38

我们正在从 Java 8 迁移到 Java 11,从而从 Spring Boot 1.5.6 迁移到 2.1.2。我们注意到,在使用 RestTemplate 时,“+”号不再编码为“%2B”(由 SPR-14828 更改)。这没关系,因为 RFC3986 没有将“+”列为保留字符,但在 Spring Boot 端点中接收时仍将其解释为“”(空格)。

我们有一个搜索查询,它可以将可选的时间戳作为查询参数。查询看起来像http://example.com/search?beforeTimestamp=2019-01-21T14:56:50%2B00:00.

如果没有双重编码,我们无法弄清楚如何发送编码的加号。查询参数2019-01-21T14:56:50+00:00将被解释为2019-01-21T14:56:50 00:00. 如果我们自己对参数进行编码 ( 2019-01-21T14:56:50%2B00:00),那么它将被接收并解释为2019-01-21T14:56:50%252B00:00

另一个约束是,我们希望在设置 restTemplate 时在别处设置基本 url,而不是在执行查询的位置。

或者,有没有办法强制“+”不被端点解释为“”?

我写了一个简短的例子,展示了一些实现更严格编码的方法,它们的缺点解释为注释:

package com.example.clientandserver;

import org.springframework.boot.CommandLineRunner;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.boot.web.client.RestTemplateBuilder;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.bind.annotation.RestController;
import org.springframework.web.client.RestTemplate;
import org.springframework.web.util.DefaultUriBuilderFactory;
import org.springframework.web.util.UriComponentsBuilder;
import org.springframework.web.util.UriUtils;

import java.nio.charset.StandardCharsets;
import java.util.HashMap;
import java.util.Map;

@SpringBootApplication
@RestController
public class ClientAndServerApp implements CommandLineRunner {

    public static void main(String[] args) {
        SpringApplication.run(ClientAndServerApp.class, args);
    }

    @Override
    public void run(String... args) {
        String beforeTimestamp = "2019-01-21T14:56:50+00:00";

        // Previously - base url and raw params (encoded automatically). 
        // This worked in the earlier version of Spring Boot
        {
            RestTemplate restTemplate = new RestTemplateBuilder()
               .rootUri("http://localhost:8080").build();
            UriComponentsBuilder b = UriComponentsBuilder.fromPath("/search");
            if (beforeTimestamp != null) {
                b.queryParam("beforeTimestamp", beforeTimestamp);
            }
            restTemplate.getForEntity(b.toUriString(), Object.class);
            // Received: 2019-01-21T14:56:50 00:00
            //       Plus sign missing here ^
        }

        // Option 1 - no base url and encoding the param ourselves.
        {
            RestTemplate restTemplate = new RestTemplate();
            UriComponentsBuilder b = UriComponentsBuilder
                .fromHttpUrl("http://localhost:8080/search");
            if (beforeTimestamp != null) {
                b.queryParam(
                    "beforeTimestamp",
                    UriUtils.encode(beforeTimestamp, StandardCharsets.UTF_8)
                );
            }
            restTemplate.getForEntity(
                b.build(true).toUri(), Object.class
            ).getBody();
            // Received: 2019-01-21T14:56:50+00:00
        }

        // Option 2 - with templated base url, query parameter is not optional.
        {
            RestTemplate restTemplate = new RestTemplateBuilder()
                .rootUri("http://localhost:8080")
                .uriTemplateHandler(new DefaultUriBuilderFactory())
                .build();
            Map<String, String> params = new HashMap<>();
            params.put("beforeTimestamp", beforeTimestamp);
            restTemplate.getForEntity(
                "/search?beforeTimestamp={beforeTimestamp}",
                Object.class,
                params);
            // Received: 2019-01-21T14:56:50+00:00
        }
    }

    @GetMapping("/search")
    public void search(@RequestParam String beforeTimestamp) {
        System.out.println("Received: " + beforeTimestamp);
    }
}
4

4 回答 4

32

我们意识到在编码完成后可以在拦截器中修改 URL。所以一个解决方案是使用一个拦截器,它对查询参数中的加号进行编码。

RestTemplate restTemplate = new RestTemplateBuilder()
        .rootUri("http://localhost:8080")
        .interceptors(new PlusEncoderInterceptor())
        .build();

一个简短的例子:

public class PlusEncoderInterceptor implements ClientHttpRequestInterceptor {

    @Override
    public ClientHttpResponse intercept(HttpRequest request, byte[] body, ClientHttpRequestExecution execution) throws IOException {
        return execution.execute(new HttpRequestWrapper(request) {
            @Override
            public URI getURI() {
                URI u = super.getURI();
                String strictlyEscapedQuery = StringUtils.replace(u.getRawQuery(), "+", "%2B");
                return UriComponentsBuilder.fromUri(u)
                        .replaceQuery(strictlyEscapedQuery)
                        .build(true).toUri();
            }
        }, body);
    }
}
于 2019-01-24T09:49:48.583 回答
3

这个问题也在这里讨论过。

RestTemplate 上 URI 变量的编码 [SPR-16202]

一个更简单的解决方案是将 URI 构建器上的编码模式设置为 VALUES_ONLY。

    DefaultUriBuilderFactory builderFactory = new DefaultUriBuilderFactory();
    builderFactory.setEncodingMode(DefaultUriBuilderFactory.EncodingMode.VALUES_ONLY);
    RestTemplate restTemplate = new RestTemplateBuilder()
            .rootUri("http://localhost:8080")
            .uriTemplateHandler(builderFactory)
            .build();

这与使用查询参数时使用 PlusEncodingInterceptor 的结果相同。

于 2019-11-14T14:59:41.077 回答
1

感谢https://stackoverflow.com/users/4466695/gregor-eesmaa,它解决了我的问题。只是想补充一点,如果您可以在调用之前格式化 URL RestTemplate,您可以立即修复 URL(而不是在 中替换它PlusEncoderInterceptor):

UriComponentsBuilder uriBuilder = UriComponentsBuilder.fromUriString("/search");
uriBuilder.queryParam("beforeTimestamp", "2019-01-21T14:56:50+00:00");
URI uriPlus = uriBuilder.encode().build(false).toUri();

// import org.springframework.util.StringUtils;
String strictlyEscapedQuery = StringUtils.replace(uriPlus.getRawQuery(), "+", "%2B");
URI uri = UriComponentsBuilder.fromUri(uriPlus)
        .replaceQuery(strictlyEscapedQuery)
        .build(true).toUri();

// prints "/search?beforeTimestamp=2019-01-21T14:56:50%2B00:00"
System.out.println(uri);

然后你可以在RestTemplate通话中使用:

RequestEntity<?> requestEntity = RequestEntity.get(uri).build();
ResponseEntity<String> responseEntity = restTemplate.exchange(requestEntity, String.class);
于 2020-12-24T09:13:36.913 回答
1

为了解决这类问题,我发现手动构建 URI 更容易。

URI uri = new URI(siteProperties.getBaseUrl()
  + "v3/elements/"
  + URLEncoder.encode("user/" + user + "/type/" + type, UTF_8)
  + "/"
  + URLEncoder.encode(id, UTF_8)
);

restTemplate.exchange(uri, DELETE, new HttpEntity<>(httpHeaders), Void.class);
于 2021-07-10T07:08:03.560 回答