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

var jsonString = `{ "schema": { "title": "User Feedback", "description": "so", "type": "object", "properties": { "name": { "type": "string" } } }, "options": { "form": { "attributes": {}, "buttons": { "submit": { "title": "It", "click": "function(){alert('hello');}" } } } } }`; var jsonData = JSON.parse(jsonString); function Iterate(data) { jQuery.each(data, function (index, value) { if (typeof value == 'object') { alert("Object " + index); Iterate(value); } else { alert(index + " : " + value); } }); } Iterate(jsonData); <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

其他回答

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

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));
});

如果不容易,请告诉我:

var jsonObject = {
  name: 'Amit Kumar',
  Age: '27'
};

for (var prop in jsonObject) {
  alert("Key:" + prop);
  alert("Value:" + jsonObject[prop]);
}

mootools的例子:

var ret = JSON.decode(jsonstr);

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

你可以使用像objx - http://objx.googlecode.com/这样的迷你库

你可以这样写代码:

var data =  [ {"id":"10", "class": "child-of-9"},
              {"id":"11", "class": "child-of-10"}];

// alert all IDs
objx(data).each(function(item) { alert(item.id) });

// get all IDs into a new array
var ids = objx(data).collect("id").obj();

// group by class
var grouped = objx(data).group(function(item){ return item.class; }).obj()

还有更多的“插件”可以让你这样处理数据,请参阅http://code.google.com/p/objx-plugins/wiki/PluginLibrary