Servlet 3 File Uplaod example using Tomcat 7
- Thursday, November 8, 2012, 17:43
- Java EE
- 19,556 views
Many java web developer aware that for uploading files on server there was no api support in servlet specifications. We have to depend on third party library to upload files. Apache common’s file upload library is very popular for this task.
Servlet 3 comes with default support of file upload. To make your servlet multipart aware developers suppose to use @MultipartConfig annotation. @MultipartConfig annotation is used to indicate that the Servlet on which it is declared expects requests to made using the multipart/form-data MIME type.
To use servlet 3.0 api’s you should deploy your application on Java EE 6 aware web/app server. I am using Apache Tomcat 7.x which comes with servlet 3 support.
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package org.techzoo.servlet3; import java.io.FileOutputStream; import java.io.IOException; import java.io.InputStream; import javax.servlet.ServletException; import javax.servlet.annotation.MultipartConfig; import javax.servlet.annotation.WebServlet; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; import javax.servlet.http.Part; @WebServlet ("/upload") @MultipartConfig public class FileUploadServlet extends HttpServlet { private static final long serialVersionUID = 1L; protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { uploadFile(request, response); } protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { uploadFile(request, response); } /** * uploadFile method used to upload file to server. * ***/ private void uploadFile(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { for(Part part : request.getParts()) { String name = part.getName(); InputStream is = request.getPart(name).getInputStream(); String fileName = getUploadedFileName(part); FileOutputStream fos = new FileOutputStream("c:/upload/"+fileName); int data = 0; while((data = is.read()) != -1) { fos.write(data); } fos.close(); is.close(); response.sendRedirect("success.jsp"); } } /** * Method used to get uploaded file name. * This will parse following string and get filename * Content-Disposition: form-data; name="content"; filename="a.txt" **/ private String getUploadedFileName(Part p) { String file = "", header = "Content-Disposition"; String[] strArray = p.getHeader(header).split(";"); for(String split : strArray) { if(split.trim().startsWith("filename")) { file = split.substring(split.indexOf('=') + 1); file = file.trim().replace("\"", ""); System.out.println("File name : "+file); } } return file; } } |
Now create a jsp page with a form tag. This form tag must have attribute enctype="multipart/form-data".
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<form action="upload" enctype="multipart/form-data" method="post"> <input type="file" name="uploadFile" /> <input type="submit" value="Upload File" /> </form> |
Output will look like as following screenshot.
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Tousif Khan has written 69 posts.
I Love Coding, Analyzing, Designing and doing it the right way. I am obsessed about learn anything new daily and sharing what I have learn each Day. This Blog is journey of my Zeal, going through step by step to the mountain of desire to become prefect. Google Wanna Hire me ?