Question

我正在尝试传递路径参数并在URL中查询参数但我得到一个奇怪的错误。下面是代码

    String url = "http://test.com/Services/rest/{id}/Identifier"
    Map<String, String> params = new HashMap<String, String>();
    params.put("id", "1234");
    UriComponentsBuilder builder = UriComponentsBuilder.fromUriString(url)
                                        .queryParam("name", "myName");
    String uriBuilder = builder.build().encode().toUriString();
    restTemplate.exchange(uriBuilder , HttpMethod.PUT, requestEntity,
                    class_p, params);

我的网址正变为http://test.com/Services/rest/%7Bid%7D/Identifier?name=myName

我该怎么做才能让它发挥作用。我期待http://test.com/Services/rest/{id}/Identifier?name=myName以便params将id添加到url

请建议。在此先感谢

Answer 1

我会使用buildAndExpand中的UriComponentsBuilder来传递所有类型的URI参数。

例如：

String url = "http://test.com/solarSystem/planets/{planet}/moons/{moon}";

// URI (URL) parameters
Map<String, String> uriParams = new HashMap<String, String>();
uriParams.put("planets", "Mars");
uriParams.put("moons", "Phobos");

// Query parameters
UriComponentsBuilder builder = UriComponentsBuilder.fromUriString(url)
        // Add query parameter
        .queryParam("firstName", "Mark")
        .queryParam("lastName", "Watney");

System.out.println(builder.buildAndExpand(uriParams).toUri());
/**
 * Console output:
 * http://test.com/solarSystem/planets/Mars/moons/Phobos?firstName=Mark&lastName=Watney
 */

restTemplate.exchange(builder.buildAndExpand(uriParams).toUri() , HttpMethod.PUT,
        requestEntity, class_p);

/**
 * Log entry:
 * org.springframework.web.client.RestTemplate Created PUT request for "http://test.com/solarSystem/planets/Mars/moons/Phobos?firstName=Mark&lastName=Watney"
 */

Answer 2

使用带有参数映射的TestRestTemplate.exchange函数的单线。

restTemplate.exchange("/someUrl?id={id}", HttpMethod.GET, reqEntity, respType, ["id": id])

像这样初始化的参数映射是 groovy 初始化程序*

Answer 3

来自 Michal Foksa 的答案的一个问题是它首先添加查询参数，然后扩展路径变量。如果查询参数包含括号，例如{foobar}，这会导致异常。

安全的方法是先扩展路径变量，然后添加查询参数：

String url = "http://test.com/Services/rest/{id}/Identifier";
Map<String, String> params = new HashMap<String, String>();
params.put("id", "1234");
URI uri = UriComponentsBuilder.fromUriString(url)
        .buildAndExpand(params)
        .toUri();
uri = UriComponentsBuilder
        .fromUri(uri)
        .queryParam("name", "myName")
        .build()
        .toUri();
restTemplate.exchange(uri , HttpMethod.PUT, requestEntity, class_p);

Answer 4

String url = "http://test.com/Services/rest/{id}/Identifier";
Map<String, String> params = new HashMap<String, String>();
params.put("id", "1234");
URI uri = UriComponentsBuilder.fromUriString(url)
        .buildAndExpand(params)
        .toUri();
uri = UriComponentsBuilder
        .fromUri(uri)
        .queryParam("name", "myName")
        .build()
        .toUri();
restTemplate.exchange(uri , HttpMethod.PUT, requestEntity, class_p);

<块引用>

安全的方法是先展开路径变量，再添加查询参数：

对我来说，这导致了重复的编码，例如空格被解码为 %2520（空格 -> %20 -> %25）。

我通过以下方式解决了：

String url = "http://test.com/Services/rest/{id}/Identifier";
Map<String, String> params = new HashMap<String, String>();
params.put("id", "1234");
UriComponentsBuilder uriComponentsBuilder = UriComponentsBuilder.fromUriString(url);
uriComponentsBuilder.uriVariables(params);
Uri uri = uriComponentsBuilder.queryParam("name", "myName");
        .build()
        .toUri();
restTemplate.exchange(uri , HttpMethod.PUT, requestEntity, class_p);

基本上我使用 uriComponentsBuilder.uriVariables(params); 添加路径参数。文档说：

<块引用>

... 与 UriComponents.expand(Map) 或 buildAndExpand(Map) 相比，当您需要提供 URI 变量而不构建 UriComponents 实例时，此方法很有用，或者可能预先扩展一些共享默认值，例如作为主机和端口。 ...

来源：https://docs.spring.io/spring-framework/docs/current/javadoc-api/org/springframework/web/util/UriComponentsBuilder.html#uriVariables-java.util.Map-

Answer 5

一种简单的方法是：

String url = "http://test.com/Services/rest/{id}/Identifier"

UriComponents uriComponents = UriComponentsBuilder.fromUriString(url).build();
uriComponents = uriComponents.expand(Collections.singletonMap("id", "1234"));

，然后添加查询参数。

RestTemplate：如何一起发送URL和查询参数

5 个答案: