根据id数组，删除对象数组中的指定项

Question

题目描述

var arr = [1,2];
var allArr = [{id:1,name:"s"},{id:2,name:"ds"},{id:3,name:"gg"}]
var finalArr = [];


// 最终结果
finalArr = [{id:3,name:"gg"}];

Answer 1

楼主，你好！
参考下：

var arr = [1,2];
var allArr = [{id:1,name:"s"},{id:2,name:"ds"},{id:3,name:"gg"}]


// 最终结果
// finalArr = [{id:3,name:"gg"}];

var finalArr = allArr.filter((item) => !arr.includes(item.id));

---

如有帮助，麻烦点击下采纳，谢谢~

Answer 2

allArr.splice(2,2)就行了

Answer 3

这个是不是你要的

var arr = [1,2];
    var allArr = [{id:1,name:"s"},{id:2,name:"ds"},{id:3,name:"gg"}]
    var finalArr = [];
    for(var i = 0, max = allArr .length; i < max; i++ ){
        var newArrid = newArr[i].id;
        if(newArrid != arr[i]){
            finalArr.push(newArr[i]);
            
        }
    }
    console.log(finalArr);

Answer 4

var finalArr = allArr.filter(i => !arr.includes(i.id))

Answer 5

var arr = [1, 3];
var allArr = [
  { id: 1, name: "s" },
  { id: 2, name: "ds" },
  { id: 3, name: "gg" }
];
// var finalArr = [];

let finalArr = allArr.filter(val => !arr.includes(val.id));

console.log(finalArr, "finalArr");

// allArr.forEach((val, index) => {
//   if (arr.includes(val.id)) allArr.splice(index, 1);
// });
// console.log(allArr, "allArr");

Answer 6

马克马克马克