重构json，其中值不存在将值设置为0

Question

我正试图从mysql查询中重构数据，我花了几天时间想要弄清楚如何做到这一点。mysql数据：

var data = [{rep:'Test1', day: 1 , binder: 500},
{rep:'Test1', day: 2 , binder: 500},{rep:'Test1', day: 3 , binder: 500},
{rep:'Test2', day: 2 , binder: 500},{rep:'Test2', day: 2 , binder: 500},
{rep:'Test1', day: 4 , binder: 500},{rep:'Test1', day: 5 , binder: 500},
{rep:'Test2', day: 4 , binder: 500},{rep:'Test1', day: 5 , binder: 500},
{rep:'Test1', day: 6 , binder: 500},{rep:'Test2', day: 7 , binder: 500}]

转换为datatables的可读对象

    var data = { 
"data": [
    [
      "Test1",
      "500", // total Binder from monday for Test1
      "500", // total Binder from Tuesday for Test1
      "500", // total Binder from wednesday for Test1
      "500", // total Binder from thursday for Test1
      "1000", // total Binder from friday for Test1
      "500" // total Binder from saturday for Test1
      "500", // total Binder from sunday for Test1
      "4000" // sum of all previous 
    ],[
      "Test2",
      "500", // total Binder from monday for Test2
      "500", // total Binder from Tuesday for Test2
      "500", // total Binder from wednesday for Test2
      "500", // total Binder from thursday for Test2
      "1000", // total Binder from friday for Test2
      "500" // total Binder from saturday for Test2
      "500", // total Binder from sunday for Test2
      "4000" // sum of all previous 
    ]
    
  ]
}

我的当前函数是

const transformObj = (repKey) => { 
const days = [ 
"Sunday", 
"Monday", 
"Tuesday", 
"Wednesday", 
"Thursday", 
"Friday", 
"Saturday" 
]; 

let obj = {}; 

const filteredDatasets = data.filter( 
(d) => d.rep.toLowerCase() === repKey.toLowerCase() 
); 

const total = filteredDatasets.reduce( 
(prevValue, currDataset) => prevValue + currDataset.binder, 
0 
); 

filteredDatasets.forEach((dataset) => { 
obj = { ...obj, rep: dataset.rep, [days[dataset.day - 1]]: dataset.binder }; 
}); 

obj.total = total; 

return obj; 
};

返回

{ rep: "jeff", Tuesday: 1500, total: 1500 }

我不知道如何做到这一点，我正在尝试自学js

Answer 1

输入数据、期望的输出和您的代码之间似乎没有什么一致性。如果我仅仅依赖于输入数据和所需输出的形状，我将建议您首先创建一个由rep值键控的映射，并将所需数组结构与其相关联的过程。

然后通过对数据进行迭代来填充这些数组。

最后提取那些数组来构建最终的数据结构：

null

let data = [{rep:'Jeff', day: 1 , binder: 500},{rep:'Test1', day: 2 , binder: 500},{rep:'Test1', day: 3 , binder: 500},{rep:'Test2', day: 2 , binder: 500},{rep:'Test2', day: 2 , binder: 500},{rep:'Test1', day: 4 , binder: 500},{rep:'Test1', day: 5 , binder: 500},{rep:'Test2', day: 4 , binder: 500},{rep:'Test1', day: 5 , binder: 500},{rep:'Test1', day: 6 , binder: 500},{rep:'Test2', day: 7 , binder: 500}];

let reps = new Map(data.map(({rep}) => [rep, [rep, 0, 0, 0, 0, 0, 0, 0, 0]]));
for (let {rep, day, binder} of data) {
    let arr = reps.get(rep);
    arr[8] += arr[day] = binder;
}
let result = { data: [...reps.values()] };

console.log(result);