如何使用拼接从数组中删除匹配的对象元素？

Question

这是我的 array1 数据：

[ { members: [ '60ee9148104cc81bec3b97ab' ] } ]

这是array2：

[{"_id": "60ee9148104cc81bec3b97ab","username": "user1", "email": "user1@gmail.com"}, {"_id": "60ee917f767bd11d687326c7","username": "user2","email": "user2@gmail.com"}]

我想从我的 array2 中删除 _id 等于60ee9148104cc81bec3b97ab

我已经尝试过

    let user = await User.find({ _id: { $ne: req.user._id } })

        const getNonfriends = (one) => {
            user.splice(user.indexOf(one.members[0]), 1)
           //user.filter(entry => entry._id !== one.members[0])
        }

   array1.map(getNonfriends)

过滤或拼接非他们带来我的解决方案。

Answer 1

我们可以使用 Array.prototype.filter() 我认为不需要找到数组和切片的索引。
另外，我们可以使用Array.prototype.map，它类似于使用过滤器功能

 const obj1 = [{ members: ['60ee9148104cc81bec3b97ab'] }] const obj2 = [{ "_id": "60ee9148104cc81bec3b97ab", "username": "user1", "email": "user1@gmail.com" }, { "_id": "60ee917f767bd11d687326c7", "username": "user2", "email": "user2@gmail.com" }] const getAnswer = (obj2) => { const res = obj2.filter(item => !obj1[0].members.includes(item._id)) return res; } console.log(getAnswer(obj2));

Answer 2

你犯的错误是你有相似的对象，但你搜索相同的对象。

var array1 = [ { members: [ '60ee9148104cc81bec3b97ab' ] } ];
var array2 = [{"_id": "60ee9148104cc81bec3b97ab","username": "user1", "email": "user1@gmail.com"}, {"_id": "60ee917f767bd11d687326c7","username": "user2","email": "user2@gmail.com"}]

    for (var x of array1) {
        for (var member of x.members) {
            var objects = array2.filter(item => item._id === member)
            for (var obj of objects) array2.splice(array2.indexOf(obj), 1)
        }
    }

Answer 3

当您使用 ID 时，我将假设第二个数组不能包含两个具有相同 ID 的项目。 如果这个假设是正确的，你可以这样做：

 const data = [{ "_id": "60ee9148104cc81bec3b97ab", "username": "user1", "email": "user1@gmail.com" }, { "_id": "60ee917f767bd11d687326c7", "username": "user2", "email": "user2@gmail.com" }]; const filter = [{ members: ['60ee9148104cc81bec3b97ab'] }]; // SOLUTION 1: using Array.prototype.findIndex() and Array.prototype.splice() const filteredDataA = [...data] // as splice() modifies the original array, I think it is safer to work on a copy of the original data filter[0].members.forEach(id => { const index = filteredDataA.findIndex(item => item._id === id); // find the index of the item with the same ID if (index > -1) filteredDataA.splice(index, 1); // (if found) remove the item }) console.log('SOLUTION 1: using Array.prototype.findIndex() and Array.prototype.splice()'); console.log(filteredDataA); console.log('--------------------'); // SOLUTION 2: using array.prototype.filter() and array.prototype.includes() const filteredDataB = data.filter(item => !filter[0].members.includes(item._id)); console.log('SOLUTION 2: using array.prototype.filter() and array.prototype.includes()'); console.log(filteredDataB); console.log('--------------------');

我更喜欢“解决方案 2”，因为我认为它更具可读性

Answer 4

从答案中，我得到了线索，这带来了我想要的结果。 有一件事是 array1 可以有多个元素，因此需要通过调用 getNonfriends 函数来映射它。 所以，

let user = await User.find({ _id: { $ne: req.user._id } })
const getNonfriends = (one) => {
           user = user.filter(item => !one.members.includes(item._id))
           return user;
        }
await array1.map(getNonfriends)