Question

我有一个问题是将两个JSON对象（每个来自另一个数组）合并到一个新的JSON对象中，在新的数组中（合并必须在给定条件下发生）。

示例：

First Array：

var array1 = [{box:123,name:xxx,amount:xxx},{box:321,....},...]
var array2 = [{_id:123,look:xxx,title:myBox1},{_id:321,.....},...]
var newArray = [{box:123,name:xxx,amount:xxx,look:xxx,title:myBox1},...]

我基本上需要什么： 迭代Array1，遍历Array2。如果框(array1) == _id (array2)将两个对象的所有属性合并为一个新的JSON，并将其放入新数组3的第一个空闲槽中。

我希望这个例子有助于了解我想要做的事情。什么是最好的解决方案？

通常情况下，两个数组中的第一个属性也是需要进行比较的属性，但是如果解决方案在我想要进行比较的情况下也可以进行比较，那么它就可以了。第一个数组中的第一个属性，第二个数组中的第三个属性。

希望有人可以帮助我！

Answer 1

您可以通过两个循环实现此目的。第一个迭代存储在array1中的对象，第二个迭代array2中匹配对象的属性。试试这个：

var newArray = [];
for (var i = 0; i < array1.length; i++) {
    var obj = array1[i];
    if (array2[i] && obj.box == array2[i]._id) {
        for (key in array2[i]) {
            obj[key] = array2[i][key];
        }
        newArray.push(obj);
    }
};

Example fiddle

Answer 2

你可以试试这个：

 array3=[]
for(item1 in array1){
          for(item2 in array2){
         if(item1.box == item2._id)
           array3.push(Merge_objects(item1,item2))
         }
      }



    function Merge_objects(obj1,obj2){
        var obj3 = {};
        for (var attrname in obj1) { obj3[attrname] = obj1[attrname]; }
        for (var attrname in obj2) { obj3[attrname] = obj2[attrname]; }
        return obj3;
    }

Answer 3

以下功能可以做到。它比其他答案略长，但更通用（您可以指定哪些属性是ID，而不是坚持box和_id）并且不会修改原始数组以任何方式。它也不依赖于两个数组中的项的顺序相同（即box:123可以在array1中的索引1中，_id:123可以在array2中的索引3中）。

/*
    a1 = the first array of objects
    id1 = the name of the property of each object in a1 that is its id
    a2 = the second array of objects
    id2 = the name of the property of each object in a2 that is its id
*/
function combine(a1, id1, a2, id2) {
    // create our new array to return
    var newArray = []
    // iterate over a1 (the outer loop)
    for(var i = 0, l = a1.length; i < l; i++) {
        var outer = a1[i]; // get the current object for this iteration
        // iterate over a2 (the inner loop)
        for(var j = 0, m = a2.length; j < m; j++) {
            var inner = a2[j]; // get the current object for this iteration of the inner loop
            if(outer[id1] == inner[id2]) { // compare the ids for the outer and inner objects
                var o = {}; // create a new blank object to add to newArray
                o[id1] = outer[id1]; // set its id (using the same property name as that of a1)
                for(var prop in outer) { // iterate over properties
                    if(prop != id1) { // ignore the id property, but copy the others
                        o[prop] = outer[prop]
                    }
                }

                for(var prop in inner) {
                    if(prop != id2) {
                        o[prop] = inner[prop]
                    }
                }

                // add this object to newArray
                newArray.push(o);
            }
        }
    }

    // return newArray
    return newArray;
}

用法：

var array1 = [{box:123,name:'Name 1',amount:1},{box:321,name:'Name 2',amount:2}]
var array2 = [{_id:123,look:'Look 1',title:'My box 1'},{_id:321,look:'Look 2',title:'My box 2'}]
var newArray = combine(array1, "box", array2, "_id");

Answer 4

您可以使用Array.prototype.map或Array.prototype.reduce。记住time complexity！如果对象是密钥（_id或box）并且数组具有相同的长度：

// TODO: implement your own comparator based on _id/box type
var first = array1.sort((a, b) => a._id - b._id);
var second = array2.sort((a, b) => a.box - b.box);
var merged = sorted_first.map((item, index) => Object.assign(item, second[index]));

您可以轻松地将其调整为不同长度的情况，但您确实需要首先了解您的数据。

Answer 5

        Creating the dummy datas, this is what the programming should be:
        var a = [{
            id: 1,
            name: 'name1',
          }, {
            id: 2,
            name: 'name2',
          }, {
            id: 3,
            name: 'name3',

          }];

          var b = [{
            id: 1,
            look: 'look1'
          }, {
            id: 321,
            look: 'look321'
          }, {
            id: 3,
            look: 'look3',

          }

          ];

        // put the object id : 321 and look : look321 into array a from array b
          for (var i = 0; i < b.length; i++) {
            var obj_b= b[i];
            var id_b= b[i].id;
            var idNotFound = true;
            a.forEach(function(rowOfA){
            var id_a = rowOfA.id;
            if(id_a == id_b ){
                idNotFound = false;
            }
            });
            if( idNotFound ){

                a.push( b[i] );
            }//if

        }
        // merge the objects with same ids into array a
            for (var i = 0; i < a.length; i++) {
                var id_a = a[i].id;
                var found = false;
                for (var j = 0; j < b.length; j++) {
                    var id_b= b[j].id;
                    if(id_a == id_b){
                      found = b[j];
                    }//if

                }//for2
                if (found) {
                    a[i]["look"] = found.look;
                }//if
            }//for1

            console.log('aa=>',a);
    /*
The output is  aa=> [ { id: 1, name: 'name1', look: 'look1' },
      { id: 2, name: 'name2' },
      { id: 3, name: 'name3', look: 'look3' },
      { id: 321, look: 'look321' } ]
   */

在两个JSON阵列的条件下合并两个JSON对象

5 个答案: