Question

我有这个JSON：

{
    "fakultaeten": [
        {
            "id": "1",
            "name": "Carl-Friedrich Gauß",
            "institut": [
                {
                    "Mathematik": [
                        {
                            "Verbrauch": "852 kWH",
                            "Effizienz": "5,5"
                        }
                    ],
                    "Informatik": [
                        {
                            "Verbrauch": "852 kWH",
                            "Effizienz": "5,5"
                        }
                    ],
                    "Wirtschaftswissenschaften": [
                        {
                            "Verbrauch": "852 kWH",
                            "Effizienz": "5,5"
                        }
                    ],
                    "Sozialwissenschaften": [
                        {
                            "Verbrauch": "852 kWH",
                            "Effizienz": "5,5"
                        }
                    ]
                }
            ],
            "verbrauch": "852 kWH"
        }
    ]
}

我想在“institut”中创建所有项目的列表，如下所示：

Mathematik
Infomratik
Wirtschaftswissenschaften
等

我正在尝试这个：

$.each(data.fakultaeten, function(key,value)
            {
                var mother = "<li id='first'>"+value.name+"<ul>";
                $.each(value.institut, function(key1, value1)
                {
                // create the list here
                });
             })

结果仅为：[object Object]

我的解决方案出了什么问题？

Answer 1

您的问题是institut是一个包含单个对象的数组，因此您可以使用institut[0]引用该单个对象。

这将有效：

$.each(data.fakultaeten, function (key, value) {
    var mother = "<li id='first'>" + value.name + "<ul>";
    $.each(value.institut[0], function (key1, value1) {
        alert(key1);
    });
});

Here is a working example

Answer 2

试试这个：

$.each(data.fakultaeten, function (key, value) {
    var mother = "<li id='first'>" + value.name + "<ul>";
    $.each(value.institut[0], function (key1, value1) {
        console.log(key1); // key1 gives you output that you want
    });
});

嵌套json与jQuery：创建嵌套项的列表

2 个答案: