带有参数的Spring RestTemplate GET

我必须进行REST调用，其中包括自定义头和查询参数。我设置我的HttpEntity只有头(没有正文)，我使用RestTemplate.exchange()方法如下:

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

Map<String, String> params = new HashMap<String, String>();
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 entity = new HttpEntity(headers);

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

这在客户端失败，因为调度程序servlet无法将请求解析到处理程序。调试之后，似乎没有发送请求参数。

当我使用请求体和没有查询参数的POST做一个交换时，它工作得很好。

有人有什么想法吗?

当前回答

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);
    }

2018-05-17 05:26:12

其他回答

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。

2011-11-29 15:07:52

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

2019-11-21 08:14:26

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);

2013-07-27 21:33:37

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();

}

2022-12-05 15:16:50

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);

2018-07-04 14:27:53

带有参数的Spring RestTemplate GET

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