不改变索引的Javascript数组拼接

I am working on a chat and using an array to hold the users. Here is my problem:

我正在处理一个聊天，并使用一个数组来保存用户。这是我的问题:

User1 joins and is given Index 0 in the array via push. User2 joins and is given Index 1 in the array via push.

User1连接并通过push在数组中给出索引0。User2连接并通过push在数组中获得索引1。

User1 disconnects and is removed via splice.

User1断开连接，通过剪接删除。

NOW User2 becomes Index 0.

现在User2变成了索引0。

User1 reconnects and is given Index 1 via push.

User1重新连接，并通过push获得索引1。

User2 disconnects, and Index 1 is removed which is now User1.

User2断开连接，索引1被删除，现在是User1。

This is of course causing a problem.

这当然会引起问题。

So my question is how can I remove the item from the array without the index of the other elements changing? Am I on the wrong track here?

我的问题是，如何在不改变其他元素的索引的情况下，从数组中删除项目?我走错路了吗?

4 个解决方案

#1

Instead of removing the items from the array with splice(), why not just set the value to null or undefined?

与其使用splice()从数组中删除项，不如将值设置为null或undefined?

Then when you're adding a new user, you can just scan through the array to find the first available slot.

然后，当您添加一个新用户时，您只需扫描数组以找到第一个可用的插槽。

javascript arrays are simply lists of items - they're not keyed to a specific key like you might be familiar with in PHP. So if you want to keep the same position in the array, you can't remove other items - you need to keep them, and just mark them as empty.

javascript数组仅仅是项目列表——它们不像PHP中常见的键那样被键化到特定的键上。所以如果你想在数组中保持相同的位置，你不能删除其他项——你需要保留它们，把它们标记为空。

You might scan through something like this:

你可以这样浏览一下:

var users = [];
function addUser(user) {
    var id = users.indexOf(null);
    if (id > -1) {
        // found an empty slot - use that
        users[id] = user;
        return id;
    } else {
        // no empty slots found, add to the end and return the index
        users.push(user);
        return users.length - 1;
    }
}
function removeUser(id) {
    users[id] = null;
}

#2

Another option is to use a javascript object instead of an array.

另一种选择是使用javascript对象而不是数组。

Something like this:

是这样的:

var users = {};

users[1] = 'user 1';
users[2] = 'user 2';

delete users[1];
alert(users[2]);        // alerts "user 2"
alert(typeof users[1]); // alerts "undefined"

You lose the array length property though, so you'll have to keep track of your max user number yourself.

但是你失去了数组长度属性，所以你必须自己跟踪你的最大用户号。

#3

Use delete instead of splice.

使用delete而不是splice。

> a = ['1', '2', '3']
< Array [ "1", "2", "3" ]

> delete a[1]
< true

> a
< Array [ "1", undefined × 1, "3" ]

> a.length
< 3

#4

remove array elements without facing re-indexing problem

删除数组元素而不面临重新索引问题

    var ind=[1,6]; //index positions of elements to remove
    var arr=['a','b','c','d','e','f','g']; // actual array
    var arr2 = arr.filter(function(item,index){
            if(ind.indexOf(index)== -1){
            return true;
    }});

now arr2 is ==========>> ['a','c','d','e','f']

现在arr2 ==========>> ['a'，'c'，'d'，'e'，'f']

#1