Spring RestTemplate GET与参数

我必须打一个REST电话,其中包括自定义标头和查询参数。我只设置HttpEntity了标题(没有正文),然后使用以下RestTemplate.exchange()方法:

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,dispatcher servlet因为无法将请求解析为处理程序。调试完成后,似乎没有发送请求参数。

当我POST使用请求正文和无查询参数进行交换时,它工作正常。

有人有什么想法吗?

回答:

为了轻松地操纵URL / path / params /等等,可以使用Spring的UriComponentsBuilder类。手动连接字符串比较干净,它会为你处理URL编码:

HttpHeaders headers = new HttpHeaders();

headers.set("Accept", MediaType.APPLICATION_JSON_VALUE);

UriComponentsBuilder builder = UriComponentsBuilder.fromHttpUrl(url)

.queryParam("msisdn", msisdn)

.queryParam("email", email)

.queryParam("clientVersion", clientVersion)

.queryParam("clientType", clientType)

.queryParam("issuerName", issuerName)

.queryParam("applicationName", applicationName);

HttpEntity<?> entity = new HttpEntity<>(headers);

HttpEntity<String> response = restTemplate.exchange(

builder.toUriString(),

HttpMethod.GET,

entity,

String.class);

以上是 Spring RestTemplate GET与参数 的全部内容, 来源链接: utcz.com/qa/434829.html

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