我刚接触Spring Boot框架的WebClient。如何仅在出错时记录4xx/5xx及其错误主体,但如果成功,则将clientResponse作为String返回,稍后我们可以使用gson对其进行序列化?
String postResponse =
post(
ENDPOINT,
token,
request,
Request.class);
Response response = gson.fromJson(postResponse, Response.class); private String post(String endpoint, String token, Mono<?> requestBody, Class<?> requestBodyClass) {
WebClient.Builder webClientBuilder = WebClient.builder();
WebClient.RequestBodySpec requestBodySpec = webClientBuilder.build().post().uri(uri);
requestBodySpec.header(
"Authorization", BEARER_TOKEN_PREFIX + this.accessTokens.get(token));
return requestBodySpec
.header("Content-Type", HEADER_VALUE_CONTENT_TYPE)
.body(Mono.just(requestBody), requestBodyClass)
.retrieve()
.bodyToMono(String.class)
.block();
}发布于 2020-10-28 10:07:48
我可以这样建议:
private String post(String endpoint, String token, Mono<?> requestBody, Class<?> requestBodyClass) {
Mono<String> stringMono = WebClient.builder().build()
.post()
.uri(endpoint)
.body(BodyInserters.fromValue(requestBodyClass))
.exchange()
.flatMap(SomeServiceImpl::getResultAsStringOrLog4xx5xx);
//get the JSON from the Mono, but you might not want to block
}(我省略了标题)
问题的实际答案是,使用类似的方法执行日志记录:
private static Mono<String> getResultAsStringOrLog4xx5xx(ClientResponse response) {
if (response.statusCode().is2xxSuccessful()) {
return response.bodyToMono(String.class);
} else if (response.statusCode().is5xxServerError()) {
Logger.error(); //up to you
} else {
//at this point, you can have your logic on error code
}
return
}还有其他方法可以做到这一点,希望这能帮助你入门。
另外,尽量不要在应用程序中阻塞()。
谢谢
https://stackoverflow.com/questions/64565064
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