Sinon Spy从未打过电话,但应该是

问题描述 投票:1回答:1

问题

我正在使用Jest和SinonJS测试自定义redux中间件,更确切地说,我想测试是否在中间件内的特殊条件下调用某些函数。

我使用SinonJS来创建间谍,我用Jest运行我的测试。我初步化了间谍以寻找我想要跟踪的特定功能,当我检查间谍是否被调用时,间谍甚至不应该(手动测试)。

这是我要测试的中间件:

import { Cookies } from 'react-cookie';
import setAuthorizationToken from './setAuthorizationToken';

let cookies = new Cookies();

export const bindTokenWithApp = (store) => (next) => (action) => {
    // Select the token before action
    const previousToken = getToken(store.getState());
    // Dispatch action
    const result = next(action);
    // Select the token after dispatched action
    const nextToken = getToken(store.getState());

    if (previousToken !== nextToken) {
        if (nextToken === '') {
            setAuthorizationToken(false);
            cookies.remove(SESSION_COOKIE_NAME, COOKIE_OPTIONS);
        } else {
            cookies.set(SESSION_COOKIE_NAME, nextToken, COOKIE_OPTIONS);
            setAuthorizationToken(nextToken);
        }
    }

    return result;
};

这是我的实际测试

import { bindTokenWithApp } from './middleware';
import { Cookies } from 'react-cookie';
import sinon, { assert } from 'sinon';
import setAuthorizationToken from './setAuthorizationToken';

describe('bindTokenWithApp', () => {
    const next = jest.fn();
    const action = jest.fn();
    let cookies = new Cookies();

    it('removes cookies when there is no token', () => {
        // My actual not working spies
        const cookieSpy = sinon.spy(cookies.remove);
        const authSpy = sinon.spy(setAuthorizationToken);

        // Stub for the specific case. This code works, 
        // I console.logged in the middleware and I'm getting the below values
        const getState = sinon.stub();
        getState.onFirstCall().returns({ auth: { token: 'a token' } });
        getState.onSecondCall().returns({ auth: { token: '' } });
        const store = { getState: getState };

        bindTokenWithApp(store)(next)(action);

        assert.calledOnce(cookieSpy);
        assert.calledOnce(authSpy);
        // Output : AssertError: expected remove to be called once but was called 0 times
        // AssertError: expected setAuthorizationToken to be called once but was called 0 times

        cookieSpy.restore(); // <= This one works
        authSpy.restore(); // TypeError: authSpy.restore is not a function 

  });
});

我已经阅读过SinonJS doc和一些StackOverFlow帖子,但没有解决方案。我也不能打电话给authSpy.restore();。我想我没有以正确的方式初始化间谍,我在SinonJS中误解了一个概念,但我找不到哪一个!

setAuthorizationToken签名是 (别名)const setAuthorizationToken :( token:any)=> void import setAuthorizationToken

我认为这是一个经典模块,所以我无法弄清楚为什么我会与authSpy.restore();斗争

javascript unit-testing sinon spy
1个回答
0
投票

你有两个间谍实际上有两个不同的修复,都有相同的潜在问题。 sinon.spy(someFunction)实际上并没有包裹someFunction本身,它为它返回间谍但不执行任何替换。

对于第一个间谍,存在一个自动包装对象方法的简写:sinon.spy(cookie, 'remove')应该做你需要的。

对于第二个间谍来说,它更复杂,因为你需要将间谍包裹在setAuthorizationToken的默认导出周围。为此你需要像proxyquire这样的东西。 Proxyquire是一种专门的需求机制,允许您使用所需的测试方法替换导入。以下是您需要做的事情的简要说明:

const authSpy = sinon.spy(setAuthorizationToken);
bindTokenWithApp = proxyquire('./middleware', { './setAuthorizationToken': authSpy});
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