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?
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