反应式java将Mono Object转换为不带block（）的Object；

Question

请帮助我转换＜code＞Mono

java prettyprint-override">public RulesEngineResponse prepareResponse(Result rulesResponse) {
    RulesEngineResponse rulesEngineResponse = rulesEngineService(data).block();
}

public Mono < RulesEngineResponse > rulesEngineService(AggregatedMessage input) {
    return WebClient.builder()
        .defaultHeader(HttpHeaders.CONTENT_TYPE, MediaType.APPLICATION_JSON_VALUE)
        .build()
        .post()
        .uri(this.rulesEngineUrl)
        .body(BodyInserters.fromObject(input))
        .accept(MediaType.APPLICATION_JSON)
        .header(HttpHeaders.CONTENT_TYPE)
        .exchange()
        .flatMap(clientResponse - > clientResponse.toEntity(RulesEngineResponse.class)
            .map(responseEntity - > responseEntity.getBody()));
}

Answer 1

您可以使用toFuture（）方法，如下所示：

RulesEngineResponse rulesEngineResponse = rulesEngineService(data).toFuture().get();