sinon:如何存根整个类，而不仅仅是一个方法 [英] sinon: How to stub an entire class, rather than just a method

问题描述

我有一个正在测试的类，它创建了另一个类的实例.我想删除整个第二个类，以便它的构造函数永远不会被调用.例如，如果我有这个设置:

I have a class under test that creates instances of another class. I want to stub out the entirety of the second class, so that its constructor never gets called. For example, If I have this setup:

Test.js

class Test {
  constructor() {
  }

  func() {
    let foo = new Foo()
    foo.hello()
  }
}

Foo.js

class Foo {
  constructor() {
    this.a = 1
    this.b = 2
    this.c = 3
    console.log('original constructor')
  }

  hello() {
    console.log('original hello')
  }

  goodbye() {
    console.log('original goodbye')
  }
}

在我的测试文件中，我想以某种方式删除整个 Foo 类，以便当我为 Test.func() 运行我的测试时不会调用原始的 Foo 构造函数，而是调用返回假 Foo 对象的存根构造函数.然后我将存根假的 Foo 对象的 hello 函数来打印 stubbed hello 而不是 original hello.

In my test file, I want to somehow stub out the entirety of the Foo class, so that when I run my test for for Test.func() it doesn't call the original Foo constructor, but rather a stubbed constructor that returns an fake Foo object. I'll then stub the hello function of the fake Foo object to print stubbed hello instead of original hello.

我怎样才能像这样存根整个班级?

How can I stub the entire class like this?

注意:我不想创建一个可以在我的测试文件中使用的存根实例.我需要存根构造函数本身，以便如果堆栈中的某些东西调用构造函数，它会返回一个存根实例.

NOTE: I do NOT want to create a stub instance that I can use inside my test file. I need to stub the constructor itself, so that if something up the stack calls the constructor, it gets back a stub instance.

推荐答案

在 sinon 文档中:

In sinon documentation:

如果您想创建 MyConstructor 的存根对象，但不想调用构造函数，请使用此实用程序函数.

If you want to create a stub object of MyConstructor, but don’t want the constructor to be invoked, use this utility function.

var stub = sinon.createStubInstance(MyConstructor)

http://sinonjs.org/releases/v1.17.7/stubs/

这篇关于sinon:如何存根整个类，而不仅仅是一个方法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！