function arrayDiff(a, b) {
      return a.concat(b).filter(val => !(b.includes(val)));
      //(or) return a.concat(b).filter(val => !(a.includes(val) && b.includes(val)));
    }

根据之前的答案…取决于你是想要一个高效的还是“漂亮的联机”解决方案。

一般有三种方法……

"manual iterative" (using indexOf) - naive with O(n2) complexity (slow) var array_diff_naive = function(a,b){ var i, la = a.length, lb = b.length, res = []; if (!la) return b; else if (!lb) return a; for (i = 0; i < la; i++) { if (b.indexOf(a[i]) === -1) res.push(a[i]); } for (i = 0; i < lb; i++) { if (a.indexOf(b[i]) === -1) res.push(b[i]); } return res; } "abstract iterative" (using filter and concat library methods) - syntactic sugar for manual iterative (looks nicer, still sucks) var array_diff_modern = function(a1,a2){ return a1.filter(function(v) { return !a2.includes(v); } ) .concat(a2.filter(function(v) { return !a1.includes(v);})); } "using hashtable" (using object keys) - much more efficient - only O(n), but has slightly limited range of input array values var array_diff_hash = function(a1,a2){ var a = [], diff = []; for (var i = 0; i < a1.length; i++) { a[a1[i]] = true; } for (var i = 0; i < a2.length; i++) { if (a[a2[i]]) { delete a[a2[i]]; } else { a[a2[i]] = true; } } for (var k in a) { diff.push(k); } return diff; }

在jsperf上可以看到 https://jsperf.com/array-diff-algo

修正了一下最佳答案

function arr_diff(a1, a2)
{
  var a=[], diff=[];
  for(var i=0;i<a1.length;i++)
    a[a1[i]]=a1[i];
  for(var i=0;i<a2.length;i++)
    if(a[a2[i]]) delete a[a2[i]];
    else a[a2[i]]=a2[i];
  for(var k in a)
   diff.push(a[k]);
  return diff;
}

这将考虑当前的元素类型。B /c当我们创建一个[a1[i]]时，它将一个值从原始值转换为字符串，因此我们失去了实际值。

Array.prototype.difference = function(e) {
    return this.filter(function(i) {return e.indexOf(i) < 0;});
};

eg:- 

[1,2,3,4,5,6,7].difference( [3,4,5] );  
 => [1, 2, 6 , 7]

这是我用来获得两个数组之间的差异的函数-它适用于数值，字符串，混合num/字符串数组。在数组/多维数组中不是对象文字

function diff(arr1, arr2) {

    var x, 
        t;

    function uniq(a, b) {
        t = b;

        if( (b === 0 && x[b+1]!==a) || 
           (t > 0 && a !== x[b+1] && a !== x[b-1]) ) {
            return  a;
        }
    }


    x = arr1.concat(arr2).sort();

    return x.filter(uniq);
}

var a1 = ['a', 'b', 'e', 'c'],
    a2 = ['b', 'a', 'c', 'f' ];

diff(a1, a2);

要从另一个数组中减去一个数组，只需使用下面的代码片段:

var a1 = ['1','2','3','4','6'];
var a2 = ['3','4','5'];

var items = new Array();

items = jQuery.grep(a1,function (item) {
    return jQuery.inArray(item, a2) < 0;
});

它将返回['1，'2'，'6']，这些是第一个数组中不存在的项。

因此，根据您的问题示例，以下代码是准确的解决方案:

var array1 = ["test1", "test2","test3", "test4"];
var array2 = ["test1", "test2","test3","test4", "test5", "test6"];

var _array = new Array();

_array = jQuery.grep(array2, function (item) {
     return jQuery.inArray(item, array1) < 0;
});