在 jquery 中读取 json 映射字符串
Read json map string in jquery
我正在尝试读取 jQuery 中的 json 字符串映射。我尝试了以下方法但得到:
Invalid operand a error
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);
}
});
期待类别 1 - A,类别 1 - B ...
使用JSON.parse解析json字符串。
试试下面的代码-
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>
使用 v[item]
因为你必须通过索引获取数组的项目。
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]);
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
@abpatil 的回答略有变化,这正是您想要的。
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>
希望对您有所帮助!
我正在尝试读取 jQuery 中的 json 字符串映射。我尝试了以下方法但得到:
Invalid operand a error
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);
}
});
期待类别 1 - A,类别 1 - B ...
使用JSON.parse解析json字符串。 试试下面的代码-
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>
使用 v[item]
因为你必须通过索引获取数组的项目。
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]);
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
@abpatil 的回答略有变化,这正是您想要的。
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>
希望对您有所帮助!