CODEWITHSUNDEEP
java
Ravi Kumar
Ravi Kumar

Reputation: 3

How to upload a file into server using servlet?

I am very new to servlet technology. I want to upload a file from the local file system (i.e., client machine) to the server, which is running on Tomcat. Can someone please tell me how to do this.

I am using html input element type file (<input type="file"...>) and form action attribute posts the data to a servlet.

plz help me.

Upvotes: 0

Views: 1436

Answers (4)

Sagar Maiyad
Sagar Maiyad

Reputation: 12733

Pass the file from JSP or HTML using file Upload Component. and set servlet.java in form action.

Servlet.java

import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
import java.sql.*;

public class UploadServlet extends HttpServlet{ 
     public void doPost(HttpServletRequest request,HttpServletResponse response) throws ServletException,IOException {
     response.setContentType("text/html");
     PrintWriter out = response.getWriter();

    String saveFile="";
    String contentType = request.getContentType();
    if((contentType != null)&&(contentType.indexOf("multipart/form-data") >= 0)){
    DataInputStream in = new DataInputStream(request.getInputStream());
    int formDataLength = request.getContentLength();
    byte dataBytes[] = new byte[formDataLength];
    int byteRead = 0;
    int totalBytesRead = 0;
    while(totalBytesRead < formDataLength){
    byteRead = in.read(dataBytes, totalBytesRead,formDataLength);
    totalBytesRead += byteRead;
   }
    String file = new String(dataBytes);
    saveFile = file.substring(file.indexOf("filename=\"") + 10);
    saveFile = saveFile.substring(0, saveFile.indexOf("\n"));
    saveFile = saveFile.substring(saveFile.lastIndexOf("\\") + 1,saveFile.indexOf("\""));
    int lastIndex = contentType.lastIndexOf("=");
    String boundary = contentType.substring(lastIndex + 1,contentType.length());
    int pos;
    pos = file.indexOf("filename=\"");
    pos = file.indexOf("\n", pos) + 1;
    pos = file.indexOf("\n", pos) + 1;
    pos = file.indexOf("\n", pos) + 1;
    int boundaryLocation = file.indexOf(boundary, pos) - 4;
    int startPos = ((file.substring(0, pos)).getBytes()).length;
    int endPos = ((file.substring(0, boundaryLocation)).getBytes()).length;
    File ff = new File(saveFile);
    FileOutputStream fileOut = new FileOutputStream(ff);
    fileOut.write(dataBytes, startPos, (endPos - startPos));
    fileOut.flush();
    fileOut.close();
    out.println("You have successfully upload the file:"+saveFile);
    Connection connection = null;
    String connectionURL = "jdbc:mysql://localhost:3306/test";
    ResultSet rs = null;
    PreparedStatement psmnt = null;
    FileInputStream fis;
    try{
    Class.forName("com.mysql.jdbc.Driver").newInstance();
    connection = DriverManager.getConnection(connectionURL, "root", "root");
    File f = new File(saveFile);
    psmnt = connection.prepareStatement("insert into file(file_data) values(?)");
    fis = new FileInputStream(f);
    psmnt.setBinaryStream(1, (InputStream)fis, (int)(f.length()));
    int s = psmnt.executeUpdate();
    if(s>0){
    System.out.println("Uploaded successfully !");
    }
    else{
    System.out.println("Error!");
    }
    }
    catch(Exception e){
        e.printStackTrace();
        }
    }
  }
}

Upvotes: 1

ŁukaszBachman
ŁukaszBachman

Reputation: 33735

The best option would be to use some external library which can handle file upload. If you are using Spring, then check out this page: http://static.springsource.org/spring/docs/3.0.0.M3/spring-framework-reference/html/ch16s08.html.

Spring is internally using Apache Commons for file upload, so if you don't use Spring, or you would like just stick to Servlet API, I suggest using Commons as well: http://commons.apache.org/fileupload/

Upvotes: 1

tofarr
tofarr

Reputation: 7841

This is not included in the servlet api, but is available through http://commons.apache.org/fileupload/

Basically this uses the IO streaming included in the servlet API to process uploaded stream data - I think there was plans to include file upload functionality out of the box in the newest version of the API, but this has always worked for me.

Upvotes: 1

Matt Ball
Matt Ball

Reputation: 359776

Use the Apache Commons File Upload.

Apache.org seems to be having some issues right now, so here's the Google cached page.

Upvotes: 3

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