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

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)

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

2018-11-29 17:47:40

其他回答

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。

2018-04-29 17:48:55

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);

2022-03-09 06:22:44

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

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