你如何设置“url"?创建新的响应()时?

How do you Set the quot;urlquot; when creating a new Response()?(你如何设置“url?创建新的响应()时?)

问题描述

我正在关注 这个为测试创建模拟响应的示例.

I'm following this example to create a mocked response for a test.

稍作修改，如下图:

 var data = { foo: 'bar'};
 var blob = new Blob([JSON.stringify(data)], {type : 'application/json'});
 var init = { "status" : 200 , "statusText" : "SuperSmashingGreat!" };
 var resp = new Response(blob,init);

console.log(resp.url)

body: (...)
bodyUsed: true
headers: Headers {}
ok: true
redirected: false
status: 200
statusText: "SuperSmashingGreat!"
type: "default"
url: ""

这在模拟数据和状态方面做得很好但是，我也想模拟 resp.url 我不知道如何使用构造函数来设置它并且 [因为它是只读的] 我不能单独设置它

this does a good job mocking the data and status however, I also want to mock resp.url I don't see how I can set that using the constructor and [since it's readonly] I can't set it on resp itself

resp.url 
>> ""
resp.url = 'www.test.com'
>> "www.test.com"
resp.url
>> ""

那么如何设置网址呢?

推荐答案

由于 url 实际上是由继承的 getter 定义的，所以可以使用 Object.definedProperty 来定义一个直接在 Response 实例上的简单值属性，它会隐藏继承的 getter 属性:

Since url is actually defined by an inherited getter, you can use Object.definedProperty to define a simple value property directly on your Response instance, which shadows the inherited getter property:

Object.defineProperty(resp, "url", { value: "foobar" });

<小时>

为了自己的理解，可以通过查看Response.prototype.url的属性描述符看到继承的getter:

For your own understanding, you can see the inherited getter by looking at the property descriptor of Response.prototype.url:

Object.getOwnPropertyDescriptor(Object.getPrototypeOf(new Response()), 'url')
> {get: ƒ, set: undefined, enumerable: true, configurable: true}

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