js 实现两个数组（一个数组包含于另一个数组中）去重的问题

Question

arr2是arr1的子数组，（也就是说arr2中的元素arr1都有）现在要得到在arr1中删除arr2中出现的元素后的结果，各位大神有什么好办法

    var arr1 = [
        {id:1,name:'a'},
        {id:2,name:'abc'},
        {id:3,name:'abbb'}, 
        {id:4,name:'abxxx'}, 
        {id:5,name:'xyz'},
        {id:6,name:'abcdef'},
        {id:7,name:'abzzzz'}];
    var arr2 = [
        {id:1,name:'a'},
        {id:5,name:'xyz'},
        {id:7,name:'abzzzz'}];

Answer 1

var result = arr1.filter(function(item1) {
    return arr2.every(function(item2) {
        return item2.id !== item1.id
    })
})

Answer 2

var out=[];
var flag=true;
for(var i=0;i<arr1.length;i++){
    flag=true;
    for(var j=0;j<arr2.length;j++){
        if(arr1[i].id==arr2[j].id && arr[i].name == arr2[j].name ){ // 如果数据能保证id和name一直配对，则可以少一个比较条件
            flag = false;
            break;
        }
    }
    if(flag){
        out.push(arr1[i]);
    }
}

Answer 3

for(var i = 0 ; i < arr1.length ; i++){

for(var j = 0 ; j < arr2.length ; j++){
    if(arr1[i].id == arr2[j].id){
    arr1.splice(i,1);
    }
 }

}
console.log(arr1)

Answer 4

遍历数组取交集