Spring WebClient - logging is2xxSuccessful HTTPStatus

Question

我正在使用Spring WebClient下载文件，如下所示:

private void dnloadFileAPI(String theId, String destination) {
log.info("Downloading file.. " + theId);
Flux<DataBuffer> dataBuffer = webClient
.get()
.uri("/some/fancy/" + theId + "/api")
.retrieve()
.onStatus(HttpStatus::is2xxSuccessful, response -> Mono.just(new CustomException("Success")))
.bodyToFlux(DataBuffer.class);
DataBufferUtils.write(dataBuffer, Paths.get(destination), StandardOpenOption.CREATE).share().block();
}

文件下载正常。我唯一的问题是，当响应是200时，我只想记录一行，像这样:

log.info("{}", theId + " - File downloaded successfully")

我也试过了，但没有得到我正在寻找的-如何记录Spring WebClient响应

简而言之，是否有某种方法可以在不编写单独的CustomException的情况下实现上述目标?感到迷茫和无助。如有任何建议，我将不胜感激。

Answer 1

我认为在bodyToFlux(DataBuffer.class)行之后添加以下内容将是您所需要的

.doOnComplete(() -> log.info("File downloaded successfully"))

像这样

private void dnloadFileAPI(String theId, String destination) {
log.info("Downloading file.. " + theId);
Flux<DataBuffer> dataBuffer = webClient
.get()
.uri("/some/fancy/" + theId + "/api")
.retrieve()
.bodyToFlux(DataBuffer.class)
.doOnComplete(() -> log.info("File downloaded successfully"));
DataBufferUtils.write(dataBuffer, Paths.get(destination), StandardOpenOption.CREATE).share().block();
}

Answer 2

我们可以使用doOnSuccess方法

.doOnSuccess(response -> logger.info("File downloaded successfully"))

webflux version 5.3.22