如何使用 Sinon JS 测试异步函数？

Question

下面是我想要实现的一个最小示例：fn3()如果异步函数已fn2()解析（调用时fn1），我想测试被调用的情况。但是我以某种方式未能使用sinon的存根语法做到这一点。我想知道我误解了什么。

// System Under Test

export function fn1() {

    fn2().then(() => {

        fn3();

    });

}

export async function fn2() {

    return new Promise((resolve, reject) => {

        // expensive work

        resolve();

    });

}

export function fn3() {

    console.log("fn3");

}

// Test

import * as Page from "xxx";

it("test async", () => {

    // stub this async to isolate the SUT

    sinon.stub(Page, "fn2").resolves();

    const stub = sinon.stub(Page, "fn3");

    Page.fn1();

    sinon.assert.calledOnce(stub);

});

/*

    AssertError: expected fn3 to be called once but was called 0 times

      276 |         Page.fn1();

      277 |

    > 278 |         sinon.assert.calledOnce(stub);

          |                      ^

      279 |     });

      280 | });

      at Object.fail (node_modules/sinon/lib/sinon/assert.js:107:21)

      at failAssertion (node_modules/sinon/lib/sinon/assert.js:66:16)

      at Object.calledOnce (node_modules/sinon/lib/sinon/assert.js:92:13)

      at Object.<anonymous> (src/test.test.tsx:278:22)

*/

Answer 1

正如@jonrsharpe 所建议的那样，根据 JavaScript 的“单线程、事件循环”性质，您sinon.assert.calledOnce(stub);将在解决（即调用）fn2内部承诺之前被调用。这里有一篇文章供参考：https ://dev.to/lydiahallie/javascript-visualized-promises-async-await-5gke 。Page.fn1.then(() => { fn3(); })您可以尝试将代码更改为：&nbsp; &nbsp; ...&nbsp; &nbsp; return Page.fn1().then(() => {&nbsp; &nbsp; &nbsp; sinon.assert.calledOnce(stub);&nbsp; &nbsp; });&nbsp; &nbsp;&nbsp;&nbsp; &nbsp; ...要么it("test async", async () => {&nbsp; &nbsp; // stub this async to isolate the SUT&nbsp; &nbsp; sinon.stub(Page, "fn2").resolves();&nbsp; &nbsp; const stub = sinon.stub(Page, "fn3");&nbsp; &nbsp; await Page.fn1();&nbsp; &nbsp; sinon.assert.calledOnce(stub);});所以在解决sinon.assert.calledOnce(stub);后会被调用Page.fn1()。