合并2个对象数组

Question

让我们看一个例子。var arr1 = new Array({name: "lang", value: "English"}, {name: "age", value: "18"});var arr2 = new Array({name : "childs", value: '5'}, {name: "lang", value: "German"});我需要合并对象的这两个数组并创建以下数组：arr3 = new Array({name: "lang", value: "German"}, {name: "age", value: "18"}, {name : "childs", value: '5'});有没有JavaScript或jQuery函数可以做到这一点？$.extend不适合我 它返回arr4 = new Array({name : "childs", value: '5'}, {name: "lang", value: "German"});

Answer 1

如果要合并JavaScript中的2个对象数组。您可以使用这一招

Array.prototype.push.apply(arr1,arr2);

例如

var arr1 = [{name: "lang", value: "English"},{name: "age", value: "18"}];

var arr2 = [{name : "childs", value: '5'}, {name: "lang", value: "German"}];

Array.prototype.push.apply(arr1,arr2); 

console.log(arr1);  // final merged result will be in arr1

输出：

[{"name":"lang","value":"English"},

{"name":"age","value":"18"},

{"name":"childs","value":"5"},

{"name":"lang","value":"German"}]

Answer 2

对于那些正在尝试现代事物的人：

var odd = [

    { name : "1", arr: "in odd" },

    { name : "3", arr: "in odd" }

];

var even = [

    { name : "1", arr: "in even" },

    { name : "2", arr: "in even" },

    { name : "4", arr: "in even" }

];

// ----

// ES5 using Array.filter and Array.find

function merge(a, b, prop){

  var reduced = a.filter(function(aitem){

      return ! b.find(function(bitem){

          return aitem[prop] === bitem[prop];

      });

  });

  return reduced.concat(b);

}

console.log( "ES5", merge(odd, even, "name") );

// ----

// ES6 arrow functions

function merge(a, b, prop){

    var reduced =  a.filter( aitem => ! b.find ( bitem => aitem[prop] === bitem[prop]) )

  return reduced.concat(b);

}

console.log( "ES6", merge(odd, even, "name") );

// ----

// ES6 one-liner

var merge = (a, b, p) => a.filter( aa => ! b.find ( bb => aa[p] === bb[p]) ).concat(b);

console.log( "ES6 one-liner", merge(odd, even, "name") );

// Results

// ( stuff in the "b" array replaces things in the "a" array )

// [

//    {

//         "name": "3",

//         "arr": "in odd"

//     },

//     {

//         "name": "1",

//         "arr": "in even"

//     },

//     {

//         "name": "2",

//         "arr": "in even"

//     },

//     {

//         "name": "4",

//         "arr": "in even"

//     }

// ]

Answer 3

var arr3 = [];

for(var i in arr1){

   var shared = false;

   for (var j in arr2)

       if (arr2[j].name == arr1[i].name) {

           shared = true;

           break;

       }

   if(!shared) arr3.push(arr1[i])

}

arr3 = arr3.concat(arr2);