JavaScript 迭代二维数组并返回不匹配项

Question

我有两个 2D 数组，我想使用 JavaScript 比较它们，忽略匹配，如果不匹配，则将整行返回到新数组中。

    var array1 = [ ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 [null,'ABC'],
                 ['6f93cfa0106f','xxx'],
                  ];

    var array2 = [ ['52a1fd0296fc','ABC'],
                   ['6f93cfa0106f','xxx'],
                   ['52a1fd0296fc','ABC'],
                   ['52a1fd0296fc','ABC'],
                   ['52a1fd0296fc','DEF'],
                   ['52a1fd0296fcasd','DEF'],  ];

我想获取此输出，即 array2 中存在而不是 array1 中存在的数组:

array3 = [['52a1fd0296fcasd','DEF'],['52a1fd0296fc','ABC']]

请问有什么想法吗？

Answer 1

只需循环两个数组:

var array1 = [ ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 ['52a1fd0296fc','DEF'],
                 [null,'ABC'],
                 ['6f93cfa0106f','xxx'],
                 ['52a1fd0296fc','ABC'] ];

var array2 = [ ['52a1fd0296fc','ABC'],
                   ['6f93cfa0106f','xxx'],
                   ['52a1fd0296fc','ABC'],
                   ['52a1fd0296fc','ABC'],
                   ['52a1fd0296fc','DEF'] ];
                   
var array3 = [];
for(var i = 0; i<array1.length; ++i) {
	var a = array1[i];
	var found = false;
	for(var j = 0; j<array2.length; ++j) {
		var b = array2[j];
		if(a[0] == b[0] && a[1] == b[1]) {
			found = true;
			break;
		}
	}
	if(!found) {
		array3.push(a);
	}
}

console.dir(array3);

我假设您想要 array1 而没有 array2，而不是相反。