Question

我有一个对象，如下所示：

dum2a

我有一个数组，其中包含该格式的多个对象。

我想从该数组创建一个新的对象数组，其中将包含以下格式的对象：

 {
    "id": 1,
    "dataLockVersion": 0,
    "auditData": {
      "createDate": "2018-09-18T11:41:28.362",
      "createUser": "XXX",
      "updateDate": null,
      "updateUser": null
    },
    "property1": 14021,
    "property2": {...},
    "property3": "Obj"
  }

这是我尝试过的：

var结果= [];

{
    "property1": 14021,
    "property2": {...},
    "property3": "Obj"
  }

有更好的方法吗？

Answer 1

使用地图和对象分解

const result = array
                   .map(
                       ({ property1, property2, property3 })
                       => ({ property1, property2, property3 }));

Answer 2

您只需为此使用Array.map()和对象destructuring：

let arr =[{ "id": 1, "dataLockVersion": 0, "auditData": { "createDate": "2018-09-18T11:41:28.362", "createUser": "XXX", "updateDate": null, "updateUser": null }, "property1": 14021, "property2": {"x" :1}, "property3": "Obj" }, { "id": 1, "dataLockVersion": 0, "auditData": { "createDate": "2018-09-18T11:41:28.362", "createUser": "XXX", "updateDate": null, "updateUser": null }, "property1": 14021, "property2": {"x" :12}, "property3": "Obj" }];
  
let result = arr.map(({property1,property2,property3})=>Object.assign({},{property1,property2,property3}));
console.log(result);

Answer 3

我会使用lodash.pick作为一种简单高效的解决方案。

通常，在应用程序的其他部分中会需要这种逻辑。

在您的情况下，它将是：

var newArrayWithPickedProperties = array.map(item => {
  return _.pick(item, ['property1', 'property2', 'property3']);    
})

如果您采用这种方式，请确保您import only lodash.pick不是整个lodash库。

Answer 4

您可以尝试以下操作：

 var data = [
    {"id": 1,"dataLockVersion": 0,"auditData": {"createDate": "2018-09-18T11:41:28.362","createUser": "XXX","updateDate": null,"updateUser": null},"property1": 14021,"property2": {},"property3": "Obj"},
    {"id": 2,"dataLockVersion": 1,"auditData": {"createDate": "2018-09-18T11:41:28.362","createUser": "YYY","updateDate": null,"updateUser": null},"property1": 140221,"property2": {},"property3": "Obj3"}
];

var res = data.map(function(m){return {property1: m.property1, property2: m.property2, property3: m.property3};})

console.log(res);

或者，如果您喜欢技巧和，所有值都是string或number或包含它们的对象，则可以使用它（不建议使用非常重的对象）：

let data = [
        {"id": 1,"dataLockVersion": 0,"auditData": {"createDate": "2018-09-18T11:41:28.362","createUser": "XXX","updateDate": null,"updateUser": null},"property1": 14021,"property2": {},"property3": "Obj"},
        {"id": 2,"dataLockVersion": 1,"auditData": {"createDate": "2018-09-18T11:41:28.362","createUser": "YYY","updateDate": null,"updateUser": null},"property1": 140221,"property2": {},"property3": "Obj3"}
    ];
var res=[];
JSON.stringify(data).replace(/"(property1)"\:(.+?),.+?"(property\d+)"\:(.+?)(?=})/gi, function(a){res.push(JSON.parse("{"+a+"}"));});
console.log(res);

Answer 5

如果数据来自JSON字符串，则JSON.parse reviver parameter可用于排除属性：

var json = '{"id":1,"dataLockVersion":0,"auditData":{"createDate":"2018-09-18T11:41:28.362","createUser":"XXX","updateDate":null,"updateUser":null},"property1":14021,"property2":"{...}","property3":"Obj"}'

var obj = JSON.parse(json, (key, value) => /id|data/i.test(key) ? void 0 : value)

console.log( obj )

Answer 6

如果您有n个具有固定的前3个键的属性，则可以进行解构分配。 https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment

let data = [
    {"id": 1,"dataLockVersion": 0,"auditData": {"createDate": "2018-09-18T11:41:28.362","createUser": "XXX","updateDate": null,"updateUser": null},"property1": 14021,"property2": {},"property3": "Obj","property4":"yo","property5":"hey"},
    {"id": 2,"dataLockVersion": 1,"auditData": {"createDate": "2018-09-18T11:41:28.362","createUser": "YYY","updateDate": null,"updateUser": null},"property1": 140221,"property2": {},"property3": "Obj3"}
];
const arr=data.map(a=>{
           let {id,dataLockVersion,auditData,...props}=a
           return props;
          }
       )
console.log(arr);

从另一个对象数组创建一个对象数组

6 个答案: