Spring WebClient-日志记录为2xxSuccessful 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();
    }

文件下载得很好。我唯一挣扎的是，当响应为200时，我只想像这样记录一行：

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

我也试过了，但没有得到我想要的-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？在这里感到迷失和无知。这方面的任何指针都将受到高度赞赏。

Answer 1

我认为在body ToFlux（DataBuffer.class）行之后添加以下内容将是您需要的

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

像这样的

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();
}