如何在javascript中“删除”?(复制)

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Why does changing an Array in JavaScript affect copies of the array? 13 answers
为什么在JavaScript中修改数组会影响数组的副本?13个答案

I'm little confused with js 'delete' operator.

我对js 'delete'操作符有点困惑。

for example

例如

var data = { 
   list : [a, b, c]
}
var temp = data;

console.log(temp.list[2]); //Result 'c'
console.log(data.list[2]); //Result 'c'

delete temp.list[2];

console.log(temp.list[2]); //Result undefined
console.log(data.list[2]); //Result undefined why?

in this case, I have just only deleted temp.list[2] but data.list[2] shows undefined as well.

在这种情况下，我只删除了temp.list[2]但是删除了数据。list[2]表示未定义。

does 'delete' operator trace array address and delete data in it? is that why data.list[2] is also undefined? (because data.list[2] has address, not the actual data itself) I think there is explanation. can you help me?

是否“删除”操作符跟踪数组地址并删除其中的数据?这就是为什么数据。[2]列表也未定义的?(因为数据。清单[2]有地址，不是实际的数据本身)我认为有解释。你能帮我吗?

4 个解决方案

#1

reference type

引用类型

var data = [] ==> instance
var temp = data; ==> data reference

get temp ==> return data

// shallow copy !!! 
var temp = Object.assign([], data); ===> instance



console.log(temp.list[2]); //Result undefined

console.log(data.list[2]); //Result c

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Object/assign

#2

Check out all the great references others have posted. But for a quick solution use the ES6 spread operator to create a copy.

查看其他人发布的所有重要参考资料。但是为了快速解决问题，使用ES6扩展操作符创建一个副本。

    var data = { 
    list : ["a", "b", "c"]
    }
    var temp = {...data};  // this is now a copy not a reference

#3

Complex types in JS are passed by reference, unless you explicitly take care of copying them.

JS中的复杂类型通过引用传递，除非您显式地处理它们的复制。

What is happening here is:

这里发生的是:

var data = {        // <--\
   list : [a, b, c] // <----------------------------------------------
}                   // <--/                                           |
var temp = data;  // this is NOT a copy, but literally a pointer to -/

Try

试一试

var temp = Object.assign({}, data);

if you want a copy.

如果你想要拷贝的话。

#4

Your question touches on several topics:

你的问题涉及几个主题:

First, delete operator removes the given property from the object, but when used to remove an element from an array, the array length is not affected:

首先，delete操作符从对象中删除给定的属性，但是当用于从数组中删除元素时，不影响数组长度:

let trees = ['redwood', 'bay', 'cedar', 'oak', 'maple'];
delete trees[3];
alert(trees.length); //shows 5

If this is not what you want, consider using splice method instead. It is really powerful and can be used not only to remove elements.

如果这不是您想要的，可以考虑使用splice方法。它非常强大，不仅可以用来删除元素。

let trees = ['redwood', 'bay', 'cedar', 'oak', 'maple'];
trees.splice(3, 1);
alert(trees.length); //shows 4

Second, as everybody has already mentioned, var temp = data; doesn't create a copy of your object, it's just copies a reference. To make a fully identical shallow clone of your object, you could write:

第二，正如大家已经提到的，var temp = data;不是创建对象的副本，而是复制引用。要对你的对象进行完全相同的浅复制，你可以这样写:

let clone = Object.create(Object.getPrototypeOf(obj), 
    Object.getOwnPropertyDescriptors(obj));

This call makes a truly exact copy of obj, including all properties: enumerable and non-enumerable, data properties and setters/getters – everything, and with the right [[Prototype]], when Object.assign() method only copies enumerable and own properties from a source object to a target object

这个调用真正精确地复制了obj，包括所有属性:可枚举和不可枚举，数据属性和setter /getter——所有的东西，以及正确的[Prototype]，当object .assign()方法只将可枚举属性从源对象复制到目标对象时

#1