2016-11-21 52 views
0

當我嘗試將對象複製/克隆到變量中,然後更改一個值而不影響另一個對象時,我遇到了引用問題。Object.Assign on Composite Object

我一直在看看Object.Assign方法的文檔,我嘗試了深層克隆的方法,但沒有成功。

我創建了一個plnkr爲了更好地表達自己,以防萬一你知道如何正確地做到這一點。

Code Sample in here

const value = 20.20; 
 
let a = { a: 10 }; 
 
let bb = { v: value }; 
 
let b = { b: 20, bb: bb }; 
 
let c = { c: 30 }; 
 
let object1 = { name: 'cloneObject', a, b, c } 
 
let object2 = Object.assign({}, object1); 
 
let object3 = Object.assign({}, object1); 
 

 
object1.name = 'it works fine'; 
 
object1.b.bb.v = 999; // this will change same property in object2 & object3 and it shouldn't 
 

 
document.getElementById('expected1').innerHTML = object1.b.bb.v; 
 
document.getElementById('expected2').innerHTML = value; 
 
document.getElementById('expected3').innerHTML = value; 
 

 

 
document.getElementById('name1').innerHTML = object1.name; 
 
document.getElementById('name2').innerHTML = object2.name; 
 
document.getElementById('name3').innerHTML = object3.name; 
 

 
document.getElementById('value1').innerHTML = object1.b.bb.v; 
 
document.getElementById('value2').innerHTML = object2.b.bb.v; 
 
document.getElementById('value3').innerHTML = object3.b.bb.v;
<table style="width:100%; text-align: center"> 
 
    <tr> 
 
    
 
    <th>Object</th> 
 
    
 
    <th>Name</th> 
 
    
 
    <th>Value</th> 
 
    
 
    <th>Expected Value</th> 
 
    </tr> 
 
    <tr> 
 
    <td>Object 1</td> 
 
    <td id='name1'></td> 
 
    <td id='value1'></td> 
 
    <td id='expected1'></td> 
 
    </tr> 
 
    <tr> 
 
    <td>Object 2</td> 
 
    <td id='name2'></td> 
 
    <td id='value2'></td> 
 
    <td id='expected2'></td> 
 
    </tr> 
 
    <tr> 
 
    <td>Object 3</td> 
 
    <td id='name3'></td> 
 
    <td id='value3'></td> 
 
    <td id='expected3'></td> 
 
    </tr> 
 
</table>

隨着@czosel答案,我得到:

JSON.parse(JSON.stringify(object1));

所以它看起來是這樣的:

let object1 = {name: 'cloneObject', a, b, c} 
    let object2 = JSON.parse(JSON.stringify(object1)); 
    let object3 = JSON.parse(JSON.stringify(object1)); 

我發現@ gon250答案更好,因爲我要在父類中實現它,並且它將默認繼承該方法。我認爲遞歸會更有效率。

謝謝! :)

回答

1

下面你可以看到一個簡單的函數來克隆對象:

function cloneObject(myObj) { 
    var temp = myObj.constructor(); 
    if (myObj === null || typeof myObj !== 'object') { 
     return myObj; 
    } 
    for (var key in myObj) { 
     temp[key] = cloneObject(myObj[key]); 
    } 
    return temp; 
} 

這裏是實現功能到您的代碼示例:

https://jsfiddle.net/gon250/4qx247th/