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Vijay
How display a Image on Servlet from file upload
1 Answer(s) 4 years and 5 months ago
Posted in : JSP-Servlet
1 Answer(s) 4 years and 5 months ago
Posted in : JSP-Servlet
View Answers
December 15, 2008 at 1:29 AM
Hi friend,
Code to help in solving the problem :
"upload.html"
<html>
<head><title>Upload page</title></head></p> <p><body>
<form action="tmp" method="post" enctype="multipart/form-data" name="form1" id="form1">
<center>
<table border="2">
<tr>
<td align="right"><b>Employee Image </td>
<td>
<input name="file" type="file" id="file">
<td>
</tr>
<tr>
<td align="center">
<input type="submit" name="Submit" value="Submit"/>
<input type="reset" name="Reset" value="Reset"/>
</td>
</tr>
</table>
</center>
</form>
</body>
</html>
"UploadFile.java"
import java.sql.*;
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
import java.util.*;
import org.apache.commons.fileupload.servlet.ServletFileUpload;
import org.apache.commons.fileupload.disk.DiskFileItemFactory;
import org.apache.commons.fileupload.*;
import org.apache.commons.fileupload.disk.*;
public class UploadFile extends HttpServlet{
public void doPost(HttpServletRequest request,HttpServletResponse response) throws ServletException, IOException{
response.setContentType("text/html");
PrintWriter out= response.getWriter();
Connection con=null;
String itemName="";
boolean fileUpload=false;
try{
boolean isMultipart = ServletFileUpload.isMultipartContent(request);
FileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload upload = new ServletFileUpload(factory);
List items = null;
try {
items = upload.parseRequest(request);
} catch (FileUploadException e) {
e.printStackTrace();
}
Iterator itr = items.iterator();
while (itr.hasNext())
{
FileItem item = (FileItem) itr.next();
String name = item.getFieldName();
if(name.equals("file"))
{
itemName = item.getName();
File savedFile = new File(request.getRealPath("/")+"image\\"+itemName);
item.write(savedFile);
fileUpload=true;
}
}
if(fileUpload==true)
{
out.println("File has been uploaded successfully");
out.println("<html>");
out.println("<body>");
out.println("<table><tr><td width='210'></td><td> <img border=2 src=image/"+itemName.toString().trim()+" width=137 height=140></td></tr></table>");
out.println("</body>");
out.println("</html>");
}
}
catch (Exception e){
e.printStackTrace();
}
}
}
web.xml:
<web-app>
<servlet>
<servlet-name>one</servlet-name>
<servlet-class>UploadFile</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>one</servlet-name>
<url-pattern>/tmp</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>Home.html</welcome-file>
</welcome-file-list>
</web-app>
For more information on Servlet visit to :
http://www.roseindia.net/servlets/
Thanks
Code to help in solving the problem :
"upload.html"
<html>
<head><title>Upload page</title></head></p> <p><body>
<form action="tmp" method="post" enctype="multipart/form-data" name="form1" id="form1">
<center>
<table border="2">
<tr>
<td align="right"><b>Employee Image </td>
<td>
<input name="file" type="file" id="file">
<td>
</tr>
<tr>
<td align="center">
<input type="submit" name="Submit" value="Submit"/>
<input type="reset" name="Reset" value="Reset"/>
</td>
</tr>
</table>
</center>
</form>
</body>
</html>
"UploadFile.java"
import java.sql.*;
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
import java.util.*;
import org.apache.commons.fileupload.servlet.ServletFileUpload;
import org.apache.commons.fileupload.disk.DiskFileItemFactory;
import org.apache.commons.fileupload.*;
import org.apache.commons.fileupload.disk.*;
public class UploadFile extends HttpServlet{
public void doPost(HttpServletRequest request,HttpServletResponse response) throws ServletException, IOException{
response.setContentType("text/html");
PrintWriter out= response.getWriter();
Connection con=null;
String itemName="";
boolean fileUpload=false;
try{
boolean isMultipart = ServletFileUpload.isMultipartContent(request);
FileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload upload = new ServletFileUpload(factory);
List items = null;
try {
items = upload.parseRequest(request);
} catch (FileUploadException e) {
e.printStackTrace();
}
Iterator itr = items.iterator();
while (itr.hasNext())
{
FileItem item = (FileItem) itr.next();
String name = item.getFieldName();
if(name.equals("file"))
{
itemName = item.getName();
File savedFile = new File(request.getRealPath("/")+"image\\"+itemName);
item.write(savedFile);
fileUpload=true;
}
}
if(fileUpload==true)
{
out.println("File has been uploaded successfully");
out.println("<html>");
out.println("<body>");
out.println("<table><tr><td width='210'></td><td> <img border=2 src=image/"+itemName.toString().trim()+" width=137 height=140></td></tr></table>");
out.println("</body>");
out.println("</html>");
}
}
catch (Exception e){
e.printStackTrace();
}
}
}
web.xml:
<web-app>
<servlet>
<servlet-name>one</servlet-name>
<servlet-class>UploadFile</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>one</servlet-name>
<url-pattern>/tmp</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>Home.html</welcome-file>
</welcome-file-list>
</web-app>
For more information on Servlet visit to :
http://www.roseindia.net/servlets/
Thanks
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