From browse button of input type=file, can I show the content of a file

Hi all,
I am using, input type=file, where browse button appears. I am having a text area. My requirement is after I select the browse button, I need to show the content of the file in the text area. As of now I 've handled by having another button 'show', which should be pressed after browse done. But, How to handle it in the browse button itself, how will I get that action event from browse button. Plz help.
Regards,
Sam

I think someone asked a very similar question here,
http://forum.java.sun.com/thread.jsp?forum=45&thread=501889
check the thread, it may help you.
-S-

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FILE UPLOAD PROBLEM SHOWING THE CONTENTS IN THE SAME BROWSER WINDOW

Hi,
This is amit Joshi
I have uploaded content using input tag of type file and posted to jsp as multipart/form-data type
in that jsp i am using following code to display the content in browser but only first content is displayed How can i modify it to show all content in the file ..
<html>
<head>
<title>File Upload Display</title>
</head>
<body>
<%
//ServletOutputStream sout=response.getOutputStream();
StringBuilder strBuilder = new StringBuilder();
int count=0;
String f;
f=request.getParameter("filedb");
DBManager dbm = new DBManager();
//dbm.createTable("mms3");
//log.info("In JSP : "+ f);
//dbm.insert_data(f,"mms3");
%>
<%
if (ServletFileUpload.isMultipartContent(request)){
ServletFileUpload servletFileUpload = new ServletFileUpload(new DiskFileItemFactory());
List fileItemsList = servletFileUpload.parseRequest(request);
strBuilder.append("<?xml version=\"1.0\" encoding=\"UTF-8\" ?>").append('\r').append('\n').append("<xpage version=\"1.0\">").append('\r').append('\n');
String optionalFileName = "";
FileItem fileItem = null;
Iterator it = fileItemsList.iterator();
ServletOutputStream outputStream=null;
while (it.hasNext()){
FileItem fileItemTemp = (FileItem)it.next();
if (fileItemTemp.isFormField()){
%>
<b>Name-value Pair Info:</b>
Field name: <%= fileItemTemp.getFieldName() %>
Field value: <%= fileItemTemp.getString() %>
<%
if (fileItemTemp.getFieldName().equals("filename"))
optionalFileName = fileItemTemp.getString();
else
fileItem = fileItemTemp;
if (fileItem!=null){
String fileName = fileItem.getName();
%>
<b>Uploaded File Info:</b>
Content type: <%= fileItem.getContentType() %>
Field name: <%= fileItem.getFieldName() %>
File name: <%= fileName %>
<%
if(fileItem.getContentType().equals("image/jpeg")) { %>
File : <p><%
     //response.setContentType("image/gif");
     byte[] bArray=fileItem.get();
     response.setContentType("image/jpeg");
     outputStream=null;
     outputStream= response.getOutputStream();
     outputStream.write(bArray);
     outputStream.flush();
     outputStream.close();
else if(fileItem.getContentType().equals("text/plain"))
     %> File : <%= fileItem.getString() %>
<%
byte[] bArray=fileItem.get();
response.setContentType("text/plain");
     outputStream = response.getOutputStream();
     out.println();
     outputStream.write(bArray);
     outputStream.flush();
     outputStream.close();
%> </p> <%
%>
</body>
</html>
Edited by: Amit_Joshi on Nov 13, 2007 10:58 PM

Well Well Well..
That would not work...
What you have to do is save the uploaded file content on to a location and then pass the fileName as a request parameter to a deidicated which displays the contents of that file.
Just as an example
<html>
<head>
<title>File Upload Display</title>
</head>
<body>
<%
//ServletOutputStream sout=response.getOutputStream();
StringBuilder strBuilder = new StringBuilder();
int count=0;
String f;
f=request.getParameter("filedb");
DBManager dbm = new DBManager();
//dbm.createTable("mms3");
//log.info("In JSP : "+ f);
//dbm.insert_data(f,"mms3");
%>
<%
if (ServletFileUpload.isMultipartContent(request)){
ServletFileUpload servletFileUpload = new ServletFileUpload(new DiskFileItemFactory());
List fileItemsList = servletFileUpload.parseRequest(request);
strBuilder.append("<?xml version=\"1.0\" encoding=\"UTF-8\" ?>").append('\r').append('\n').append("<xpage version=\"1.0\">").append('\r').append('\n');
String optionalFileName = "";
FileItem fileItem = null;
Iterator it = fileItemsList.iterator();
ServletOutputStream outputStream=null;
while (it.hasNext()){
   FileItem fileItemTemp = (FileItem)it.next();
%>
Name-value Pair Info:
Field name: <%= fileItemTemp.getFieldName() %><br/>
Field value: <%= fileItemTemp.getString() %><br/>
<%
if (fileItemTemp.getFieldName().equals("filename"))
    optionalFileName = fileItemTemp.getString();
if(!fileTempItem.isFormFiled()){
   String fileName = fileItem.getName();
   fileItem.write(optionalFileName);
%>
Uploaded File Info:
Content type: <%= fileItem.getContentType() %><br/>
Field name: <%= fileItem.getFieldName() %><br/>
File name: <%= fileName %><br/>
<%
if(fileItem.getContentType().equals("image/jpeg") || fileItem.getContentType().equals("image/pjeg")) {
%>
<img src="FileServlet?fileName=<%=optionalFileName%>"
<%
%>
</body>
</html>a sample code snippet for FileServlet.
String fileName = request.getParameter(fileName);
File file = new File(fileName);
   if(!file.exists())
     return;
// If JSP
String mimeType = application.getMimeType("fileName");
       If you are using servlet
       String mimeType = this.getServletContext().getMimeType(fileName);
     response.setContentType(mimeType);
     response.setHeader("Content-Disposition","inline;filename=\\"+fileName+"\\");
BufferedOutputStream out1 = null;
InputStream in = null;
if(mimeType == null)
     mimeType = "application/octet-stream";
try{
     in = new FileInputStream(f);
     response.setContentLength(in.available());
     BufferedOutputStream out1 = new BufferedOutputStream(response.getOutputStream(),1024);
     int size = 0;
     byte[] b = new byte[1024];
     while ((size = in.read(b, 0, 1024)) > 0)
        out1.write(b, 0, size);
}catch(Exception exp){
}finally{
      if(out1 != null){
         try{
            out1.flush();
            out1.close();
         }catch(Exception e){}
      if(in != null){
        try{in.close();}catch(Exception e){}
} Hope that might answer your question :)
However,this is not the recommended way of doing this make use of MVC pattern.Would be a better approach.
you might think of googling on this and can findout what is the best practise followed for problems of this sort
REGARDS,
RaHuL

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import java.io.*;
public class FileReader
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throws IOException
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new FileInputStream("text.txt");
FilterInputStream myfile=
new FilterInputStream(theFile);
int c;
while ((c = myfile.read()) != -1)
{System.out.print(c);}
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new FilterInputStream(theFile);
^
1 error
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import java.io.*;
public class FileReader
public static void main(String[] args)
throws IOException
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symbol : method readLine ()
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From browse button of input type=file, can I show the content of a file

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