Question

copySource(copySource)，而我的脚本{ "data": [ {"firstName": "Achmad"}, {"lastName": "a"} ] }来自var body = request.body;for(var i = 0;i < body.data.length;i++){var obj = body.data[i];var keyes = Object.keys(obj);}的问题响应就是这样的列表var keyes = Object.keys(obj);，我想这样[ 'firstName' ] [ 'lastName' ]

谢谢。

Answer 1

已编辑：您可以concat数组中的每一项：

const body = {
  "data": [
    {"firstName": "Achmad"},
    {"lastName": "a"}
  ]
};

let result = [];
for (item of body.data) {
  result = result.concat(Object.keys(item));
}

console.log(result); // -> ['firstName', 'lastName']

Answer 2

假设每个数组都是父数组的元素，则可以使用Array.prototype.reduce来实现此目的：

const flat = [
  ["aku"],
  ["dia"],
  ["ia"]
].reduce((accum, el) => accum.concat(el), [])

console.log(flat);

Answer 3

也许您想做这样的事情

var body = request.body;
var keyes = [];
for(var i = 0; i < body.data.length; i++){
    var obj = body.data[i];
    keyes.push( Object.keys(obj) );
}

合并多个列表到单个数组

3 个答案: