我有以下JSON结构:

[{ "id":"10", "class": "child-of-9" }, { "id": "11", "classd": "child-of-10" }]

我如何使用JavaScript迭代它?


当前回答

如果这是你的数据数组:

var dataArray = [{"id":28,"class":"Sweden"}, {"id":56,"class":"USA"}, {"id":89,"class":"England"}];

然后:

$(jQuery.parseJSON(JSON.stringify(dataArray))).each(function() {  
         var ID = this.id;
         var CLASS = this.class;
});

其他回答

mootools的例子:

var ret = JSON.decode(jsonstr);

ret.each(function(item){
    alert(item.id+'_'+item.classd);
});

对于嵌套对象,可以通过递归函数检索:

function inside(events)
  {
    for (i in events) {
      if (typeof events[i] === 'object')
        inside(events[i]);
      else
      alert(events[i]);
    }
  }
  inside(events);

其中as events是json object。

取自jQuery文档:

var arr = [ "one", "two", "three", "four", "five" ];
var obj = { one:1, two:2, three:3, four:4, five:5 };

jQuery.each(arr, function() {
  $("#" + this).text("My id is " + this + ".");
  return (this != "four"); // will stop running to skip "five"
});

jQuery.each(obj, function(i, val) {
  $("#" + i).append(document.createTextNode(" - " + val));
});

从http://www.w3schools.com复制和粘贴,不需要JQuery开销。

var person = {fname:"John", lname:"Doe", age:25};

var text = "";
var x;
for (x in person) {
    text += person[x];
}

结果:无名氏25人

另一个在JSON文档中导航的解决方案是JSONiq(在Zorba引擎中实现),在那里你可以写这样的东西:

let $doc := [
  {"id":"10", "class": "child-of-9"},
  {"id":"11", "class": "child-of-10"}
]
for $entry in members($doc) (: binds $entry to each object in turn :)
return $entry.class         (: gets the value associated with "class" :)

您可以在http://public.rumbledb.org:9090/public.html上运行它