我真是个白痴，我把查询参数和url参数搞混了。我有点希望有一个更好的方式来填充我的查询参数，而不是一个丑陋的连接字符串，但我们有。这只是一个用正确的参数构建URL的例子。如果你把它作为一个字符串传递，Spring也会为你处理编码。

uriVariables也在查询字符串中展开。例如，下面的调用将展开account和name的值:

restTemplate.exchange("http://my-rest-url.org/rest/account/{account}?name={name}",
    HttpMethod.GET,
    httpEntity,
    clazz,
    "my-account",
    "my-name"
);

实际的请求url是

http://my-rest-url.org/rest/account/my-account?name=my-name

查看HierarchicalUriComponents.expandInternal(UriTemplateVariables)了解更多细节。 Spring的版本是3.1.3。

我也尝试过类似的东西，RoboSpice的例子帮助我解决了这个问题:

HttpHeaders headers = new HttpHeaders();
headers.set("Accept", "application/json");

HttpEntity<String> request = new HttpEntity<>(input, createHeader());

String url = "http://awesomesite.org";
Uri.Builder uriBuilder = Uri.parse(url).buildUpon();
uriBuilder.appendQueryParameter(key, value);
uriBuilder.appendQueryParameter(key, value);
...

String url = uriBuilder.build().toString();

HttpEntity<String> response = restTemplate.exchange(url, HttpMethod.GET, request , String.class);

为了方便地操作URL / path / params /等等，您可以使用Spring的UriComponentsBuilder类创建一个URL模板，其中包含参数占位符，然后在RestOperations.exchange(…)调用中提供这些参数的值。它比手动连接字符串更干净，它会为你处理URL编码:

HttpHeaders headers = new HttpHeaders();
headers.set(HttpHeaders.ACCEPT, MediaType.APPLICATION_JSON_VALUE);
HttpEntity<?> entity = new HttpEntity<>(headers);

String urlTemplate = UriComponentsBuilder.fromHttpUrl(url)
        .queryParam("msisdn", "{msisdn}")
        .queryParam("email", "{email}")
        .queryParam("clientVersion", "{clientVersion}")
        .queryParam("clientType", "{clientType}")
        .queryParam("issuerName", "{issuerName}")
        .queryParam("applicationName", "{applicationName}")
        .encode()
        .toUriString();

Map<String, ?> params = new HashMap<>();
params.put("msisdn", msisdn);
params.put("email", email);
params.put("clientVersion", clientVersion);
params.put("clientType", clientType);
params.put("issuerName", issuerName);
params.put("applicationName", applicationName);

HttpEntity<String> response = restOperations.exchange(
        urlTemplate,
        HttpMethod.GET,
        entity,
        String.class,
        params
);

我采取不同的方法，你可能同意或不同意，但我想从.properties文件控制，而不是编译的Java代码

内部应用程序。属性文件

endpoint。url = https://host/resource?

Java代码在这里，你可以写if或切换条件，以找出端点URL在.properties文件中是否有@PathVariable(包含{})或@RequestParam (yourURL?key=value)等…然后调用相应的方法…这样它是动态的，不需要在未来的一站式商店更改代码…

我试图在这里给出比实际代码更多的想法……试着为@RequestParam和@PathVariable等编写泛型方法…然后在需要时进行相应的调用

  @Value("${endpoint.url}")
  private String endpointURL;
  // you can use variable args feature in Java
  public String requestParamMethodNameHere(String value1, String value2) {
    RestTemplate restTemplate = new RestTemplate();
    restTemplate
           .getMessageConverters()
           .add(new MappingJackson2HttpMessageConverter());

    HttpHeaders headers = new HttpHeaders();
    headers.set("Accept", MediaType.APPLICATION_JSON_VALUE);
    HttpEntity<String> entity = new HttpEntity<>(headers);

    try {
      String formatted_URL = MessageFormat.format(endpointURL, value1, value2);
      ResponseEntity<String> response = restTemplate.exchange(
                    formatted_URL ,
                    HttpMethod.GET,
                    entity,
                    String.class);
     return response.getBody();
    } catch (Exception e) { e.printStackTrace(); }

至少从Spring 3开始，许多RestTemplate方法不是使用UriComponentsBuilder来构建URL(这有点啰嗦)，而是在参数路径中接受占位符(不仅仅是交换)。

从文档中可以看到:

Many of the RestTemplate methods accepts a URI template and URI template variables, either as a String vararg, or as Map<String,String>. For example with a String vararg: restTemplate.getForObject( "http://example.com/hotels/{hotel}/rooms/{room}", String.class, "42", "21"); Or with a Map<String, String>: Map<String, String> vars = new HashMap<>(); vars.put("hotel", "42"); vars.put("room", "21"); restTemplate.getForObject("http://example.com/hotels/{hotel}/rooms/{room}", String.class, vars);

参考:https://docs.spring.io/spring/docs/current/spring-framework-reference/integration.html # rest-resttemplate-uri

如果你查看JavaDoc for RestTemplate并搜索“URI Template”，你可以看到哪些方法可以使用占位符。

public static void main(String[] args) {
         HttpHeaders httpHeaders = new HttpHeaders();
         httpHeaders.set("Accept", MediaType.APPLICATION_JSON_VALUE);
         final String url = "https://host:port/contract/{code}";
         Map<String, String> params = new HashMap<String, String>();
         params.put("code", "123456");
         HttpEntity<?> httpEntity  = new HttpEntity<>(httpHeaders); 
         RestTemplate restTemplate  = new RestTemplate();
         restTemplate.exchange(url, HttpMethod.GET, httpEntity,String.class, params);
    }

如果您为RestTemplate传递非参数参数，那么考虑到参数，您将为传递的每个不同URL都有一个Metrics。你想要使用参数化url:

http://my-url/action?param1={param1}&param2={param2}

而不是

http://my-url/action?param1=XXXX&param2=YYYY

第二种情况是使用UriComponentsBuilder类得到的结果。

实现第一个行为的方法如下:

Map<String, Object> params = new HashMap<>();
params.put("param1", "XXXX");
params.put("param2", "YYYY");

String url = "http://my-url/action?%s";

String parametrizedArgs = params.keySet().stream().map(k ->
    String.format("%s={%s}", k, k)
).collect(Collectors.joining("&"));

HttpHeaders headers = new HttpHeaders();
headers.set("Accept", MediaType.APPLICATION_JSON_VALUE);
HttpEntity<String> entity = new HttpEntity<>(headers);

restTemplate.exchange(String.format(url, parametrizedArgs), HttpMethod.GET, entity, String.class, params);

在Spring Web 4.3.6中我也看到了

public <T> T getForObject(String url, Class<T> responseType, Object... uriVariables)

这意味着你不必创建一个丑陋的地图

如果你有这个url

http://my-url/action?param1={param1}&param2={param2}

你可以选择

restTemplate.getForObject(url, Response.class, param1, param2)

or

restTemplate.getForObject(url, Response.class, param [])

    String uri = http://my-rest-url.org/rest/account/{account};

    Map<String, String> uriParam = new HashMap<>();
    uriParam.put("account", "my_account");

    UriComponents builder = UriComponentsBuilder.fromHttpUrl(uri)
                .queryParam("pageSize","2")
                        .queryParam("page","0")
                        .queryParam("name","my_name").build();

    HttpEntity<String> requestEntity = new HttpEntity<>(null, getHeaders());

    ResponseEntity<String> strResponse = restTemplate.exchange(builder.toUriString(),HttpMethod.GET, requestEntity,
                        String.class,uriParam);

    //final URL: http://my-rest-url.org/rest/account/my_account?pageSize=2&page=0&name=my_name

RestTemplate:使用UriComponents (URI变量和请求参数)构建动态URI

将哈希映射转换为查询参数字符串:

Map<String, String> params = new HashMap<>();
params.put("msisdn", msisdn);
params.put("email", email);
params.put("clientVersion", clientVersion);
params.put("clientType", clientType);
params.put("issuerName", issuerName);
params.put("applicationName", applicationName);

UriComponentsBuilder builder = UriComponentsBuilder.fromHttpUrl(url);
for (Map.Entry<String, String> entry : params.entrySet()) {
    builder.queryParam(entry.getKey(), entry.getValue());
}

HttpHeaders headers = new HttpHeaders();
headers.set("Accept", "application/json");

HttpEntity<String> response = restTemplate.exchange(builder.toUriString(), HttpMethod.GET, new HttpEntity(headers), String.class);

如果您的url是http://localhost:8080/context path?msisdn = {msisdn}电子邮件= {email}

然后

Map<String,Object> queryParams=new HashMap<>();
queryParams.put("msisdn",your value)
queryParams.put("email",your value)

适用于您所描述的resttemplate交换方法

我提供了一个RestTemplate GET方法的代码片段与路径参数示例

public ResponseEntity<String> getName(int id) {
    final String url = "http://localhost:8080/springrestexample/employee/name?id={id}";
    Map<String, String> params = new HashMap<String, String>();
    params.put("id", id);
    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.APPLICATION_JSON);
    HttpEntity request = new HttpEntity(headers);
    ResponseEntity<String> response = restTemplate.exchange(url, HttpMethod.GET, String.class, params);
    return response;
}

可以对String使用follow代码。

URL_EXAMPLE="http://{domain}/Index.php?Username={user}&password={password}";

String domain = "example.com";
String user = "user";
String password = "password";

String data=this.restTemplate.getForObject(URL_EXAMPLE,String.class,domain,user,password);

嗨，我建立url与查询参数使用这段代码:

UriComponentsBuilder.fromHttpUrl(url)
                .queryParam("bikerPhoneNumber", "phoneNumberString")
                .toUriString();

还有一个解决方法:

private String execute(String url, Map<String, String> params) {
    UriComponentsBuilder uriBuilder = UriComponentsBuilder.fromUriString(url)
    // predefined params
            .queryParam("user", "userValue")
            .queryParam("password", "passwordValue");
    params.forEach(uriBuilder::queryParam);
    HttpHeaders headers = new HttpHeaders() {{
        setContentType(MediaType.APPLICATION_FORM_URLENCODED);
        setAccept(List.of(MediaType.APPLICATION_JSON));
    }};
    ResponseEntity<String> request = restTemplate.exchange(uriBuilder.toUriString(), 
                HttpMethod.GET, new HttpEntity<>(headers), String.class);
    return request.getBody();

}