我正在尝试更新或创建xml文件(如果不存在)。然后我使用下面的代码使用服务的PUT方法发送文件。
public void importClusterProperties(RestManPropertyHolder propertyHolder,File file,String id) throws RestManServiceException {
testRestTemplate = new TestRestTemplate(propertyHolder.getSbusUserName(), propertyHolder.getSbusUserPassword());
String sbusUrl = utils.prepareGatewayURI(propertyHolder);
try {
HttpHeaders requestHeaders = new HttpHeaders();
List <MediaType> mediaTypeList = new ArrayList<MediaType>();
mediaTypeList.add(MediaType.APPLICATION_ATOM_XML);
requestHeaders.setAccept(mediaTypeList);
requestHeaders.setContentType(MediaType.APPLICATION_ATOM_XML);
HttpEntity<String> requestEntity = new HttpEntity<String>(requestHeaders);
// Create the HTTP PUT request,
ResponseEntity<String> response = testRestTemplate.exchange(sbusUrl + "/clusterproperty?",HttpMethod.PUT, requestEntity,String.class);
if (null != response) {
System.out.println("RESPONSE::" + response.toString());
}
} catch (RestClientException rce) {
System.out.println("REST EXCEPTION:::" + rce.getMessage());
}
}
如何将原始xml文件传递给RestTemplate而不先将其转换为java对象?
答案 0 :(得分:0)
将文件转换为字节数组并使用ByteArrayHttpMessageConverter
发送。
RestTemplate restTemplate = new RestTemplate();
// Add ByteArrayHttpMessageConverter if not present by default.
restTemplate.getMessageConverters().add(
new ByteArrayHttpMessageConverter());
String fileName = "path + file name";
// FileUtils is from Apache Commons IO
// import org.apache.commons.io.FileUtils;
byte[] requestBody = FileUtils.readFileToByteArray(new File(fileName));
HttpEntity<byte[]> requestEntity = new HttpEntity<byte[]>(requestBody , requestHeaders);
// Create the HTTP PUT request,
ResponseEntity<byte[]> response =
restTemplate.exchange("URL ...." , HttpMethod.PUT ,
requestEntity , byte[].class);