Angular 如果使用es6存在第二个数组，如何循环响应并连接该数组

我正在使用选择器获取响应，如果存在第二个数组，我希望以以下格式显示

[
  {
    "details": {
      "name": "john",
      "point": "20"
    },

    "list": {
      "number": "30",

    }
  },
  {
    "details": {
      "name": "doe",
      "point": "25"
    },
    "list": {
      "number": "30",

    }
  }
]

这就是我要做的，请提前表示感谢

{
  "details": [
    {
      "name": "john",
      "point": "20"
    },
    {
      "name": "doe",
      "point": "25"
    }
  ]
}

---

使用rest操作符解析它，并以所需格式返回数据

this.store.pipe(
        select(someselector), 
        filter(result => !!result), 
        map(r => { 
             if (result) { 
              //concat the array if exist;
              return result;
             } 
           }
        ));

这很简单 让我们考虑一下你的第一个数组是

if (result) {
  // concat if arr exist. let say array name is arr.
  result = {
    details: [...result, ...arr].map(d => {
      return {
        name: d.details.name,
        point: d.details.point,
      }
    })
  }
  return result;
}

第二个数组是

let arry1 = [{"details":{"name":"john","point":"20"},"list":{"number":"30",}},{"details":{"name":"doe","point":"25"},"list":{"number":"30",}}];

现在将细节从第一个阵列推送到第二个阵列

let newarr = [];

那么newarr包含

newarr.push(arrlist.map(ob=>{return ob.details}));

希望这对你有帮助

快乐编码

[[{"name":"john","point":"20"},{"name":"doe","point":"25"}]]