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
);

嗨，我建立url与查询参数使用这段代码:

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

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);
    }

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);

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。

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 [])