在这里，我尝试使用 react 器进行异步和非阻塞调用，对于每个请求，我可能必须按顺序调用两个服务(在我下面的例子中，getAccountInfoFromAAA 和 getAccountInfoFromBBB).

这是我的 ItemRequest 对象:

public class ItemRequest {
    private Account account;
    private Result firstServiceResult;
    private Result secondServiceResult;
    private PostingParameterCode postingParameterCode; //enum 
    //...
    //...
    //getters and setters
}

因此，我的请求输入将包含多个 itemRequest，对于每个 itemRequest，我正在执行异步调用:

public void getAccountData(List<ItemRequest> itemRequests) {
    ImmutableList<ItemRequest> list = ImmutableList.copyOf(itemRequests);
    Flux.fromIterable(list).flatMap(this::callBothSors).blockLast();
}

public Mono<ItemRequest> callBothSors(ItemRequest itemRequest) {
    return getAccountDataService.getAccountDataFromAAAandBBB(itemRequest); 
    //here, it will enter into a sequential call for each itemRequest
}

这是我的第一个服务调用接口(interface):

public Mono<ItemRequest> getAccountDataFromAAA(ItemRequest itemRequest);

这是我的第二个服务调用接口(interface):

public Mono<ItemRequest> getAccountDataFromBBB(ItemRequest itemRequest);

此方法将根据条件依次调用最多两次:

public Mono<ItemRequest> getAccountDataFromAAAandBBB(ItemRequest itemRequest){
    Mono<ItemRequest> firstCallResult = Mono.empty();
    Mono<ItemRequest> secondCallResult = Mono.empty();

if(isFirstServiceCallRequired(itemRequest)){
    firstCallResult = this.firstServiceCallImpl.getAccountDataFromAAA(itemRequest); 
//basically, firstService call will update the accountKey information and
//will also set the result status to OK which is required to decide 
//whether to make secondService call.
} else {
    //Account key is already present, so just update the result status which I need later.
    Result result = new Result();
    result.setStatus(Result.Status.OK);
    result.setMessageText("First call not required as account info is set for item request");
    itemRequest.setFirstServiceResult(result);
}

//Now, before calling the second service, I need to check the following:

if(null!= itemRequest.getFirstServiceResult() && 
    itemRequest.getFirstServiceResult().getStatus().equals(Result.Status.OK) && 
    itemRequest.getPostingParameterCode().equals(PostingParameterCode.MOBILECREDIT)){ 
        secondCallResult = this.secondServiceCallImpl.getAccountDataFromBBB(itemRequest);
    }

    return firstCallResult.then(secondCallResult);  //attaching the
    //firstCallResult and secondCallResult to produce a single Mono

这在不需要 firstCallResult 时工作正常。但是当需要第一次调用时，这个条件检查不会通过，因为我不会更新第一次调用结果对象:

if(null != itemRequest.getFirstServiceResult() && 
    itemRequest.getFirstServiceResult().getStatus().equals(Result.Status.OK) && 
    itemRequest.getPostingParameterCode().equals(PostingParameterCode.MOBILECREDIT))) { ... } 
 //this condition check will not pass because first service call is not actually executing

如果我输入以下语句，这两种情况都可以正常工作:

if(isFirstServiceCallRequired(itemRequest)){
        firstCallResult = this.firstServiceCallImpl.getAccountDataFromAAA(itemRequest); 
        firstCallResult.block(); //adding this case will work on both cases 
    }

但是，我认为我不会以这种方式让 react 堆受益。 我想有这样的逻辑:

Mono<ItemRequest> result = firstService.call(...)
    .doOnNext(/*do something */)
    .then( ... secondService.call())

但无法找到将 secondService 与 firstService 链接起来以获得单声道结果​​并进行这些条件检查的方法。 条件检查很重要，因为我并不总是想执行第二个服务。有什么方法可以将 secondService 与 firstService 链接起来以获得结果并进行这些条件检查吗？

很抱歉问了这么长的问题。任何建议/帮助将不胜感激。

最佳答案

在为这个问题提供了赏金点之后，我真的很兴奋并期待一些答案。 但无论如何，我能够改进我的初始解决方案并进行这些条件检查。

我做了以下事情: 我将返回类型从 Mono<ItemRequest> 更改为至 Mono<Void>在两次服务调用中，因为我基本上是将数据更新为 ItemRequest列表:

在这里处理并行调用(每个并行调用都有一个顺序调用):

public void getAccountData(List<ItemRequest> itemRequests) {
        ImmutableList<ItemRequest> list = ImmutableList.copyOf(itemRequests);
        Flux.fromIterable(list).flatMap(this::callBothSors).blockLast();
    }

    public Mono<Void> callBothSors(ItemRequest itemRequest) {
        return getAccountDataService.getAccountDataFromAAAandBBB(itemRequest);
        //here, it will enter into a sequential call for each itemRequest
    }

这些是我的 firstServiceCall和 secondServiceCall界面变化:

public Mono<Void> getAccountDataFromAAA(ItemRequest itemRequest);

public Mono<Void> getAccountDataFromBBB(ItemRequest itemRequest);

然后我链接了 secondServiceCall与 firstServiceCall获得单声道结果​​并进行这些条件检查:

public Mono<Void> getAccountDataFromAAAandBBB(ItemRequest itemRequest){
    Mono<Void> callSequence = Mono.empty();

    if(isFirstServiceCallRequired(itemRequest)){
        callSequence = this.firstServiceCallImpl.getAccountDataFromAAA(itemRequest);
    } else {
        //Account key is already present, so just update the result status which I need later.
        Result result = new Result();
        result.setStatus(Result.Status.OK);
        result.setMessageText("First call not required as account info is set for item request");
        itemRequest.setFirstServiceResult(result);
    }

    return callSequence.thenEmpty(Mono.defer(() -> {
        //note: Mono.defer ==>> Create a Mono provider that will supply a target Mono to subscribe to 
        //for each subscriber downstream.
        //only if the firstServiceCall result is successful & other condition check successful,
        // I am calling secondServiceCall:  
        if(shouldCallSecondService(itemRequest)){
            return this.secondServiceCallImpl.getAccountDataFromAAAandBBB(itemRequest);
        } else {
            return Mono.empty();
        }
    }))

关于java - react 器中基于条件检查的异步顺序调用，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/53014975/