我有pay
方法,我应该调用initiatePayment和onSuccess
我应该调用confirmPayment
。如果两个调用中的任何一个都有异常,它应该发出异常
public Single<PayResponse> pay(PayRequest apiRequest) {
return client.initiatePayment(apiRequest)
.doOnSuccess(initiatePaymentResponse -> {
client.confirmPayment(initiatePaymentResponse.getPaymentId())
.doOnSuccess(confirmPaymentResponse -> doConfirmationLogic(confirmPaymentResponse ))
.doOnError(ex -> {ex.printStackTrace();logError(ex);});
})
.doOnError(ex -> {ex.printStackTrace();logError(ex);});
}
在我引用的代码中,confirmPayment
发生错误,但initiatePayment
正常继续。
如何将异常从内部doOnError
传播到外部doOnError
?
doOnXxx()方法仅用于回调目的,它们不涉及流式传输管道,这就是它们被称为“副作用方法”的原因。所以没有办法将错误从doOnXxx()传播到上游。
错误始终是Rx世界中的终端事件,每当发生错误时管道都会被取消,因此不需要在doOnSuccess()方法中执行某些操作以确保到目前为止一切都“正常”。因此,您可以简单地以这种方式编写代码,而不是将代码嵌套到doOnSuccess()链中:
/*
you can deal with errors using these operators:
onErrorComplete
onErrorResumeNext
onErrorReturn
onErrorReturnItem
onExceptionResumeNext
retry
retryUntil
retryWhen
*/
return client.initiatePayment(apiRequest)
//if in initiatePayment was error this will send cancel upstream and error downstream
.map(initiatePaymentResponse -> { client.confirmPayment(initiatePaymentResponse.getPaymentId());})
//if in confirmPayment was error this never happens
.map(confirmPaymentResponse -> doConfirmationLogic(confirmPaymentResponse))
//every error in this pipeline will trigger this one here
.doOnError(ex -> {
ex.printStackTrace();
logError(ex);
});