json stringify结果获取特定键的值

Question

[{"displayorder":"1","menuname":"DashBoard","menuid":"5","menuurl":"dashboard.php"},{"displayorder":"3","menuname":"Accounting Module","menuid":"3","menuurl":""},{"displayorder":"4","menuname":"My Profile","menuid":"4","menuurl":"myprofile.php"},{"displayorder":"6","menuname":"HR Module","menuid":"2","menuurl":""},{"displayorder":"9","menuname":"Administrator","menuid":"1","menuurl":""}]

我在这里有一个stringify json我想知道如何在这个json对象中获取所有菜单名称的值任何想法赞赏

UPDATE:我试过这个here，但我在控制台中得到了不确定

UPDATE

[{"displayorder":"1","menuname":"Menu Management","menuid":"1","submenuurl":"","parentid":"1"},{"displayorder":"1","menuname":"hr sub menu","menuid":"7","submenuurl":"error.php","parentid":"2"},{"displayorder":"2","menuname":"Role Management","menuid":"2","submenuurl":"","parentid":"1"},{"displayorder":"2","menuname":"menu 2 management2","menuid":"8","submenuurl":"","parentid":"2"},{"displayorder":"3","menuname":"hrsubmenu","menuid":"3","submenuurl":"contactus.php","parentid":"2"},{"displayorder":"3","menuname":"submenuaccounting","menuid":"4","submenuurl":"imagegallery.php","parentid":"3"}];

如何使用parentid获取第二个json中的所有细节，具体取决于上面的menuid？

Answer 1

工作范例：

http://jsfiddle.net/0866pay3/

var json = [{"displayorder":"1","menuname":"DashBoard","menuid":"5","menuurl":"dashboard.php"},{"displayorder":"3","menuname":"Accounting Module","menuid":"3","menuurl":""},{"displayorder":"4","menuname":"My Profile","menuid":"4","menuurl":"myprofile.php"},{"displayorder":"6","menuname":"HR Module","menuid":"2","menuurl":""},{"displayorder":"9","menuname":"Administrator","menuid":"1","menuurl":""}];


json.forEach(function(el, idx){
    console.log(el.menuname);
});

Documentation Update

如果你看看this article，你会看到以下内容：

使用三个参数调用回调：

• 元素值
• 元素索引
• 正在遍历的数组

因此，idx只是表示元素索引的常用方法。无论你喜欢什么，都可以称之为theIndex，myRandomName等。

Answer 2

var myjson = [{"displayorder":"1","menuname":"DashBoard","menuid":"5","menuurl":"dashboard.php"},{"displayorder":"3","menuname":"Accounting Module","menuid":"3","menuurl":""},{"displayorder":"4","menuname":"My Profile","menuid":"4","menuurl":"myprofile.php"},{"displayorder":"6","menuname":"HR Module","menuid":"2","menuurl":""},{"displayorder":"9","menuname":"Administrator","menuid":"1","menuurl":""}]; 


var menu_names = []; 

for (var x = 0 ; x < myjson.length; x++){
    if(myjson[x].hasOwnProperty('menuname')){
        // do something usefull here 
        console.log(myjson[x]['menuname']);
        // add value to new array 
        menu_names.push(myjson[x]['menuname'])
    }
}

console.log(menu_names);