Recipe 13.4 Sending an Audio File

Problem

You want to send an audio file such as an MPEG layer 3 (MP3) media type.

Solution

Use a java.io.BufferedInputStream to fetch the audio data, and the javax.servlet.ServletOutputStream from the javax.servlet.http.HttpServletResponse object to send the data to the client.

Discussion

The code in Example 13-4 uses the same approach as the prior recipes, except for the MIME type, which is specified as audio/mpeg .

Web browsers associate a number of other MIME types for MP3 files, including audio/x-mpeg , audio/mp3 , and audio/x-mp3 .

The user requests a filename in the URL, as in:

 http://localhost:8080/home/sendmp3?file=song_name

The deployment descriptor ( web.xml ) maps the servlet path /sendmp3 to the servlet class of Example 13-4: com.jspservletcookbook.SendMp3 . If the requested file does not already have the .mp3 suffix, then the code adds that file extension. A context-param element in the deployment descriptor specifies the directory where the audio files are kept:

 <context-param>     <param-name>mp3-dir</param-name>     <param-value>h:/home/mp3s</param-value> </context-param>

Example 13-4 uses this directory name, plus the filename, as the constructor parameter to create a new java.io.File object, which is the source for a java.io.FileInputStream . A BufferedInputStream buffers the bytes from the song file, which the ServletOutputStream response reads.

Example 13-4. Sending an MP3 file

 package com.jspservletcookbook;  import java.io.FileInputStream; import java.io.BufferedInputStream; import java.io.File;  import java.io.IOException; import javax.servlet.*; import javax.servlet.http.*; public class SendMp3 extends HttpServlet {   public void doGet(HttpServletRequest request,      HttpServletResponse response) throws ServletException, IOException {              //get the filename from the "file" parameter         String fileName = (String) request.getParameter("file");         if (fileName == null  fileName.equals(""))             throw new ServletException(             "Invalid or non-existent file parameter in SendMp3 servlet.");  // add the .mp3 suffix if it doesn't already exist         if (fileName.indexOf(".mp3") == -1)            fileName = fileName + ".mp3";                    //where are MP3 files kept?         String mp3Dir = getServletContext( ).getInitParameter("mp3-dir");         if (mp3Dir == null  mp3Dir.equals(""))             throw new ServletException(             "Invalid or non-existent mp3-Dir context-param.");  ServletOutputStream stream = null;         BufferedInputStream buf = null;         try{                   stream = response.getOutputStream( );             File mp3 = new File(mp3Dir + "/" + fileName);  //set response headers             response.setContentType("audio/mpeg");  response.addHeader(             "Content-Disposition","attachment; filename="+fileName );             response.setContentLength( (int) mp3.length( ) );                    FileInputStream input = new FileInputStream(mp3);             buf = new BufferedInputStream(input);             int readBytes = 0;             //read from the file; write to the ServletOutputStream             while((readBytes = buf.read( )) != -1)                 stream.write(readBytes);         } catch (IOException ioe){                   throw new ServletException(ioe.getMessage( ));                   } finally {                   //close the input/output streams             if(stream != null)                 stream.close( );             if(buf != null)                 buf.close( );           }        } //doGet       public void doPost(HttpServletRequest request,      HttpServletResponse response) throws ServletException, IOException {                doGet(request,response);   }  }

Review Recipe 13.1 for a further explanation of this code, including the warning at the end of the "Discussion" section about logged exceptions that may occur with Internet Explorer.

Problem

Solution

Discussion

Example 13-4. Sending an MP3 file

See Also