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[英]JavaScript Transferable Objects: Why doesn't the engine preserve the original instance?
[英]Why don't nested objects in Javascript preserve their values?
我正在运行两个测试,以检查对象在javascript中的工作方式:
测试一个:
//Method 1 var Player = { name: "", id: "", action: { action1: "", action2: "", action3: "" } } var player1 = Object.create(Player); player1.name = " hack"; player1.id = 1; player1.action.action1 = "aaa"; player1.action.action2 = "aaa"; player1.action.action3 = "aaa"; console.log(JSON.stringify(player1.action)); var player2 = Object.create(Player); player2.name = " Jason"; player2.id = 2; player2.action.action1 = "bbb"; player2.action.action2 = "bbb"; player2.action.action3 = "bbb"; console.log(JSON.stringify(player2.action)); console.log(JSON.stringify(player1.action));
结果是:
{"action1":"aaa","action2":"aaa","action3":"aaa"}
VM174:29 {"action1":"bbb","action2":"bbb","action3":"bbb"}
VM174:30 {"action1":"bbb","action2":"bbb","action3":"bbb"}
您可以看到通过创建player2更改了player1的动作对象。
如果我希望操作对象保留其价值怎么办?
我能想到的唯一方法是:
//Medthod 2 var actionManager = { action1: "", action2: "", action3: "" } var Player = { name: "", id: "", action: null } var player1 = Object.create(Player); var actions1 = Object.create(actionManager); actions1.action1 = "aaa"; actions1.action2 = "aaa"; actions1.action3 = "aaa"; player1.name = " hack"; player1.id = 1; player1.action = actions1; console.log(JSON.stringify(player1)); var player2 = Object.create(Player); player2.name = " Jason"; player2.id = 2; var actions2 = Object.create(actionManager); actions2.action1 = "bbb"; actions2.action2 = "bbb"; actions2.action3 = "bbb"; player2.action = actions2; console.log(JSON.stringify(player2)); console.log(JSON.stringify(player1));
在这种情况下,输出为:
{"name":"hack","id":1,"action:{"action1":"aaa","action2":"aaa","action3":"aaa"}}
{"name":" Jason","id":2,"action:{"action1":"bbb","action2":"bbb","action3":"bbb"}}
{"name":" hack","id":1,"action":{"action1":"aaa","action2":"aaa","action3":"aaa"}}
有什么更好的方法可以使用方法1,但不更改操作对象?
有什么更好的方法可以使用方法1,但不更改操作对象?
使用构造函数为您初始化action
属性,这样您就不必手动将其写出。 您无法避免需要它们 ,除非将它们弄平并将.action*
直接放在播放器上。
因为当您在Javascript中替换数组或对象时,Javascript会引用数组/对象而不是复制。 因此,当您执行player1.action.action1 = "aaa";
或player2.action.action1 = "bbb";
,您正在更改Player
的值。 由于玩家1或玩家2只是指向Player
对象。
Object.create()方法使用指定的原型对象和属性创建一个新对象。
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Object/create
这意味着您正在创建一个对数组/对象具有相同引用的新对象。 因为甚至Player
都没有Array作为其属性,而是对Array / Object的内存地址的引用,所以Object.create()
创建的新Object也只有引用。
如果同时检查“播放器”对象的内容,则会注意到其值也已更改。
console.log(Player)
action:
action1: "bbb"
action2: "bbb"
action3: "bbb"
为了避免这种情况。 将对象用作原型时,需要深度复制该对象。
//Method 1
var Player = {
name: "",
id: "",
action: {
action1: "",
action2: "",
action3: ""
}
}
objectCreate = function(obj){
return jQuery.extend(true, {},obj)
};
var player1 = objectCreate(Player);
player1.name = " hack";
player1.id = 1;
player1.action.action1 = "aaa";
player1.action.action2 = "aaa";
player1.action.action3 = "aaa";
console.log(JSON.stringify(player1.action));
var player2 = objectCreate(Player);
player2.name = " Jason";
player2.id = 2;
player2.action.action1 = "bbb";
player2.action.action2 = "bbb";
player2.action.action3 = "bbb";
console.log(JSON.stringify(player2));
console.log(JSON.stringify(player1));
console.log(Player)
你需要jQuery
objectCreate = function(obj){
return jQuery.extend(true, {},obj)
};
这为您提供了完美的深度复制对象。 并且上面的代码按预期工作。
请在此处查看jQuery.extend详细信息http://api.jquery.com/jquery.extend/
这里的问题是两个播放Player
都共享与原型相同的Player
对象。 当一个玩家更改了该对象中的字段时,当寻找另一个对象时,它们也会被更改。
您可以为此使用构造函数:
// constructor function
function Player () {
this.name = "";
this.id = "";
this.action = {
action1: "",
action2: "",
action3: ""
};
}
var player1 = new Player();
player1.name = " hack";
player1.id = 1;
player1.action.action1 = "aaa";
player1.action.action2 = "aaa";
player1.action.action3 = "aaa";
console.log(JSON.stringify(player1.action));
var player2 = new Player();
player2.name = " Jason";
player2.id = 2;
player2.action.action1 = "bbb";
player2.action.action2 = "bbb";
player2.action.action3 = "bbb";
console.log(JSON.stringify(player2.action));
console.log(JSON.stringify(player1.action));
当您调用new Player()
,将实例化一个新对象,并为此对象创建新属性,例如name
和id
,并且Player
多个实例不会彼此冲突。
Object.create
作用是,将一个对象作为输入,并将其所有属性复制到新对象。 现在,如果属性指向可变值,则可以使用新创建的对象对其进行突变。 因此,您可以使用Player.action
中指向它的值的任何引用变量, player1.action
和player2.action
来player1.action
player2.action
的值。
并且根据您的代码, Player
应该是类而不是对象。 因此,我们可以使它成为一个构造函数(JS没有类,但是有对象构造函数),如下所示。
//New Method 1
var Player = function() {
this.name = "";
this.id = "";
this.action = {
action1: "",
action2: "",
action3: ""
};
};
var player1 = new Player(); // Create a new object using `Player` constructor function
player1.name = " hack";
player1.id = 1;
player1.action.action1 = "aaa";
player1.action.action2 = "aaa";
player1.action.action3 = "aaa";
console.log(JSON.stringify(player1.action));
var player2 = new Player(); // Create another object using `Player` constructor function
player2.name = " Jason";
player2.id = 2;
player2.action.action1 = "bbb";
player2.action.action2 = "bbb";
player2.action.action3 = "bbb";
console.log(JSON.stringify(player2.action));
console.log(JSON.stringify(player1.action));
在这里,对象player1
和player2
是使用构造函数Player
创建的。 new
关键字实际上启动了一个新对象。 从传统的OOP角度讲,您可以说Player
类的player1
和player2
对象是使用new
关键字创建的。
输出将是
{"name":"hack","id":1,"action":{"action1":"aaa","action2":"aaa","action3":"aaa"}}
{"name":" Jason","id":2,"action":{"action1":"bbb","action2":"bbb","action3":"bbb"}}
{"name":" hack","id":1,"action":{"action1":"aaa","action2":"aaa","action3":"aaa"}}
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