这种怎么转，按照arr1 中detail个数 分成多个组 最终转成arr2 这种

Question

var arr = [
  {
    name:"张三",
    value:111,
    totalfee: 60,
    detail:[
      { 
        cost: 25,
        endtime: "2021-10-11",
        reason: "测试1"
      },
      { 
        cost: 35,
        endtime: "2021-10-22",
        reason: "测试2"
      },
    ]
  },
  {
    name:"李四",
    value:222,
    totalfee: 150,
    detail:[
      { 
        cost: 10,
        endtime: "2021-10-11",
        reason: "测试1"
      },
      { 
        cost: 20,
        endtime: "2021-10-12",
        reason: "测试2"
      },
      { 
        cost: 50,
        endtime: "2021-10-13",
        reason: "测试4"
      },
      { 
        cost: 70,
        endtime: "2021-10-14",
        reason: "测试3"
      },
    ]
  },
];

转成

var arr2 = [
    {
      name:"张三",
      value:111,
      totalfee: 60,
      cost: 25,
      endtime: "2021-10-11",
      reason: "测试1"
    },
    {
      name:"张三",
      value:111,
      totalfee: 60,
      cost: 35,
      endtime: "2021-10-22",
      reason: "测试2"
    },
    {
      name:"李四",
      value:222,
      totalfee: 150,
      cost: 10,
      endtime: "2021-10-11",
      reason: "测试1"
    },
    {
      name:"李四",
      value:222,
      totalfee: 150,
      cost: 20,
      endtime: "2021-10-12",
      reason: "测试2"
    },
    {
      name:"李四",
      value:222,
      totalfee: 150,
      cost: 50,
      endtime: "2021-10-13",
      reason: "测试4"
    },
    {
      name:"李四",
      value:222,
      totalfee: 150,
      cost: 70,
      endtime: "2021-10-14",
      reason: "测试3"
    },
];

Answer 1

arr.flatMap(({detail,...item})=>{
    return detail.map(detailItem=>{
        return Object.assign({},item,detailItem);
    })
})

es5

arr.flatMap(function(item){
    var detail = item.detail;
    delete item.detail;
    return detail.map(function(detailItem){
        return Object.assign({},item,detailItem);
    })
})

ie可运行

arr.reduce(function(acc, item){
    var detail = item.detail;
    delete item.detail;
    return acc.concat(detail.map(function(detailItem){
        var obj={};
        for(prop in item){
          obj[prop] = item[prop];
         }
        for(prop in detailItem){
          obj[prop] = detailItem[prop];
         }
        return obj;
    }))
},[])

arr.reduce(function(acc, item){
    var detail = item.detail;
    delete item.detail;
    detail.forEach(function(detailItem){
        var obj={};
        for(prop in item){
          obj[prop] = item[prop];
         }
        for(prop in detailItem){
          obj[prop] = detailItem[prop];
         }
        acc.push(obj);
    })
    return acc;
},[])

也可以用 babel babeljs 打补丁的方式，支持es6

https://www.babeljs.cn/repl

Answer 2

这个问题其实没那么复杂，你看下原数组有几个元素？2个；detail有几个？一共有6个，而你要的结果数组刚好就是6个元素，detail又是数组元素中某个属性的数组，所以要写双重循环，第一重循环两次，第二重一共循环6次，把内外属性值合并到一起放入空数组就行了。用各种数组方法实现也行，但不太利于理解逻辑，用最原始的写法实现如下：

function transform(arr) {
    var ret = [];
    for (var i = 0; i < arr.length; ++i) {
        var item = arr[i];
        var name = item.name;
        var value = item.value;
        var totalfee = item.totalfee;
        var detail = item.detail;
        for (var j = 0; j < detail.length; ++j) {
            item = detail[j];
            var obj = {
                name: name,
                value: value,
                totalfee: totalfee,
                cost: item.cost,
                endtime: item.endtime,
                reason: item.reason
            };
            ret.push(obj);
        }
    }
    return ret;
}
console.dir(transform(arr));