如何使用restTemplate Spring-mvc发送Multipart表单数据 [英] How to send Multipart form data with restTemplate Spring-mvc
问题描述
我正在尝试将带有RestTemplate的文件上传到带有Jetty的Raspberry Pi。在Pi上有一个运行的servlet：
I am trying to upload a file with RestTemplate to Raspberry Pi with Jetty. On Pi there is a servlet running:
protected void doPost(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
PrintWriter outp = resp.getWriter();
StringBuffer buff = new StringBuffer();
File file1 = (File) req.getAttribute("userfile1");
String p = req.getParameter("path");
boolean success = false;
if (file1 == null || !file1.exists()) {
buff.append("File does not exist\n");
} else if (file1.isDirectory()) {
buff.append("File is a directory\n");
} else {
File outputFile = new File(req.getParameter("userfile1"));
if(isValidPath(p)){
p = DRIVE_ROOT + p;
final File finalDest = new File(p
+ outputFile.getName());
success = false;
try {
copyFileUsingFileChannels(file1, finalDest);
finalDest.setWritable(true);
success = true;
} catch (Exception e) {
e.printStackTrace();
}
if (success){
buff.append("File successfully uploaded.\n");
}
else{
buff.append("Failed to save file.");
}
}
else{
buff.append("Invalid path.\n");
}
}
outp.write(buff.toString());
}
我能用curl成功完成
I am able to successfully do it with curl
curl --form userfile1=@/home/pi/src/CreateNewFolderServlet.java --form press = OK localhost：2222 / pi / GetFileServlet？path =/ media /
这是在webapp上具有相同功能的方法。
This is the method that is supposed to have the same functionality on webapp.
@ResponseBody
@RequestMapping(value="/upload/",method=RequestMethod.POST ,produces = "text/plain")
public String uploadFile(MultipartHttpServletRequest request2, HttpServletResponse response2){
Iterator<String> itr = request2.getFileNames();
MultipartFile file = request2.getFile(itr.next());
System.out.println(file.getOriginalFilename() +" uploaded!");
System.out.println(file.toString());
MultiValueMap<String, Object> parts = new LinkedMultiValueMap<String, Object>();
parts.add("userfile1",file);
//reqEntity.addPart("userfile1", file);
String path="/public/";
RestTemplate restTemplate = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.MULTIPART_FORM_DATA);
System.out.println("1");
HttpEntity<MultiValueMap<String, Object>> request = new HttpEntity<MultiValueMap<String, Object>>(parts, headers);
String url = url2+"/pi/GetFileServlet?path="+path;
System.out.println("2");
/* restTemplate.getMessageConverters().add(new FormHttpMessageConverter());
restTemplate.getMessageConverters().add(
new MappingJackson2HttpMessageConverter());*/
System.out.println("3");
ResponseEntity<String> response = restTemplate.exchange(url, HttpMethod.POST, request,String.class);
System.out.println("4");
System.out.println("response : " +response);
if(response==null||response.getBody().trim()==""){
return "error";
}
return response.getBody();
}
这是我得到的输出：
ui-elements.html已上传！
ui-elements.html uploaded!
org.springframework.web.multipart.support.StandardMultipartHttpServletRequest$StandardMultipartFile@47e7673e
org.springframework.web.multipart.support.StandardMultipartHttpServletRequest$StandardMultipartFile@47e7673e
1
2
3
正如您所见，数字4未打印
控制台中没有异常。
调试期间发现异常：
As you can see number 4 is not printed No exception in console. Exceptions found during debugging:
org.springframework.http.converter.HttpMessageNotWritableException: Could not write JSON: No serializer found for class java.io.ByteArrayInputStream and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationFeature.FAIL_ON_EMPTY_BEANS) ) (through reference chain: org.springframework.web.multipart.support.StandardMultipartFile["inputStream"]); nested exception is com.fasterxml.jackson.databind.JsonMappingException: No serializer found for class java.io.ByteArrayInputStream and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationFeature.FAIL_ON_EMPTY_BEANS) ) (through reference chain: org.springframework.web.multipart.support.StandardMultipartFile["inputStream"])
推荐答案
您正在获取异常，因为RestTemplate的默认MessageConverters都不知道如何序列化MultipartFile文件包含的InputStream。通过RestTemplate发送对象时，在大多数情况下，您希望发送POJO。你可以通过将MultipartFile的字节添加到MultiValueMap而不是MultipartFile本身来解决这个问题。
You are getting the exception because none of RestTemplate's default MessageConverters know how to serialize the InputStream contained by the MultipartFile file. When sending objects via RestTemplate, in most cases you want to send POJOs. You can fix this by adding the bytes of the MultipartFile to the MultiValueMap instead of the MultipartFile itself.
我认为你的servlet部分也有问题。例如
I think there is also something wrong with your servlet part. For instance
File file1 = (File) req.getAttribute("userfile1");
应始终返回null，因为ServletRequest的getAttribute方法不返回请求/表单参数但是由servlet上下文设置的属性。你确定它实际上正在使用你的curl示例吗？
should always return null, as ServletRequest's getAttribute method does not return request/form parameters but attributes set by the servlet context. Are you sure it is actually working with your curl example?
这是一个Spring MVC方法将文件转发到servlet的例子：
Here is an example of a Spring MVC method forwarding a file to a servlet:
Servlet（虽然我测试了它在Spring MVC容器中运行），改编自此处：
Servlet (though I tested it running in a Spring MVC container), adapted from here:
@RequestMapping("/pi")
private void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
final String path = request.getParameter("destination");
final Part filePart = request.getPart("file");
final String fileName = request.getParameter("filename");
OutputStream out = null;
InputStream fileContent = null;
final PrintWriter writer = response.getWriter();
try {
out = new FileOutputStream(new File(path + File.separator
+ fileName));
fileContent = filePart.getInputStream();
int read = 0;
final byte[] bytes = new byte[1024];
while ((read = fileContent.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
writer.println("New file " + fileName + " created at " + path);
} catch (FileNotFoundException fne) {
writer.println("You either did not specify a file to upload or are "
+ "trying to upload a file to a protected or nonexistent "
+ "location.");
writer.println("<br/> ERROR: " + fne.getMessage());
} finally {
if (out != null) {
out.close();
}
if (fileContent != null) {
fileContent.close();
}
if (writer != null) {
writer.close();
}
}
}
Spring MVC方法：
Spring MVC method:
@ResponseBody
@RequestMapping(value="/upload/", method=RequestMethod.POST,
produces = "text/plain")
public String uploadFile(MultipartHttpServletRequest request)
throws IOException {
Iterator<String> itr = request.getFileNames();
MultipartFile file = request.getFile(itr.next());
MultiValueMap<String, Object> parts =
new LinkedMultiValueMap<String, Object>();
parts.add("file", new ByteArrayResource(file.getBytes()));
parts.add("filename", file.getOriginalFilename());
RestTemplate restTemplate = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.MULTIPART_FORM_DATA);
HttpEntity<MultiValueMap<String, Object>> requestEntity =
new HttpEntity<MultiValueMap<String, Object>>(parts, headers);
// file upload path on destination server
parts.add("destination", "./");
ResponseEntity<String> response =
restTemplate.exchange("http://localhost:8080/pi",
HttpMethod.POST, requestEntity, String.class);
if (response != null && !response.getBody().trim().equals("")) {
return response.getBody();
}
return "error";
}
使用这些我可以通过MVC方法成功将文件上传到servlet以下卷曲：
Using these I can succesfully upload a file through the MVC method to the servlet by the following curl:
curl --form file=@test.dat localhost:8080/upload/
这篇关于如何使用restTemplate Spring-mvc发送Multipart表单数据的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
