可以使用es7数组新增的includes方法。遍历数组a, 然后看数组b是否includes。
functionisEquals(a, b){returnJSON.stringify(a.sort()) ===JSON.stringify(b.sort());}
改进算法
缺陷：比较数组的时候必须按顺序相等，因为数级里的 object 没法排序
constcompare =(() =>{functioncompareArray(a, b){console.log("array", a, b);if(a.length !== b.length) {returnfalse; }constlength = a.length;for(leti =0; i < length; i++) {if(!compare(a[i], b[i])) {returnfalse; } }returntrue; }functioncompareObject(a, b){console.log("object", a, b);constkeya =Object.keys(a);constkeyb =Object.keys(b);if(keya.length !== keyb.length) {returnfalse; }returnkeya.every(key=>{if(!compare(a[key], b[key])) {returnfalse; }returntrue; }); }functioncompare(a, b){if(a === b) {returntrue; }if(typeofa !==typeofb || a ===null|| b ===null) {returnfalse; }if(Array.isArray(a)) {if(!Array.isArray(b)) {returnfalse; }returncompareArray(a, b); }if(typeofa ==="object") {returncompareObject(a, b); }console.log("value", a, b);returnfalse; }returncompare;})();varaa = [{Name:"YuanXP",Id:9}, {Name:"YuanX",Id:9}];varbb = [{Name:"YuanXP",Id:9}, {Id:9,Name:"YuanX"}];varcc = [{Name:"YuanXP",Id:19}, {Id:9,Name:"YuanX"}];console.log(compare(aa, bb));console.log(compare(aa, cc));
function isEquals(a, b) {
let_a= a.sort();let_b= b.sort();return JSON.stringify(_a) === JSON.stringify(_b);
}
你好，貌似compare缺少typeof相等 值不等时候的判断
var aa = [{Name:"YuanXP",Id:[9] }, {Name:"YuanX",Id:9}];var bb = [{Name:"YuanXP",Id:[10]}, {Id:9,Name:"YuanX"}];
http://underscorejs.org/#differencedifference_.difference(array, *others)Similar to without, but returns thevaluesfromarraythat arenotpresentinthe otherarrays._.difference([1,2,3,4,5], [5,2,10]);=> [1,3,4]if(_.difference(arr1, arr2).length==0&&_.difference(arr2, arr1).length==0) { //两个数组相同, 数组元素有obj之类的不知行不行}
连接两个数组，用hash判断重复
https://github.com/hanzichi/u... 不知道有没有帮助
来源https://segmentfault.com/q/1010000006845104?_ea=1150060
