捕获身份验证异常

Question

我使用此代码发送 XML 请求：

RestClient client = RestClientBuilder.builder()
                        .gatewayUrl(URL) 
                        .build();
Mono<AuthorizeResponse> result = client.executeAndReceiveAuthorize(request);
                response = result.block();

 public RestClient(String gatewayUrl, String token, String username, String password, SslContext sslContext) {
        this.token = token;
        this.gatewayUrl = gatewayUrl;
        WebClient.Builder builder = WebClient.builder().baseUrl(gatewayUrl);
        if (sslContext != null) {
            HttpClient httpClient = HttpClient.create().secure(sslContextSpec -> sslContextSpec.sslContext(sslContext));
            ClientHttpConnector httpConnector = new ReactorClientHttpConnector(httpClient);
            builder.clientConnector(httpConnector);
        }
        if (username != null && password != null) {
            builder.filter(basicAuthentication(username, password));
        }
        client = builder.build();
    }

    public Mono<AuthorizeResponse> executeAndReceiveAuthorize(AuthorizeRequest transaction) {
        Mono<AuthorizeRequest> transactionMono = Mono.just(transaction);
        return client.post().uri(checkTrailingSlash(gatewayUrl) + token)
                .header(HttpHeaders.USER_AGENT, "Mozilla/5.0")
                .accept(MediaType.APPLICATION_XML)
                .contentType(MediaType.APPLICATION_XML)
                .body(transactionMono, AuthorizeRequest.class)
                .retrieve()
                .bodyToMono(AuthorizeResponse.class);
    }

但有时我会收到此错误：

org.springframework.web.reactive.function.client.WebClientResponseException$Unauthorized: 401 Unauthorized
 at java.base/java.lang.Thread.run(Thread.java:834)
2019-08-27 22:05:45,720 ERROR [stderr] (processingTransactionGenesisAuthorizeContainer-1)       Suppressed: java.lang.Exception: #block terminated with an error
 Caused by: java.lang.NullPointerException: null

当我收到错误 401 时如何捕获和处理异常？ 如果可能的话，我想在response = result.block(); 行之后处理异常。

Answer 1

首先，在响应式代码中显式使用block 并不是一个好主意。这背后的原因是，它将代码变成了阻塞，这不是反应流应该如何工作的。所以，请删除response = result.block();

现在，如果您想处理特定状态，则可以修改您的 executeAndReceiveAuthorize 方法以使用 WebClient 的 onStatus(Predicate<HttpStatus> var1, Function<ClientResponse, Mono<? extends Throwable>> var2)

public Mono<AuthorizeResponse> executeAndReceiveAuthorize (AuthorizeRequest transaction){

    Mono<AuthorizeRequest> transactionMono = Mono.just(transaction);

    return client.post().uri(checkTrailingSlash(gatewayUrl) + token)
        .header(HttpHeaders.USER_AGENT, "Mozilla/5.0").accept(MediaType.APPLICATION_XML)
        .contentType(MediaType.APPLICATION_XML).body(transactionMono, AuthorizeRequest.class)
        .retrieve().onStatus(status -> status == HttpStatus.UNAUTHORIZED, clientResponse -> Mono
            .error(new SomeCustomAuthorizationException("Some failure exception")))
        .bodyToMono(AuthorizeResponse.class);
}

Answer 2

在mono 上定义doOnError 怎么样？

例如：

Mono<AuthorizeResponse> result = client.executeAndReceiveAuthorize(request);
response = result.doOnError(e -> {
                      throw new RuntimeException(e);
                  })
                 .block();

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如果您需要访问 json 响应，请检查此answer。