Javascript “调试”;在jasmine指定的超时内未调用异步回调。默认值为“u timeout”u INTERVAL;错误
我正在运行一个测试,基本上是这样做的:打开窗口,等待加载程序消失,点击菜单并多次填写表单 一切似乎都很好,表单填写得很好,但大约60秒后,我的测试失败,并显示一条愤怒的错误消息:Javascript “调试”;在jasmine指定的超时内未调用异步回调。默认值为“u timeout”u INTERVAL;错误,javascript,typescript,asynchronous,jasmine,protractor,Javascript,Typescript,Asynchronous,Jasmine,Protractor,我正在运行一个测试,基本上是这样做的:打开窗口,等待加载程序消失,点击菜单并多次填写表单 一切似乎都很好,表单填写得很好,但大约60秒后,我的测试失败,并显示一条愤怒的错误消息: Message: Error: Timeout - Async callback was not invoked within timeout specified by jasmine.DEFAULT_TIMEOUT_INTERVAL. Stack: Error: Timeout - Async
Message:
Error: Timeout - Async callback was not invoked within timeout specified by jasmine.DEFAULT_TIMEOUT_INTERVAL.
Stack:
Error: Timeout - Async callback was not invoked within timeout specified by jasmine.DEFAULT_TIMEOUT_INTERVAL.
at listOnTimeout (internal/timers.js:549:17)
at processTimers (internal/timers.js:492:7)
Message:
Error: Timeout - Async callback was not invoked within timeout specified by jasmine.DEFAULT_TIMEOUT_INTERVAL.
Stack:
Error: Timeout - Async callback was not invoked within timeout specified by jasmine.DEFAULT_TIMEOUT_INTERVAL.
at listOnTimeout (internal/timers.js:549:17)
at processTimers (internal/timers.js:492:7)
对于这个输出,我不知道如何调试它,特别是因为一切似乎都运行得很好。有没有办法让它更冗长?我需要知道在超时时间内没有调用哪个异步回调