按属性搜索JSON的最简单方法是什么？

Question

"names": [
    {
        "id": 17,
        "user_id": 9,
        "code": "de",
        "name": "Ich bin Hans",
        "created_at": "2017-07-31 12:43:19",
        "updated_at": "2017-07-31 12:43:19"
    },
    {
        "id": 18,
        "user_id": 9,
        "code": "en",
        "name": "My name is Hans",
        "created_at": "2017-07-31 12:43:19",
        "updated_at": "2017-07-31 12:43:19"
    }
]

是否有任何方法可以code='en' 从jQuery中的上述JSON数组中获取JSON对象？

我可以用for循环来做,但我想也许有一种更简单的方法.

Answer 1

使用Array.prototype.filtervanilla JS中的函数来过滤掉对象 - 请参阅下面的演示:

var obj={names:[{id:17,user_id:9,code:"de",name:"Ich bin Hans",created_at:"2017-07-31 12:43:19",updated_at:"2017-07-31 12:43:19"},{id:18,user_id:9,code:"en",name:"My name is Hans",created_at:"2017-07-31 12:43:19",updated_at:"2017-07-31 12:43:19"}]};

var result = obj.names.filter(function(e){return e.code == 'en'})
console.log(result);

ES6版本更简单:

var obj={names:[{id:17,user_id:9,code:"de",name:"Ich bin Hans",created_at:"2017-07-31 12:43:19",updated_at:"2017-07-31 12:43:19"},{id:18,user_id:9,code:"en",name:"My name is Hans",created_at:"2017-07-31 12:43:19",updated_at:"2017-07-31 12:43:19"}]};

var result = obj.names.filter(e => e.code == 'en');
console.log(result);

Answer 2

使用Jquery grep函数.实际上你没有搜索对象,你正在搜索一组对象(名称).

查找满足过滤函数的数组元素.原始数组不受影响.

 var input = {names:[{id:17,user_id:9,code:"de",name:"Ich bin Hans",created_at:"2017-07-31 12:43:19",updated_at:"2017-07-31 12:43:19"},{id:18,user_id:9,code:"en",name:"My name is Hans",created_at:"2017-07-31 12:43:19",updated_at:"2017-07-31 12:43:19"}]};
 
var result = $.grep(input.names, function(obj) {
    return obj.code === "en";
});
console.log(result);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

另一个是Array#find功能

var input = {names:[{id:17,user_id:9,code:"de",name:"Ich bin Hans",created_at:"2017-07-31 12:43:19",updated_at:"2017-07-31 12:43:19"},{id:18,user_id:9,code:"en",name:"My name is Hans",created_at:"2017-07-31 12:43:19",updated_at:"2017-07-31 12:43:19"}]};
 
var result = input.names.find(item => {
   return item.code == 'en'
})
console.log(result);