对象内的对象转换为单个对象[复制]

Question

这个问题在这里已有答案：

• Partially flattening a JavaScript object 2回答

是否有可能将嵌套数组对象展平为单个对象。在我的查询中，我想删除源对象并将结果对象作为一个对象（我也提到了我的输出）。

    var result = [
    {"_id":"12345",
    "_type":"feeds",
    "_source":{
              "title": "hi all solve it",
              "link": "www.face.com",
              "content": "Hi thewwewewedwe asdasdasdasd",
              "createdAt": "2018-08-08T11:42:40.073Z",
              "updatedAt": "2018-08-08T11:42:40.073Z",
              "reply": []
                }
    }]

 //resultant array

     var newResult = [
        {
            "_id":"12345",
            "_type":"feeds",
            "title": "hi all solve it",
            "link": "www.face.com",
            "content": "Hi thewwewewedwe asdasdasdasd",
            "createdAt": "2018-08-08T11:42:40.073Z",
            "updatedAt": "2018-08-08T11:42:40.073Z",
            "reply": []
        }];

Answer 1

使用普通JS的最简单版本

处理多个条目

var result = [{ "_id": "12345", "_type": "feeds", "_source": { "title": "hi all solve it", "link": "www.face.com", "content": "Hi thewwewewedwe asdasdasdasd", "createdAt": "2018-08-08T11:42:40.073Z", "updatedAt": "2018-08-08T11:42:40.073Z", "reply": [] } },{ "_id": "12346", "_type": "feeds", "_source": { "title": "hi all solve it", "link": "www.face.com", "content": "Hi thewwewewedwe asdasdasdasd", "createdAt": "2018-08-08T11:42:40.073Z", "updatedAt": "2018-08-08T11:42:40.073Z", "reply": [] } }]

result = result.map(function(item) {
  var obj = item._source;
  for (var o in item) {
    if (o != "_source") obj[o] = item[o];
  }
  return obj;
})
console.log(result)

Answer 2

你可以使用...spread

var result = [{
  "_id":"12345",
  "_type":"feeds",
  "_source": {
    "title": "hi all solve it",
    "link": "www.face.com",
    "content": "Hi thewwewewedwe asdasdasdasd",
    "createdAt": "2018-08-08T11:42:40.073Z",
    "updatedAt": "2018-08-08T11:42:40.073Z",
    "reply": []
  }
}];
    

const { _source, ...rest } = result[0];

const flattenResult = [{
  ...rest,
  ..._source,
}];

console.log(flattenResult);

将result.length> 1的解决方案作为练习留给您。

Answer 3

您可以首先遍历数组以获取数组中的每个对象，然后遍历对象键以获取key名称。然后，如果遇到名为key的_source，则使用Object.assign()将这些对象内容分配到展平对象中。这适用于带有一个或多个对象的result数组。

var result = [{
  "_id": "12345",
  "_type": "feeds",
  "_source": {
    "title": "hi all solve it",
    "link": "www.face.com",
    "content": "Hi thewwewewedwe asdasdasdasd",
    "createdAt": "2018-08-08T11:42:40.073Z",
    "updatedAt": "2018-08-08T11:42:40.073Z",
    "reply": []
   }
},
{
  "_id": "1234567",
  "_type": "feeds123",
  "_source": {
    "title": "hi all solve it 123",
    "link": "www.face.com",
    "content": "Hi thewwewewedwe asdasdasdasd",
    "createdAt": "2018-08-08T11:42:40.073Z",
    "updatedAt": "2018-08-08T11:42:40.073Z",
    "reply": []
  }
}];
var newArray = [];
result.forEach(function(obj){
  var tempObj = {};
  Object.keys(obj).forEach(function(key){
    if(key !== '_source'){
      tempObj[key] = obj[key];
    } else {
      tempObj = Object.assign(tempObj,  obj[key]);
    }
  });
  newArray.push(tempObj);
});
console.log(newArray);