String xmlfile = "C:\\xmldocs\\" +request.getParameter("sample1.xml");

Hi All,
I am getting the following error while executing the following line in my jsp...
String xmlfile = "C:\\xmldocs\\" +request.getParameter("sample1.xml");
error: ..........
java.io.FileNotFoundException: C:\xmldocs\null (The system cannot find the file specified)...
What could be the problem ?
Please reply..
Thanks
Harish Pathak

"sample1.xml" is not recognised as a parameter (for obvious reasons). It should probably look something like:
String xmlfile = "C:\\xmldocs\\" +request.getParameter("filename");

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"Ashish Jain" <[email protected]> wrote in message
news:[email protected]..
Hi
I have one full working application (containing WAR + JAR) files.
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<SCRIPT Language="JavaScript">
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<form name="Login" Border="0" TopMargin="0" LeftMargin="0" Marginheight="0" Marginwidth="0">
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Thanks
suresh
null

How are you retrieving the parameter values? I am able to retrieve parameters successfully using the following simple testcase:
untitled1.jsp:
<form method="post" action="servlet1">
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<input type="text" name="un"/>
</P>
<P>Password:
<input type="text" name="pw"/></P>
<P>
<input type="submit" value="Submit"/>
</P>
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String un = request.getParameter("un").toString();
String pw = request.getParameter("pw").toString();
out.println(un);
out.println(pw);
Please test this and let me know your results and the differences with what you are trying to do.
Regards,
Lynn
Java Tools Team

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String strLine;
String[] tokens;
while ((strLine = br.readLine()) != null) {
tokens = strLine.split(",");
for (int i = 0; i < tokens.length; i++) {
out.println("<h1>Servlet TestServlet at " + tokens[i] + "</h1>");
out.println("----------------");
out.println("</body>");
out.println("</html>");
} finally {
out.close();
protected void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
processRequest(request, response);
Needed Output :
FILE NAME : TEST
FILE PATH : C:\\Temp\\test.txt
Any suggestions Plz??

As the [HTML specification|http://www.w3.org/TR/html401/interact/forms.html] states, you should be setting the enctype to multipart/form-data to be able to upload files to the server. In the server side, you need to process the multipart/form-data binary stream by parsing the HttpServletRequest#getInputStream(). It is a lot of work to get it to work flawlessly. There are 3rd party API's around which can do that for you, such as [Apache Commons FileUpload|http://commons.apache.org/fileupload] (carefully read the User Guide how to use it).
You can also consider to use a Filter which makes use of the FileUpload API to preprocess the request so that you can continue writing the servlet code as usual. Here is an example: [http://balusc.blogspot.com/2007/11/multipartfilter.html].

String xmlfile = "C:\\xmldocs\\" +request.getParameter("sample1.xml");

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