Question

我是reactjs和javascript的新手。

在这里，我有一个对象数组，

[
  {
    "id": "CSS",
    "questionCount": [
      {
        "level": "TOUGH",
        "type": "NON_CODE",
        "count": "1"
      }
    ]
  },
  {
    "id": "Backbone",
    "questionCount": [
      {
        "level": "TOUGH",
        "type": "CODE",
        "count": "2"
      },
      {
        "level": "TOUGH",
        "type": "NON_CODE",
        "count": "5"
      },
      {
        "level": "MEDIUM",
        "type": "NON_CODE",
        "count": "7"
      },
      {
        "level": "EASY",
        "type": "NON_CODE",
        "count": "6"
      }
    ]
  },
]

现在，我想要的是拥有一个对象数组，该对象数组将包含questionCount数组中存在的所有对象。所以，就像

[  {
        "level": "TOUGH",
        "type": "NON_CODE",
        "count": "1"
      },  {
        "level": "TOUGH",
        "type": "CODE",
        "count": "2"
      },
      {
        "level": "TOUGH",
        "type": "NON_CODE",
        "count": "5"
      },
      {
        "level": "MEDIUM",
        "type": "NON_CODE",
        "count": "7"
      },
      {
        "level": "EASY",
        "type": "NON_CODE",
        "count": "6"
      } ]

那么，有人可以帮我吗？

Answer 1

您可以将Array.prototype.reduce()与Array.prototype.concat()结合使用：

const temp = [{"id": "CSS","questionCount": [{"level": "TOUGH","type": "NON_CODE","count": "1"}]},{"id": "Backbone","questionCount": [{"level": "TOUGH","type": "CODE","count": "2"},{"level": "TOUGH","type": "NON_CODE","count": "5"},{"level": "MEDIUM","type": "NON_CODE","count": "7"},{"level": "EASY","type": "NON_CODE","count": "6"}]},];
const result = temp.reduce((a, c) => a.concat(c.questionCount), []);

console.log(result);

Answer 2

您可以遍历每个元素，并将其与结果连接。尝试以下操作：

var jsonObject = [{"id": "CSS","questionCount": [{"level": "TOUGH","type": "NON_CODE","count": "1"}]},{"id": "Backbone","questionCount": [{"level": "TOUGH","type": "CODE","count": "2"},{"level": "TOUGH","type": "NON_CODE","count": "5"},{"level": "MEDIUM","type": "NON_CODE","count": "7"},{"level": "EASY","type": "NON_CODE","count": "6"}]}]

var result = [];
for(var t of jsonObject){

  if(t.questionCount){
    //As t.questionCount is an array, we need to add property to each of the element present in that array
    t.questionCount.forEach(obj => obj.Id = t.id);
    result = result.concat(t.questionCount);
   }
   
}
console.log(result);

Answer 3

让我们说给temp一个数组，那么您可以使用以下代码对问题计数数组：-

finalArr=[]
for(var i=0;i<temp.length;i++) 
    finalArray.push(temp[i].questionCount)

Answer 4

这点代码的和平就可以做到

var newData = [];
for (var i of data) {
  newData = newData.concat(i['questionCount'])
}

Answer 5

通过ES6减少和传播，您可以执行以下操作

const x = arr.reduce((accum, curr) => {
    let newArr = [...accum, ...curr.questionCount];
    return newArr
}, [])

Answer 6

// -- Use the swiss-knife-tool :-) (Lodash)
// -- Taking x as your array in the question, 
// -- you can use lodash to map specific attribute 
// -- and then flatten the output array.

const y = _
  .chain(x)
  .map(item => item.questionCount)
  .flatten()
  .value();

console.log(y);

将所有对象放入对象数组中

6 个答案: