download link in jsp ?

[manojdesai sompadi]

i have to put a doanload link in a jsp page
<a href="/test/xyz.zip"></a>

but i have the xyz file in local drive c:\\download\xyz.zip
i am generating the file in local folder i can't put it in the server root.
how to map the download file from local drive to server path without copying the file into server path ?
thanks in advance

[Ben Souther]

You would need to write a servlet that sets the content-type, sets the content-disposition, and then streams the file to the browser.

[manojdesai sompadi]

thanks for the response
fallowing code modified from the example you have reffered when i am running this servlet its opening a download dialog box asking for downloading the servlet.java file instead of java.zip file ?

 


String uri = request.getRequestURI();
 
filename = "C:\\java.zip";
 
            

String contentType;
 contentType = "zip";
      
            

BufferedInputStream in = null;
try{
in = new BufferedInputStream(
getServletContext().
getResourceAsStream("WEB-INF" + filename));
 
response.setContentType(contentType);
response.setHeader("Content-Disposition",  
   " inline; filename=" + filename);
 
ServletOutputStream out = response.getOutputStream();
            
byte[] buffer = new byte[4 * 1024];
 
int data;
while((data = in.read(buffer)) != -1){
out.write(buffer, 0, data);
}
out.flush();
}catch(Exception e){
   return;
}
    

[Ben Souther]

Is your code even finding the file?

                                                                                                                                                  
      2         String uri = request.getRequestURI();
      3
      4                 filename = "C:\\java.zip";
      5
      6
      7
      8         String contentType;
      9          contentType = "zip";
     10
     11
     12
     13         BufferedInputStream in = null;
     14         try{
     15                 in = new BufferedInputStream(
     16         getServletContext().
     17                                 getResourceAsStream("WEB-INF" + filename));
     18
     19                 response.setContentType(contentType);
     20                 response.setHeader("Content-Disposition",
     21                                                    " inline; filename=" + filename);
     22
     23                 ServletOutputStream out = response.getOutputStream();
     24
     25                 byte[] buffer = new byte[4 * 1024];
     26
     27                 int data;
     28                 while((data = in.read(buffer)) != -1){
     29                         out.write(buffer, 0, data);
     30                 }
     31                 out.flush();
     32         }catch(Exception e){
     33    return;
     34         }
  

On line 4, you're saying: filename = "c:\\file.zip".
On line 17, you're saying: getResourceAsStream("WEB-INF" + filename));

This would mean that your looking for a file with a path and filename of:
"WEB-INFC:\\file.zip"

If the file is not inside your webapp, you would get the input stream with:
FileInputStream in = new FileInputStream(new File(filename));

"context.getResourceAsStream" is for files that are inside your webapp.

[kamesh aru]

thanks for the answer .
its working now but the save dialogbox i am getting the servlet name as default name to save ,instead of it how to get the test.zip file name as deafult name .

[Ben Souther]

The content-disposition header is where you set the desired name for the file.

[kamesh aru]

i will try it
thanks for your prompt responces.
i appreciate your help
thanks once again

[kamesh aru]

i have set the name in Content-Disposition
as fallows
resp.setHeader(filename,"inline;filename=\""+filename +"\"");
by this i could get the file extenison correctly but still name of the file it takes as name of servlet.
advice

[kamesh aru]

i commented the line
resp.setHeader(filename,"inline;filename=\""+filename +"\"");
i am getting same result .
why application is not picking up setHeader ?

[Ben Souther]

  
response.setHeader("Content-Disposition"," inline; filename=" + filename);

[kamesh aru]

i changed the content type to
contentType = "application/x-zip";
resp.setHeader("Content-Disposition",
" inline; filename=" + filename);
then its working

thanks for the responces till now they are very helpful to me