Versuchen, MultipartFile mit Postman

Question

Ich versuche, eine Multipart-Datei mit PostMan hochladen und Fehler bekommen. Hier ist der Code und Screenshots:

http://imgur.com/pZ5cXrh

http://imgur.com/NaWQceO

@RequestMapping(value = "/upload", method = RequestMethod.POST) 
public void uploadFileHandler(@RequestParam("name") String name, 
     @RequestParam("name") MultipartFile file) { 

    if (!file.isEmpty()) { 
     try { 
      byte[] bytes = file.getBytes(); 

      // Creating the directory to store file 
      //String rootPath = System.getProperty("catalina.home"); 
      String rootPath = "C:\\Desktop\\uploads"; 
      File dir = new File(rootPath + File.separator + "tmpFiles"); 
      if (!dir.exists()) 
       dir.mkdirs(); 

      // Create the file on server 
      File serverFile = new File(dir.getAbsolutePath() 
        + File.separator + name); 
      BufferedOutputStream stream = new BufferedOutputStream(
        new FileOutputStream(serverFile)); 
      stream.write(bytes); 
      stream.close(); 

      System.out.println("Server File Location=" 
        + serverFile.getAbsolutePath()); 

      System.out.println("You successfully uploaded file=" + name); 
     } catch (Exception e) { 
      System.out.println("You failed to upload " + name + " => " + e.getMessage()); 
     } 
    } else { 
     System.out.println("You failed to upload " + name 
       + " because the file was empty."); 
    } 
} 

Answer 1

Sie sollten etwas wie dieses:

@RequestMapping(value = "/upload", method = RequestMethod.POST, consumes = "multipart/form-data") 
    public void uploadFileHandler(@RequestParam("name") String name, 
            @RequestParam("file") MultipartFile file) { 

     if (!file.isEmpty()) { 
      try { 
       byte[] bytes = file.getBytes(); 

       // Creating the directory to store file 
       //String rootPath = System.getProperty("catalina.home"); 
       String rootPath = "C:\\Users\\mworkman02\\Desktop\\uploads"; 
       File dir = new File(rootPath + File.separator + "tmpFiles"); 
       if (!dir.exists()) 
        dir.mkdirs(); 

       // Create the file on server 
       File serverFile = new File(dir.getAbsolutePath() 
         + File.separator + name); 
       BufferedOutputStream stream = new BufferedOutputStream(
         new FileOutputStream(serverFile)); 
       stream.write(bytes); 
       stream.close(); 

       System.out.println("Server File Location=" 
         + serverFile.getAbsolutePath()); 

       System.out.println("You successfully uploaded file=" + name); 
      } catch (Exception e) { 
       System.out.println("You failed to upload " + name + " => " + e.getMessage()); 
      } 
     } else { 
      System.out.println("You failed to upload " + name 
        + " because the file was empty."); 
     } 
    } 

Beachten Sie consumes = "multipart/form-data". Es ist notwendig für Ihre hochgeladene Datei, weil Sie einen mehrteiligen Anruf haben sollten. Sie sollten @RequestParam("file") MultipartFile file anstelle von @RequestParam("name") MultipartFile file) haben.

Natürlich sollten Sie einen multipartview konfiguriert haben Resolver die eingebaute Unterstützung für Apache-Commons-Datei-Upload und nativen Servlet 3.