在GAE/J的FileService API設置文件名

Question

我寫了成功上傳的斑點一個servlet，但它是一個沒有名字上傳如何在servlet

在這裏設置上傳文件的名稱是servlet的段代碼

public void doGet(HttpServletRequest req, HttpServletResponse resp) 
      throws IOException { 



     FileService fileService = FileServiceFactory.getFileService(); 

      // Create a new Blob file with mime-type "text/plain" 

      String url="http://www.cbwe.gov.in/htmleditor1/pdf/sample.pdf"; 
      URL url1=new URL(url); 
      HttpURLConnection conn=(HttpURLConnection) url1.openConnection(); 
      String content_type=conn.getContentType(); 
      InputStream stream =conn.getInputStream(); 
      AppEngineFile file = fileService.createNewBlobFile("application/pdf"); 

      file=new AppEngineFile(file.getFullPath()); 
     Boolean lock = true; 
      FileWriteChannel writeChannel = fileService.openWriteChannel(file, lock); 

      // This time we write to the channel directly 
      String s1=""; 
      String s2=""; 

      byte[] bytes = IOUtils.toByteArray(stream); 


      writeChannel.write(ByteBuffer.wrap(bytes)); 
      writeChannel.closeFinally(); 

Answer 1

將文件名作爲第二個參數傳遞給fileService.createNewBlobFile("application/pdf", "filename.pdf")。

Answer 2

您可以從您的網址獲取文件名

字符串fileNameWithoutExtn = url.substring（0，url.lastIndexOf（） ''）;

然後通過文件名作爲參數

FileWriteChannel writeChannel = fileService.openWriteChannel(file, lock,fileNameWithoutExtn);