Question

我试图在jQuery中读取json字符串映射。我尝试了以下方式，但得到了：

无效的操作数错误

感谢任何帮助。

var data = '{"Category1":["A","B","C"],"Category2":["P","Q","R"]}';

$.each(data, function(k,v) {
    console.log(k + " " + v);

    $.each(v, function(item) {
        console.log(k + "-" + item);
    }
});

期望Category1 - A，Category1 - B ......

Answer 1

使用JSON.parse解析json字符串。请尝试以下代码 -

var data='{"Category1":["A","B","C"],"Category2":["P","Q","R"]}';
//use JSON.parse to parse a JSON string
$.each(JSON.parse(data), function( k,v) {
    console.log(k +" "+v);
    $.each(v, function(item) {
            console.log(k +"-"+item);
    });//here you missed closing bracket 
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Answer 2

使用v[item]，因为你必须按索引获取数组项目。

＆＃13;

var data={"Category1":["A","B","C"],"Category2":["P","Q","R"]};

$.each(data, function( k,v) {
    console.log(k +" "+v);
    $.each(v, function(item) {
      console.log(k +"-"+v[item]);
    });
});

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

＆＃13;

Answer 3

@abpatil回答几乎没有变化，这正是你想要的。

var data='{"Category1":["A","B","C"],"Category2":["P","Q","R"]}';

$.each(JSON.parse(data), function (k, v) {
    console.log(k + " "+ v);
    $.each(v, function (i, data) {
        console.log(k + ' - ', data);
    })
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

希望有所帮助！

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3 个答案: