-28",
                "time": "09:05",
                "temperature": "18.90",
                "humidity": "50.70"
            },
            {
                "date": "2020-02-28",
                "time": "09:06",
                "temperature": "18.90",
                "humidity": "50.70"
            }

如何再次将嵌套的JSON数据解析为JQUERY变量？

Answer 1

由于data现在是JSON响应的属性，因此您可以通过ES6 destructuring访问它。请注意，data现在是一个数组，因此您必须使用forEach，如下所示：

$.post('data.php?id=0', ({name, data}) => {
    console.log(data);
    const date = [];
    const time = [];
    const temperature = [];
    const humidity = [];

    data.forEach(element => {
        date.push(element.date);
        time.push(element.time);
        temperature.push(element.temperature);
        humidity.push(element.humidity);
    });
});