Question

这是我的结构

 [{
"date": "2019-01-10T18:30:00.000Z",
"time": "2019-01-11T04:37:49.587Z",
"abc_Info": {
  "_id": "5c381da651f18d5040611eb2",
  "abc": 2.5,
  "guardian": "XYZ"
  }
}]

我想要的是

[{
"date": "2019-01-10T18:30:00.000Z",
"time": "2019-01-11T04:37:49.587Z",
"abc": 2.5,
"guardian": "XYZ"
}]

代码

this._model.find(params, (err, docs) => {
  if (err) {
    var response = this.errorResponse("error", 500, null);
    res.send(response);
  } else {
    for (var i = 0; i < docs.length; i++) {
      const abc= {
        "date": docs[i].date,
        "time": docs[i].time,
      "abc_Info":docs[i].abc_Info //this is object, i couldn't select value separately from this object
      }
      if (docs[i].abc_Info != undefined) {
        abcArray.push(abc);
      }
    }
    res.send(abc);
  }
});

我正在尝试从“ abc_Info”：docs [i] .abc_Info.abc 之类的对象中选择值，但是我无法做到这一点，它的抛出错误。我可以通过两种方式实现这一目标。

直接从对象中选择值并存储在变量中。那对我来说是个错误
将日期和时间与abc_Info合并。我不知道该怎么做

Answer 1

选择这个。

const abc= {
    "date": docs[i].date,
    "time": docs[i].time,
    "abc":docs[i].abc_Info.abc,
    "guardian":docs[i].abc_Info.guardian
}

Answer 2

希望这会有所帮助

var d=[{
"date": "2019-01-10T18:30:00.000Z",
"time": "2019-01-11T04:37:49.587Z",
"abc_Info": {
  "_id": "5c381da651f18d5040611eb2",
  "abc": 2.5,
  "guardian": "XYZ"
  }
}]

d[0]=Object.assign(d[0],d[0].abc_Info);
delete d[0]['abc_Info'];
delete d[0]['_id'];
console.log(d);

Answer 3

let x = {

      "date": "2019-01-10T18:30:00.000Z",
      "time": "2019-01-11T04:37:49.587Z",
}
let abc_Info = {
      "_id": "5c381da651f18d5040611eb2",
      "abc": 2.5,
      "guardian": "XYZ"
}
// to concat 
let z = {...x, ...abc_Info}
console.log(z)

您可以将以上ES6 spread operator用作原因 Object.assign的性能要好得多，因此如下所示：或let z = Object.assign(x, abc_info);

在您的代码中应该是这样的：

const abc= {...{
        "date": docs[i].date,
        "time": docs[i].time,
      }, ...docs[i].abc_Info}

Answer 4

使用Spread运算符，而不是使用for循环和推送，请使用Array.prototype.map

let docs= [{
"date": "2019-01-10T18:30:00.000Z",
"time": "2019-01-11T04:37:49.587Z",
"abc_Info": {
  "_id": "5c381da651f18d5040611eb2",
  "abc": 2.5,
  "guardian": "XYZ"
  }
}]

let abcArray = docs.map(({date, time, abc_Info}) => {
  delete abc_Info._id;
  return {date, time, ...abc_Info }
});

console.log(abcArray)

Answer 5

对不起，这是我的错误。我应该在if语句中声明对象。现在可以正常工作

      if (docs[i].abc_Info != undefined) {
      const abc= {
        "date": docs[i].date,
        "time": docs[i].time,
       "abc_Info":docs[i].abc_Info
        }
        abcArray.push(abc);
      }

合并节点中的两个对象

5 个答案: