我正在尝试同时发出9个api调用。所有这些调用都会返回不同的响应对象。在此之前,我们有8个api调用,由于这些mono是不同的类型,所以我使用了Mono.zip,如下所示。
Mono<ResponseEntity<Service1Response>> monoService1 = callService1();
Mono<ResponseEntity<Service2Response>> monoService2 = callService2();
...
Mono<ResponseEntity<Service3Response>> monoService7 = callService7();
Mono<ResponseEntity<Service4Response>> monoService8 = callService8();
MixResponse mix = Mono.zip(monoService1, monoService2, monoService3, monoService4, monoService5, monoService6, monoService7, monoService8).flatMap(response -> {
MixResponse mixResp = new MixResponse();
mixResp.setResponse1(response.getT1().getBody());
mixResp.setResponse2(response.getT2().getBody());
mixResp.setResponse3(response.getT3().getBody());
mixResp.setResponse4(response.getT4().getBody());
mixResp.setResponse5(response.getT5().getBody());
mixResp.setResponse6(response.getT6().getBody());
mixResp.setResponse7(response.getT7().getBody());
mixResp.setResponse8(response.getT8().getBody());
return Mono.just(mixResp);
})).block();但是现在我们又有了一个服务,Mono.zip只支持最多8个monos。除了Mono.zip之外,我还可以在我的情况下使用它吗?如果这个问题看起来很蠢的话,很抱歉。我是春天的新人。提前谢谢。
发布于 2022-05-26 05:50:41
您可以使用以Mono.zip作为参数和组合器函数的Iterable来组合结果并转换为特定类型:
<R> Mono<R> zip(final Iterable<? extends Mono<?>> monos, Function<? super Object[], ? extends R> combinator)
List<Mono<? extends ResponseEntity<?>>> requests = List.of(
monoService1,
monoService2,
monoService3,
monoService4
);
Mono.zip(requests, responses -> {
MixResponse mixResp = new MixResponse();
mixResp.setResponse1(((ResponseEntity<Service1Response>) responses[0]).getBody());
mixResp.setResponse2(((ResponseEntity<Service2Response>) responses[1]).getBody());
mixResp.setResponse3(((ResponseEntity<Service3Response>) responses[2]).getBody());
mixResp.setResponse4(((ResponseEntity<Service4Response>) responses[3]).getBody());
return mixResp;
});https://stackoverflow.com/questions/72386882
复制相似问题