遍历对象+数组+对象[重复]

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这个问题已经有了答案:

Access / process (nested) objects, arrays or JSON 18 answers
访问/处理(嵌套)对象、数组或JSON 18答案

Get guys,

的家伙,

I've tried looking through a bunch of tutorials but I can't seem to find one that covers this in detail. Maybe somebody could point me int he right direction.

我已经试着看了很多教程，但是我似乎找不到一个能够详细介绍这些内容的教程。也许有人能给我指明正确的方向。

I have a .json file that looks like this:

我有一个。json文件是这样的:

{
    "users": [{
        "firstName": "John",
        "lastName": "Doe"
    }, {
        "firstName": "Sabrina",
        "lastName": "Doe"
    }]
}

I want to run a .each loop and grab all of the users. Here's what I've tried:

我想运行一个。each循环并获取所有的用户。这是我试过:

$.getJSON("database.json", function(data) {
    $.each(data.users, function(key, val) {
        $('.dbUL').append('<li>' + key + ' : ' + val + '</li>');
    });
});

So it looks like it actually does spit out the array since I get an output of 0 : [object Object] and 1 : [object Object].

看起来它确实输出了数组因为我得到了0:[object]和1:[object]的输出。

My question is, how do I dig into the array and actually spit out the objects that I have stored in my array?

我的问题是，如何对数组进行深入挖掘，并将我存储在数组中的对象全部吐出来?

2 个解决方案

#1

Knowing the properties what you want get (fistName, lastName): jsBin

知道要获取的属性(fistName, lastName): jsBin

$.getJSON("database.json", function(data) {
    $.each(data.users, function(idx, val) {
        $('.dbUL').append('<li>' + val.firstName + ' : ' + val.lastName + '</li>');
    });
});

Not knowing the properties you want to get: jsBin

不知道要获取的属性:jsBin

$.getJSON("database.json", function(data) {

    var LIhtml = "";

    $.each(data.users, function(idx, obj) { 

      LIhtml += "<li>";

      for(var key in obj){
        if(obj.hasOwnProperty(key)){
          LIhtml += obj[key]+" ";
        }
      } 

      LIhtml += '</li>';

    });

    $('.dbUL').append(LIhtml);
});

#2

As far as I know there isn't a fancy way to grab the values of a "Hash" in JS like you do in Ruby, for example.

就我所知，没有一种方法可以像在Ruby中那样使用JS中“散列”的值。

Why not just using something like this below?

为什么不使用下面这样的东西呢?

$('.dbUL').append('<li>' + key + ' : ' + val.firstName + ' ' + val.lastName + '</li>');

#1