js中的高级过滤器对象答案

【问题标题】：Advanced filter object in jsjs中的高级过滤器对象
【发布时间】：2020-04-15 15:30:22
【问题描述】：

我正在尝试使用 Object 过滤数组对象，但我不知道该怎么做。示例：

{
  245: [
    {
      id: "12",
      name: "test",
      status: "new"
    },
    {
      id: "15",
      name: "test2",
      status: "old"
    },
    {
      id: "12",
      name: "test2",
      status: "old"
    }],
  2815: [
    {
      id: "19",
      name: "test",
      status: "new"
    },
    {
      id: "50",
      name: "test2",
      status: "old"
    },
    {
      id: "120",
      name: "test2",
      status: "new"
    }]
}

如果 status = "new" 需要过滤，但结构不能改变：

{
  245: [{
    id: "12",
    name: "test",
    status: "new"
  }],
  2815: [{
    id: "19",
    name: "test",
    status: "new"
  },
  {
    id: "120",
    name: "test2",
    status: "new"
  }]
}

【问题讨论】：

你保存数据的对象有名字吗？

标签： javascript arrays object filter

【解决方案1】：

遍历条目并使用过滤值创建一个新对象

const obj = {
245:[
{id:"12",name:"test",status:"new"},{id:"15",name:"test2",status:"old"},{id:"12",name:"test2",status:"old"}],
2815:[
{id:"19",name:"test",status:"new"},{id:"50",name:"test2",status:"old"},{id:"120",name:"test2",status:"new"}]
}

console.log(filter(obj, item => item.status === "new"))

function filter(obj, pred) {
  return Object.fromEntries(Object.entries(obj).map(([name, value]) => [name, value.filter(pred)]))
}

【讨论】：

【解决方案2】：

您需要先映射对象键，然后再映射数组元素以过滤掉最终结果

var obj = {
245:[
{id:"12",name:"test",status:"new"},{id:"15",name:"test2",status:"old"},{id:"12",name:"test2",status:"old"}],
2815:[
{id:"19",name:"test",status:"new"},{id:"50",name:"test2",status:"old"},{id:"120",name:"test2",status:"new"}]
}

var res = Object.entries(obj).reduce((acc, [key, value]) => {
  acc[key] = value.filter(item => item.status === "new");
  return acc;
}, {})
console.log(res);

【讨论】：

【解决方案3】：

对于这种特殊情况，您可以这样做：

const myObj = {
  245:[
        {id:"12",name:"test",status:"new"},
        {id:"15",name:"test2",status:"old"},
        {id:"12",name:"test2",status:"old"}
      ],
  2815:[
        {id:"19",name:"test",status:"new"},
        {id:"50",name:"test2",status:"old"},
        {id:"120",name:"test2",status:"new"}
       ]
};

const filteredObj = filterMyObj(myObj);
console.log(filteredObj);

function filterMyObj(myObj){
  const myObjCopy = {...myObj};
  for (const key in myObjCopy){
    const myArrCopy = [...myObjCopy[key]];
    myObjCopy[key] = myArrCopy.filter(item => item.status == "new");
  }
  return myObjCopy;
}

【讨论】：

【解决方案4】：

您可以使用filter 来做到这一点：

  for(let key in obj){
     obj[key] = obj[key].filter(el => el.status == "new")
  }

【讨论】：