这种JSON怎么遍历,each嵌套了还是弄不出

Question

var data = {
    "T1": [
        {
            "A": "1",
            "B": "2",
            "C": "3"
        },
        {
            "A": "1",
            "B": "2",
            "C": "3"
        },
        {
            "A": "1",
            "B": "2",
            "C": "3"
        }
    ],
    "T2": [
        {
            "A": "1",
            "B": "2",
            "C": "3"
        },
        {
            "A": "1",
            "B": "2",
            "C": "3"
        },
        {
            "A": "1",
            "B": "2",
            "C": "3"
        }
    ],
    "T3": [
        {
            "A": "1",
            "B": "2",
            "C": "3"
        },
        {
            "A": "1",
            "B": "2",
            "C": "3"
        },
        {
            "A": "1",
            "B": "2",
            "C": "3"
        }
    ]
};
console.log(data);
$.each(data,function(tmp,tb){
    $.each(tb,function(index,col){
        alert(col.A);
    });
    alert("tb");
});如果你的data是字符串形式的，先调用 var json = eval(data);

Answer 1

  $.each(data, function(na) {
    $(this).each(function(i, t) {
      alert([na, i, t.A].join(':'));
    });
  });

Answer 2

谢谢
另外想问我想取到key值应该怎么办,比如我最后想要的结果是
T1
A->1
B->2
C->3T2
A->1
B->2
C->3

Answer 3

$.each(data,function(tmp,tb){//遍历第一层tmp是key，tb是值 是数组
    document.write(tmp);
    $.each(tb,function(index,col){//遍历数组，index是数组的下标 ，col是数组的值
         $.each(col[i],function(key,val){//遍历数组元素
             document.write(key+"--->"+val);
        });
    });
   
});

Answer 4

$.each(data,function(tmp,tb){//遍历第一层tmp是key，tb是值 是数组
    document.write(tmp);
    $.each(tb,function(index,col){//遍历数组，index是数组的下标 ，col是数组的值
         $.each(col[index],function(key,val){//遍历数组元素
             document.write(key+"--->"+val);
        });
    });
   
});

Answer 5

第一层看做属性，用for in 遍历，第二层是数组，可以直接用$.each遍历，代码如下：
var data = {
                    "T1": [{
                        "A": "1",
                        "B": "2",
                        "C": "3"
                    }, {
                        "A": "1",
                        "B": "2",
                        "C": "3"
                    }, {
                        "A": "1",
                        "B": "2",
                        "C": "3"
                    }],
                    "T2": [{
                        "A": "1",
                        "B": "2",
                        "C": "3"
                    }, {
                        "A": "1",
                        "B": "2",
                        "C": "3"
                    }, {
                        "A": "1",
                        "B": "2",
                        "C": "3"
                    }],
                    "T3": [{
                        "A": "1",
                        "B": "2",
                        "C": "3"
                    }, {
                        "A": "1",
                        "B": "2",
                        "C": "3"
                    }, {
                        "A": "1",
                        "B": "2",
                        "C": "3"
                    }]
                };                for (var o1 in data) {
                    $.each(data[o1], function (i, obj) {
                        alert(obj.A + "-" + obj.B + "-" + obj.C);
                    });
                }

Answer 6

for(var key in data){
   $.each(data[key],function(i){
         alert(this.A+"-"+this.B+"-"+this.C);
   });
}