假设我有my-file.ts
:
async function run() {
await new Promise(resolve => {
setTimeout(() => resolve(), 3000);
});
setTimeout(() => console.log('hello'), 5000);
}
export default run();
我想用Jest测试它。如果我这样做:
it('should wait until all code has executed', async () => {
const start = new Date();
console.log('Waiting...', new Date());
await require('./my-file').default;
const stop = new Date();
console.log(`done after ${(stop.getTime() - start.getTime()) / 1000} s`);
});
结果是:
console.log test/tmp.test.ts:3
Waiting... 2020-01-03T11:56:59.822Z
console.log test/tmp.test.ts:8
done after 3.009 s
这很有道理。但是当我编译并运行my-file.ts
时,节点等待所有超时清除,并且正确显示了所有控制台日志:
$ tsc my-file.ts && node my-file.js
hello
$
我想重现Node的行为,即等待所有异步代码执行。我怎么能开玩笑呢?
还有,任何人都可以解释为什么以下内容根本不等待吗?
it('should wait until all code has executed', async () => {
await require('./my-file');
});
答案 0 :(得分:0)
问题在于3秒钟后您的Promise resolves
和来自笑话的测试会运行,因为您用await require('./my-file').default;
阻止了执行,直到Promise为resolved
或rejected
。您的第二次超时只会在5秒钟后向事件队列添加超时回调,该回调将运行。
要复制等待5秒钟,请在第二次超时后解决Promise。另外,如果您返回新的承诺,则不需要async/await
。
function run() {
return new Promise(resolve => {
setTimeout(() => console.log('hello after 3 sec'), 3000);
setTimeout(() => resolve(), 5000);
});
}
export default run();