Формирование объекта циклом

Question

Уже который час сижу и в голову не приходит как осуществить реализацию объекта такого типа. имея лишь только массив r_arr c объектами, которые выгружаются с базы данных access. Объектов в return может быть не 3, а бесконечно.(столько сколько одинаковых дат в массиве), так же и в plannedReferences: в каждой дате,тоже может быть бесконечное количество объектов. Помогите получить этот return пожалуйста.

r_arr =[
{workShiftDate:'2020-12-18',workShiftStringName: 'Late shift', referenceID:24, plannedQty: 5},
{workShiftDate:'2020-12-18',workShiftStringName: 'Day shift', referenceID:25, plannedQty: 7},
{workShiftDate:'2020-12-19',workShiftStringName: 'Day shift', referenceID:27, plannedQty: 9},
{workShiftDate:'2020-12-19',workShiftStringName: 'Day shift', referenceID:28, plannedQty: 11},
];
 
//что тут надо сделать?
  
 return[{
                workShiftDate: '2020-12-18',
                workShiftStringName: 'Late shift',
                plannedReferences: [
                    {
                        referenceID: 24,
                        plannedQty: 5
                    }]
            },{
                workShiftDate: '2020-12-18',
                workShiftStringName: 'Day shift',
                plannedReferences: [
                    {
                        referenceID: 25,
                        plannedQty: 7
                    }]
            },{
                workShiftDate: '2020-12-19',
                workShiftStringName: 'Day shift',
                plannedReferences: [
                    {
                        referenceID: 27,
                        plannedQty: 4
                    },{
                        referenceID: 28,
                        plannedQty: 11
                    }]
            }]

Answer 1

Вот так, как вариант

// https://tc39.github.io/ecma262/#sec-array.prototype.find
if (!Array.prototype.find) {
  Object.defineProperty(Array.prototype, 'find', {
    value: function(predicate) {
     // 1. Let O be ? ToObject(this value).
      if (this == null) {
        throw new TypeError('"this" is null or not defined');
      }

      var o = Object(this);

      // 2. Let len be ? ToLength(? Get(O, "length")).
      var len = o.length >>> 0;

      // 3. If IsCallable(predicate) is false, throw a TypeError exception.
      if (typeof predicate !== 'function') {
        throw new TypeError('predicate must be a function');
      }

      // 4. If thisArg was supplied, let T be thisArg; else let T be undefined.
      var thisArg = arguments[1];

      // 5. Let k be 0.
      var k = 0;

      // 6. Repeat, while k < len
      while (k < len) {
        // a. Let Pk be ! ToString(k).
        // b. Let kValue be ? Get(O, Pk).
        // c. Let testResult be ToBoolean(? Call(predicate, T, « kValue, k, O »)).
        // d. If testResult is true, return kValue.
        var kValue = o[k];
        if (predicate.call(thisArg, kValue, k, o)) {
          return kValue;
        }
        // e. Increase k by 1.
        k++;
      }

      // 7. Return undefined.
      return undefined;
    },
    configurable: true,
    writable: true
  });
}


let r_arr =[
{workShiftDate:'2020-12-18', referenceID:24, plannedQty: 5},
{workShiftDate:'2020-12-18', referenceID:25, plannedQty: 7},
{workShiftDate:'2020-12-19', referenceID:27, plannedQty: 9},
{workShiftDate:'2020-12-19', referenceID:28, plannedQty: 11},
{workShiftDate:'2020-12-19', referenceID:22, plannedQty: 13},
{workShiftDate:'2020-12-20', referenceID:33, plannedQty: 4},
];

const result = r_arr.reduce(function(acc, value) {
  // Group initialization
  if (!acc.find(function(item){return item.workShiftDate == value.workShiftDate})) {
    acc.push({workShiftDate: value.workShiftDate, plannedReferences: []});    
  }
 
  // Grouping
  acc.find(function(item) { return item.workShiftDate == value.workShiftDate}).plannedReferences.push({
  referenceID:value.referenceID, plannedQty: value.referenceID});
 
  return acc;
}, []);

console.log(result);

Answer 2

const r_arr =[
{workShiftDate:'2020-12-18', referenceID:24, plannedQty: 5},
{workShiftDate:'2020-12-18', referenceID:25, plannedQty: 7},
{workShiftDate:'2020-12-19', referenceID:27, plannedQty: 9},
{workShiftDate:'2020-12-19', referenceID:28, plannedQty: 11},
{workShiftDate:'2020-12-19', referenceID:22, plannedQty: 13},
{workShiftDate:'2020-12-20', referenceID:33, plannedQty: 4},
];

console.log([...new Set(r_arr.map(v=>v.workShiftDate))].map(d=>({workShiftDate:d,plannedReferences:r_arr.filter(r=>r.workShiftDate == d).map(r=>({referenceID:r.referenceID,plannedQty:r.plannedQty}))})));

Answer 3

Вот ещё вариант, совместимый с IE 9

let r_arr =[
{workShiftDate:'2020-12-18', referenceID:24, plannedQty: 5},
{workShiftDate:'2020-12-18', referenceID:25, plannedQty: 7},
{workShiftDate:'2020-12-19', referenceID:27, plannedQty: 9},
{workShiftDate:'2020-12-19', referenceID:28, plannedQty: 11},
{workShiftDate:'2020-12-19', referenceID:22, plannedQty: 13},
{workShiftDate:'2020-12-20', referenceID:33, plannedQty: 4},
];

const result = r_arr.reduce(function(acc, value) {
  // Group initialization
  if (acc.every(function(item){return item.workShiftDate != value.workShiftDate})) {
    acc.push({workShiftDate: value.workShiftDate, plannedReferences: []});    
  }
 
  // Grouping
  acc.filter(function(item) { return item.workShiftDate == value.workShiftDate})[0].plannedReferences.push({
  referenceID:value.referenceID, plannedQty: value.referenceID});
 
  return acc;
}, []);

console.log(result);

Updated.

тут прос100

r_arr =[
{workShiftDate:'2020-12-18',workShiftStringName: 'Late shift', referenceID:24, plannedQty: 5},
{workShiftDate:'2020-12-18',workShiftStringName: 'Day shift', referenceID:25, plannedQty: 7},
{workShiftDate:'2020-12-19',workShiftStringName: 'Day shift', referenceID:27, plannedQty: 9},
{workShiftDate:'2020-12-19',workShiftStringName: 'Day shift', referenceID:28, plannedQty: 11},
];

const result = r_arr.reduce(function(acc, value) {
  // Group initialization
  if (acc.every(function(item){return item.workShiftDate != value.workShiftDate || item.workShiftStringName != value.workShiftStringName})) {
    acc.push({workShiftDate: value.workShiftDate, workShiftStringName: value.workShiftStringName, plannedReferences: []});    
  }
 
  // Grouping
  acc.filter(function(item) { return item.workShiftDate == value.workShiftDate && item.workShiftStringName == value.workShiftStringName})[0].plannedReferences.push({
  referenceID:value.referenceID, plannedQty: value.referenceID});
 
  return acc;
}, []);

console.log(result);

Answer 4

Вот простой и рабочий вариант для IE.

const array = [
  {workShiftDate:'2020-12-18',workShiftStringName: 'Day shift 1', referenceID:24, plannedQty: 5},
  {workShiftDate:'2020-12-18',workShiftStringName: 'Day shift 2', referenceID:25, plannedQty: 7},
  {workShiftDate:'2020-12-19',workShiftStringName: 'Day shift 1', referenceID:27, plannedQty: 9},
  {workShiftDate:'2020-12-19',workShiftStringName: 'Day shift 2' , referenceID:28, plannedQty: 11},
  {workShiftDate:'2020-12-19',workShiftStringName: 'Day shift 1', referenceID:22, plannedQty: 13},
  {workShiftDate:'2020-12-20',workShiftStringName: 'Day shift 2', referenceID:33, plannedQty: 4},
];

function groupByDate(array) {
  const pre = {};

  array.forEach(function (el) {
    const key = el.workShiftDate + '@' + el.workShiftStringName;
    if (!pre[key]) {
      pre[key] = [];
    }

    pre[key].push({
      referenceID: el.referenceID,
      plannedQty: el.plannedQty,
    });
  });

  return Object.keys(pre).map(function (k) {
    const keys = k.split('@');

    return {
      workShiftDate: keys[0],
      workShiftStringName: keys[1],
      plannedReferences: pre[k],
    };
  });
}

console.log(groupByDate(array));

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Формирование объекта циклом

Ответы (4 шт):