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。
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! :)