基於一個公共鍵合並/合並兩個對象（按？拼接？）

Question

我一直在嘗試使用push和splice等，但沒有成功。 我希望您能給我一個（或兩個）線索以使其起作用。 在我的問題下面，這似乎並不難... :(

我有2個對象：

對象＃1：

houses:      [ { _id: 1,
                 KLLS: '72797-194155',
                 date : '01/01/1984'},

               { _id: 2,
                 KLLS: '84773-949399',
                 date : '01/01/1984'}
             ]

對象2：

works:       [ { _id: 27,
                 KLLS: '72797-194155',
                 stuff : 'some stuff'},

               { _id: 28,
                 KLLS: '72797-194155', // Note that KLLS key is the same as id:27
                 stuff : 'some stuff'},

               { _id: 29,
                 KLLS: '84773-949399',
                 stuff : 'some stuff'},

               { _id: 30,
                 KLLS: '84773-949399', // Note that KLLS key is the same as id:29
                 stuff : 'some stuff'},

]

我想要實現的是這樣的：

[ { _id: 1,
    KLLS: '72797-194155',
    date : '01/01/1984',
    stuff: 
         [ { _id: 27,
             KLLS: '72797-194155',
             stuff : 'some stuff'},

           { _id: 28,
             KLLS: '72797-194155',
             stuff : 'some stuff'}
         ]
   },

  { _id: 2,
    KLLS: '84773-949399',
    date : '01/01/1984',
    stuff: 
         [ { _id: 27,
             KLLS: '72797-194155',
             stuff : 'some stuff'},

           { _id: 28,
             KLLS: '72797-194155',
             stuff : 'some stuff'}
         ]}
]

其實我在這一點上（使用Lodash）：

for(var i = 0; i < houses.length; i++) {

      // get the KLLS key for each house[i]
      var klls = houses[i].KLLS

     // retrieve the works corresponding to KLLS key for house[i]
     var trvx = _.filter(works, ['KLLS', klls])

    // Merge the two to get the wanted output...
    --> Whatever I try here, nothing works...
        even using splice or push...

}

你有什么線索讓我步入正軌嗎？

Answer 1

你可以這樣做...

 var a = [{ _id: 1, KLLS: '72797-194155', date: '01/01/1984' }, { _id: 2, KLLS: '84773-949399', date: '01/01/1984' } ] var b = [{ _id: 27, KLLS: '72797-194155', stuff: 'some stuff' }, { _id: 28, KLLS: '72797-194155', // Note that KLLS key is the same as id:27 stuff: 'some stuff' }, { _id: 29, KLLS: '84773-949399', stuff: 'some stuff' }, { _id: 30, KLLS: '84773-949399', // Note that KLLS key is the same as id:29 stuff: 'some stuff' }, ] var newobj = [] var n = a.map(function(el) { var temp = [] b.map(function(el2) { if (el.KLLS === el2.KLLS) { temp.push(el2) } }) el.stuffs = temp newobj.push(el) }) console.log(newobj) 

Answer 2

遺憾的是，有破折號而不使用任何方便的方法來解決。

作為houses的JSON標記為“住宅”和works標JSON的“作品”，你可以做如下，讓您的解決方案：

var res = _.reduce(houses, function (memo, house) {
    return memo.concat(_.extend({}, house, {
        stuff: _.filter(works, {KLLS :house.KLLS})
    }));
}, []);

console.log(res);

想法是將houses陣列reduce為新的房屋陣列，並對其進行extended 。

我更喜歡創建一個新對象，而不是修改原始對象，以便使函數盡可能pure 。

Answer 3

合並數組，按KLLS將其分組，然后減少每個組數組以產生所需的對象：

 function combine(arr1, arr2) { return _(arr1) .concat(arr2) // concat the 2nd array .groupBy('KLLS') // group the items by KLLS .map(function(group) { return group.reduce(function(result, obj) { if (obj.date) { // if it's an object with date, merge it with current result and return return _.merge(result, obj); } result.stuff.push(obj); return result; }, { stuff: [] }); // initial value is an object with stuff array }) .value(); // end the chain } var houses = [{ _id: 1, KLLS: '72797-194155', date: '01/01/1984' }, { _id: 2, KLLS: '84773-949399', date: '01/01/1984' }]; var works = [{ _id: 27, KLLS: '72797-194155', stuff: 'some stuff' }, { _id: 28, KLLS: '72797-194155', stuff: 'some stuff' }, { _id: 29, KLLS: '84773-949399', stuff: 'some stuff' }, { _id: 30, KLLS: '84773-949399', stuff: 'some stuff' }, ]; var result = combine(houses, works); console.log(result); 

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