Question

我想得到＆＃34;姓名＆＃34;来自json数据的值（即xyz，abc，mno，mxc），但由于父节点id不同，我无法解析它。??

"all": {
    "id55": {
        "Tid": "1",
        "Name": "xyz",
        "TypeName": "author"
    },
    "id56": {
        "Tid": "2",
        "Name": "abc",
        "TypeName": "author"
    },
    "id57": {
        "Tid": "3",
        "Name": "mno",
        "TypeName": "author"
    },
    "id58": {
        "Tid": "4",
        "Name": "mzc",
        "TypeName": "author"
    },
}

Answer 1

var all = {
            "id55": {
                "Tid": "1",
                "Name": "xyz",
                "TypeName": "author"
            },
            "id56": {
                "Tid": "2",
                "Name": "abc",
                "TypeName": "author"
            },
            "id57": {
                "Tid": "3",
                "Name": "mno",
                "TypeName": "author"
            },
            "id58": {
                "Tid": "4",
                "Name": "mzc",
                "TypeName": "author"
            }
        }
for (var a in all) {
  console.log(all[a].Name);
}

<强>输出

xyz
abc
mno
mzc

Answer 2

您可以使用 $.each() 来迭代您的对象并获得Name值：

$.each(all, function(i,val) {
   console.log(all[i].Name);
});

<强> Fiddle Demo

使用JQuery解析嵌套的JSON

2 个答案: