javascript - How to retain prototype with angular.copy()? - Stack Overflow

admin管理员组

文章数量:1305793

I am using Angular.js 1.3.x. In previous versions of Angular (including 1.3.0-beta5), the following code would copy the properties from the prototype directly to the new object:

function x() {};
x.prototype.logIt = function() {console.log("it")};
var src = new x(); // x has custom properties on the prototype
var dest = {};
angular.copy(src, dest);
dest.logIt(); // "TypeError" in Angular 1.3.0+ 

However, in Angular.js 1.3.0+, the properties from the prototype are totally lost, despite the fact that the migration guide for 1.2 to 1.3 says:

This changes angular.copy so that it applies the prototype of the original object to the copied object. Previously, angular.copy would copy properties of the original object's prototype chain directly onto the copied object.

How do I keep the properties from the prototype?

I am using Angular.js 1.3.x. In previous versions of Angular (including 1.3.0-beta5), the following code would copy the properties from the prototype directly to the new object:

function x() {};
x.prototype.logIt = function() {console.log("it")};
var src = new x(); // x has custom properties on the prototype
var dest = {};
angular.copy(src, dest);
dest.logIt(); // "TypeError" in Angular 1.3.0+ 

However, in Angular.js 1.3.0+, the properties from the prototype are totally lost, despite the fact that the migration guide for 1.2 to 1.3 says:

This changes angular.copy so that it applies the prototype of the original object to the copied object. Previously, angular.copy would copy properties of the original object's prototype chain directly onto the copied object.

How do I keep the properties from the prototype?

• javascript
• angularjs

Share Improve this question Follow asked Feb 3, 2015 at 18:35 JohannJohann 4,37333 gold badges4242 silver badges4141 bronze badges

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 13

While the ment in the mit linked to from the migration guide says:

This changes angular.copy so that it applies the prototype of the original object to the copied object.

this is only true when the destination argument to angular.copy(source, [destination]); is not provided. When destination is provided, only the direct properties of the object are copied.

The solution is to provide only the source object to the angular.copy function, leaving off the destination parameter:

function x() {};
x.prototype.logIt = function() {console.log("it")};
var src = new x();
var dest = angular.copy(src); // no second parameter
dest.logIt(); // logs "it"

Update: This appears to still be relevant, as angular.copy in v1.6.5 only calls getPrototypeOf(source) if destination is undefined.

本文标签： javascriptHow to retain prototype with angularcopy()Stack Overflow