我想用Spring RestTemplate做一个简单的HTTP POST。 的WESB服务接受JSON的参数,例如:{"name":"mame","email":"[email protected]"}
在Spring Rest中使用JSON的HTTP POST
public static void main(String[] args) {
final String uri = "url";
RestTemplate restTemplate = new RestTemplate();
// Add the Jackson message converter
restTemplate.getMessageConverters().add(new MappingJackson2HttpMessageConverter());
// create request body
String input = "{ \"name\": \"name\", \"email\": \"[email protected]\" }";
JsonObject request = new JsonObject();
request.addProperty("model", input);
// set headers
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_JSON);
headers.set("Authorization", "Basic " + "xxxxxxxxxxxx");
HttpEntity<String> entity = new HttpEntity<String>(request.toString(), headers);
// send request and parse result
ResponseEntity<String> response = restTemplate
.exchange(uri, HttpMethod.POST, entity, String.class);
System.out.println(response);
}
当我测试这个代码,我得到这个错误:
Exception in thread "main" org.springframework.web.client.HttpClientErrorException: 400 Bad Request
当我打电话web服务与卷曲我有正确的结果:
curl -X POST -H "Authorization: Basic xxxxxxxxxx" --header "Content-Type: application/json" --header "Accept: application/json" -d "{ \"name\": \"name\", \"email\": \"[email protected]\" } " "url"
添加你要访问web服务代码和邮递员或正确执行卷曲表达的截图...... –
I'haven't访问web服务。通过使用curl,我可以使用以下命令调用ws:curl -X POST -H“Authorization:Basic xxxxxxxxxx”--header“Content-Type:application/json”--header“Accept:application/json”-d“{\ “name \”:\“name \”,\“email \”:\“[email protected] \”} “”url“ –