Spring WebClient - 记录 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();
    }

文件可以正常下载。我唯一苦苦挣扎的是，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东西 - How to log Spring WebClient response

简而言之，有没有什么方法可以在不单独写CustomException的情况下实现？在这里感到迷茫和不知所措。任何这方面的指针都将受到高度赞赏。

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"))

类似的东西

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();
}