如何使用 Jest 监视默认的导出函数? [英] How to spy on a default exported function with Jest?

问题描述

假设我有一个导出默认函数的简单文件:

Suppose I have a simple file exporting a default function:

// UniqueIdGenerator.js
const uniqueIdGenerator = () => Math.random().toString(36).substring(2, 8);

export default uniqueIdGenerator;

我会像这样使用:

import uniqueIdGenerator from './UniqueIdGenerator';
// ...
uniqueIdGenerator();

我想在我的测试中断言在保持原始功能的同时调用了这个方法.我会用 jest.spyOn 但是，它需要一个对象以及一个函数名称作为参数.你怎么能以干净的方式做到这一点?对于任何感兴趣的人，jasmine 都有一个类似的 GitHub 问题.

I want to assert in my test that this method was called while keeping the original functionality. I'd do that with jest.spyOn however, it requires an object as well as a function name as parameters. How can you do this in a clean way? There's a similar GitHub issue for jasmine for anyone interested.

推荐答案

我最终放弃了默认导出:

I ended up ditching the default export:

// UniqueIdGenerator.js
export const uniqueIdGenerator = () => Math.random().toString(36).substring(2, 8);

然后我可以像这样使用和监视它:

And then I could use and spy it like this:

import * as UniqueIdGenerator from './UniqueIdGenerator';
// ...
const spy = jest.spyOn(UniqueIdGenerator, 'uniqueIdGenerator');

有些人推荐将它们包装在一个 const 对象中，并出口.我想你也可以使用一个类来包装.

Some recommend wrapping them in a const object, and exporting that. I suppose you can also use a class for wrapping.

但是，如果您不能修改类，仍然有一个(不太好的)解决方案:

However, if you can't modify the class there's still a (not-so-nice) solution:

import * as UniqueIdGenerator from './UniqueIdGenerator';
// ...
const spy = jest.spyOn(UniqueIdGenerator, 'default');

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