Question

我有一个具有以下结构的对象：

mymain = { 
name1: [ {key1: number, key2: "string"},{key1: number, key2: "string"} ],
name2: [ {key1: number, key2: "string"},{key1: number, key2: "string"} ],
name3: [ {key1: number, key2: "string"},{key1: number, key2: "string"} ]
}

顶级项（名称1，名称2，名称3）的数量和键名会有所不同。

我想将所有对象（数量和值也可变，但都具有相同的键结构）在这些名称下合并为一个大对象，如下所示：

Allconcat = [{key1: number, key2: "string"},{key1: number, key2: "string"},{key1: number, key2: "string"},{key1: number, key2: "string"},{key1: number, key2: "string"},{key1: number, key2: "string"}]

我尝试提取顶级项目的名称，然后将它们循环以连接对象，但这给出了错误的项目数：

var names = Object.keys(mymain)
    var firstname = names[0]
    names.shift() // should remove first item
    var firstObj = mymain[firstname] // name1 object
    var Allconcat
    console.log(firstObj) // correctly shows name1 with 2 keys
    for (var r = 0, rlen = names.length; r < rlen; r++) {
      Allconcat = firstObj.concat(mymain[names[r]])
    }
    console.log(Allconcat) // shows 4 objects, missing 2 in the middle, first and last are there

转移或不转移都没有什么区别，而且我总是最终在中间缺少一些元素。

Answer 1

一种选择是获取Object.values对象的mymain，并展平数组：

const mymain = { 
  name1: [ {key1: 'number', key2: "string"},{key1: 'number', key2: "string"} ],
  name2: [ {key1: 'number', key2: "string"},{key1: 'number', key2: "string"} ],
  name3: [ {key1: 'number', key2: "string"},{key1: 'number', key2: "string"} ]
};

const Allconcat = Object.values(mymain).flat();
console.log(Allconcat);

或者，如果您不能使用.flat()，请传播到concat中：

const mymain = { 
  name1: [ {key1: 'number', key2: "string"},{key1: 'number', key2: "string"} ],
  name2: [ {key1: 'number', key2: "string"},{key1: 'number', key2: "string"} ],
  name3: [ {key1: 'number', key2: "string"},{key1: 'number', key2: "string"} ]
};

const Allconcat = [].concat(...Object.values(mymain));
console.log(Allconcat);

合并其父母具有可变键名的js对象

1 个答案: