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合并2個(gè)對(duì)象數(shù)組

JavaScript JQuery

Cats萌萌 2019-10-09 17:40:29

讓我們看一個(gè)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"});我需要合并對(duì)象的這兩個(gè)數(shù)組并創(chuàng)建以下數(shù)組：arr3 = new Array({name: "lang", value: "German"}, {name: "age", value: "18"}, {name : "childs", value: '5'});有沒(méi)有JavaScript或jQuery函數(shù)可以做到這一點(diǎn)？$.extend不適合我它返回arr4 = new Array({name : "childs", value: '5'}, {name: "lang", value: "German"});

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3 回答

縹緲止盈

TA貢獻(xiàn)2041條經(jīng)驗(yàn) 獲得超4個(gè)贊

如果要合并JavaScript中的2個(gè)對(duì)象數(shù)組。您可以使用這一招

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

輸出：

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

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

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

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

反對(duì) 回復(fù) 2019-10-09

慕神8447489

TA貢獻(xiàn)1780條經(jīng)驗(yàn) 獲得超1個(gè)贊

對(duì)于那些正在嘗試現(xiàn)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"

// }

// ]

反對(duì) 回復(fù) 2019-10-09

茅侃侃

TA貢獻(xiàn)1842條經(jīng)驗(yàn) 獲得超21個(gè)贊

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);

反對(duì) 回復(fù) 2019-10-09

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合并2個(gè)對(duì)象數(shù)組

合并2個(gè)對(duì)象數(shù)組

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