使用订阅成功条件的单元测试 Angular 显示错误

Question

嗨，我需要为这个函数编写单元测试角，如下所示：

  resendOTP() {
      this.authProvider.resendOTP(this.tempId, this.tempToken).subscribe((result) => {
      this.tempId = result.data.resendOtp.id;
      this.tempToken = result.data.resendOtp.access_token;
      this.showResendInfo = true;
      this.showResend = false;
      this.messageBoxService.showSuccess('SMS has been sent', 2000);
      this.startTimer();
    }, (error) => {
      if (error.graphQLErrors[0].message.includes('Maximum limit exceeded')) {
        this.validError = true;
        this.loginState = 'login';
        this.propertyAlert.message = `You\'ve requested OTP Code after the 5th attempt and your account has been
        suspended. Please contact the administrator to recover it.`;
        // setTimeout(() => {
        //   this.reloadCaptcha();
        // }, 100);
      }
    });
  }

我成功地为错误条件创建

它（'显示帐户暂停后，使5 otp atemps', () =

但是当我为成功条件而创造时

it('show account success after make 5 otp attemps', () => {
  let response = [];
  spyOn(providerService, 'resendOTP').and.returnValue(of(response));
  component.resendOTP();
  expect(component.showResend).toBeFalsy()
  expect(component.showResendInfo).toBeTruthy()    
});

它显示了“可观察”类型的参数

Answer 1

它说响应对象的类型错误。将响应的值更改为FetchResault

考虑在resendOTP()方法中处理订阅，因为现在您可能会引入内存泄漏。例如。

ngOnInit() {
  this.subscription = this.yourObservable.subscribe(() => {
   //some code
   });
}

ngOnDestroy() {
  this.subscription.unsubscribe();
}   

显式声明返回类型也是一种很好的做法，例如

 resendOTP(): void {
    //your code
  }