again java

Hi,
i will ask

iam using servlets ,in one program my requirement is if there is a image(if u like) in anywhere and we browse that image and it is store in u r account like orkut.
my doubt is if the image is stored in database it is not the good idea. my requirement is the image is stored in special folder and the path is stored in database. when sometime one image is displayed in u r account how it is possible.

Some friend has sent this code:

please observe the code:

)page.jsp:
<%@ page language="java" %>
<HTML>
<HEAD><TITLE>Display file upload form to the user</TITLE></HEAD>
<BODY> <FORM ENCTYPE="multipart/form-data" ACTION="../UploadImage" METHOD=POST>
<br><br><br>
<center>
<table border="0" bgcolor=#ccFDDEE>
<tr><center><td colspan="2" align="center"><B>UPLOAD THE FILE</B><center></td>
</tr>
<tr><td colspan="2" align="center"> </td>
</tr>
<tr><td><b>Choose the file To Upload:</b></td><td><INPUT NAME="file" TYPE="file"></td>
</tr>
<tr><td colspan="2" align="center"> </td>
</tr>
<tr><td colspan="2" align="center"><input type="submit" value="Send File"> </td>
</tr>
<table>
</center>
</FORM>
</BODY>
</HTML>

2)UploadImage.java:
import java.io.*;
import java.sql.*;
import java.util.*;
import java.text.*;
import java.util.regex.*;
import org.apache.commons.fileupload.servlet.ServletFileUpload;
import org.apache.commons.fileupload.disk.DiskFileItemFactory;
import org.apache.commons.fileupload.*;

import javax.servlet.*;
import javax.servlet.http.*;

public class UploadImage extends HttpServlet{
public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
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;
saveFile="C:/UploadedFiles/"+saveFile;
File f = new File(saveFile);
FileOutputStream fileOut = new FileOutputStream(f);
fileOut.write(dataBytes, startPos, (endPos - startPos));
fileOut.flush();
fileOut.close();
out.println(saveFile);
Connection connection = null;
String connectionURL = "jdbc:mysql://localhost:3306/test";;
ResultSet rs = null;
PreparedStatement psmnt = null;
try {
Class.forName("com.mysql.jdbc.Driver").newInstance();
connection = DriverManager.getConnection(connectionURL, "root", "root");
psmnt = connection.prepareStatement("insert into file(file) values(?)");
psmnt.setString(1,f.getPath() );
int s = psmnt.executeUpdate();
if(s>0) {
System.out.println("Uploaded successfully !");
}
else{
System.out.println("unsucessfull to upload file.");
}
}
catch(Exception e){e.printStackTrace();}
}
}
}

my problem is

how to compile,run and where to see o/p please tell me step by step procedure.

pls help me.

View Answers

July 23, 2010 at 3:43 PM

Hi Friend,

Do the servlet-mapping in web.xml.Compile your servlet on command prompt.Restart your tomcat. Type the following url on the browser and click it:

http://localhost:8080/examples/page.jsp

[Here examples is out web-application folder.]

After selecting the image, when you will click the Send File button, servlet is called which will upload your selected image.

For more information, visit the following link:

http://www.roseindia.net/servlets/directory-structure.shtml
http://www.roseindia.net/servlets/introductiontoconfigrationservlet.shtml
http://www.roseindia.net/servlets/HowToRunAServlet.shtml

Thanks