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Posted to user@commons.apache.org by sim123 <si...@gmail.com> on 2008/10/30 01:10:38 UTC
File Upload and request parameters
Hello,
I am using FileUpload for multipart request handling, my request
contains three different parameter and one file, I want to read those
parameters first and then read the file as those paramters construct
directory where this file needs to be stored, here is the code
protected void doPost(HttpServletRequest request, HttpServletResponse
response) throws ServletException {
int dir1 = 0 ;
int dir2 = 0 ;
int dir3 = 0;
String uploadLocation = null;
String fileName = null;
// parse request
if(ServletFileUpload.isMultipartContent(request)){
ServletFileUpload upload = new ServletFileUpload();
try {
FileItemIterator iterator =
upload.getItemIterator(request);
while(iterator.hasNext()){
FileItemStream item =
iterator.next();
String name = item.getFieldName();
InputStream stream =
item.openStream();
if(item.isFormField()){
if("dir1".equals(name)){
dir1 =
Integer.parseInt(Streams.asString(stream));
}else
if("dir2".equals(name)){
dir2 =
Integer.parseInt(Streams.asString(stream));
}else
if("dir3".equals(name)){
dir3 =
Integer.parseInt(Streams.asString(stream));
}
uploadLocation =
server_url+File.separatorChar+tenantId+File.separatorChar+collectionId+File.separatorChar+rowId
;
new File(uploadLocation).mkdirs();
}else {
System.out.println("File field" + name + "with file name" +
item.getName());
fileName =
item.getName();
//perform file
storing operations
File uploadedFile = new
File(uploadLocation+File.separatorChar+fileName);
FileOutputStream outputstream = new
FileOutputStream(uploadedFile);
long number = Streams.copy(testStream,
outputstream, true);
System.out.println("result from outpoutstream
operation" + number);
}
}
} catch (FileUploadException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch(NumberFormatException e){
e.printStackTrace();
}
}
}
however for some reason FileItem is being read first so I can not
store this file at desired location it is (/usr/sim/0/0/0/filename),
can some one please suggest how can I make sure that I read form item
first then file items.
I do appreciate all the help and time. Thanks
-sim
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Re: File Upload and request parameters
Posted by Martin Cooper <ma...@apache.org>.
On Wed, Oct 29, 2008 at 5:10 PM, sim123 <si...@gmail.com> wrote:
>
> Hello,
>
> I am using FileUpload for multipart request handling, my request
> contains three different parameter and one file, I want to read those
> parameters first and then read the file as those paramters construct
> directory where this file needs to be stored
The order of the fields in the form you are submitting determines the order
in which those fields occur in the byte stream that FileUpload is reading.
If you need the file to be last, you'll need to reorder the fields in your
HTML form.
--
Martin Cooper
> , here is the code
>
> protected void doPost(HttpServletRequest request, HttpServletResponse
> response) throws ServletException {
> int dir1 = 0 ;
> int dir2 = 0 ;
> int dir3 = 0;
> String uploadLocation = null;
> String fileName = null;
> // parse request
> if(ServletFileUpload.isMultipartContent(request)){
> ServletFileUpload upload = new ServletFileUpload();
> try {
> FileItemIterator iterator =
> upload.getItemIterator(request);
> while(iterator.hasNext()){
> FileItemStream item =
> iterator.next();
> String name = item.getFieldName();
> InputStream stream =
> item.openStream();
> if(item.isFormField()){
> if("dir1".equals(name)){
> dir1 =
> Integer.parseInt(Streams.asString(stream));
> }else
> if("dir2".equals(name)){
> dir2 =
> Integer.parseInt(Streams.asString(stream));
> }else
> if("dir3".equals(name)){
> dir3 =
> Integer.parseInt(Streams.asString(stream));
> }
> uploadLocation =
>
> server_url+File.separatorChar+tenantId+File.separatorChar+collectionId+File.separatorChar+rowId
> ;
> new File(uploadLocation).mkdirs();
> }else {
>
> System.out.println("File field" + name + "with file name" +
> item.getName());
> fileName =
> item.getName();
> //perform file
> storing operations
>
> File uploadedFile = new
> File(uploadLocation+File.separatorChar+fileName);
>
> FileOutputStream outputstream = new
> FileOutputStream(uploadedFile);
> long number = Streams.copy(testStream,
> outputstream, true);
> System.out.println("result from outpoutstream
> operation" + number);
> }
>
> }
>
>
>
> } catch (FileUploadException e) {
> // TODO Auto-generated catch block
> e.printStackTrace();
> } catch (IOException e) {
> // TODO Auto-generated catch block
> e.printStackTrace();
> } catch(NumberFormatException e){
> e.printStackTrace();
> }
> }
>
> }
>
>
> however for some reason FileItem is being read first so I can not
> store this file at desired location it is (/usr/sim/0/0/0/filename),
> can some one please suggest how can I make sure that I read form item
> first then file items.
>
> I do appreciate all the help and time. Thanks
>
> -sim
> --
> View this message in context:
> http://www.nabble.com/File-Upload-and-request-parameters-tp20238511p20238511.html
> Sent from the Commons - User mailing list archive at Nabble.com.
>
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: user-unsubscribe@commons.apache.org
> For additional commands, e-mail: user-help@commons.apache.org
>
>
Re: File Upload and request parameters
Posted by sim123 <si...@gmail.com>.
Thank you very much for quick reply Martin, it worked, I appreciate your
help.
sim123 wrote:
>
> Hello,
>
> I am using FileUpload for multipart request handling, my request
> contains three different parameter and one file, I want to read those
> parameters first and then read the file as those paramters construct
> directory where this file needs to be stored, here is the code
>
> protected void doPost(HttpServletRequest request, HttpServletResponse
> response) throws ServletException {
> int dir1 = 0 ;
> int dir2 = 0 ;
> int dir3 = 0;
> String uploadLocation = null;
> String fileName = null;
> // parse request
> if(ServletFileUpload.isMultipartContent(request)){
> ServletFileUpload upload = new ServletFileUpload();
> try {
> FileItemIterator iterator =
> upload.getItemIterator(request);
> while(iterator.hasNext()){
> FileItemStream item =
> iterator.next();
> String name = item.getFieldName();
> InputStream stream =
> item.openStream();
> if(item.isFormField()){
> if("dir1".equals(name)){
> dir1 =
> Integer.parseInt(Streams.asString(stream));
> }else
> if("dir2".equals(name)){
> dir2 =
> Integer.parseInt(Streams.asString(stream));
> }else
> if("dir3".equals(name)){
> dir3 =
> Integer.parseInt(Streams.asString(stream));
> }
> uploadLocation =
> server_url+File.separatorChar+tenantId+File.separatorChar+collectionId+File.separatorChar+rowId
> ;
> new File(uploadLocation).mkdirs();
> }else {
>
> System.out.println("File field" + name + "with file name" +
> item.getName());
> fileName =
> item.getName();
> //perform file
> storing operations
>
> File uploadedFile = new
> File(uploadLocation+File.separatorChar+fileName);
>
> FileOutputStream outputstream = new
> FileOutputStream(uploadedFile);
> long number = Streams.copy(testStream,
> outputstream, true);
> System.out.println("result from
> outpoutstream operation" + number);
> }
>
> }
>
>
>
> } catch (FileUploadException e) {
> // TODO Auto-generated catch block
> e.printStackTrace();
> } catch (IOException e) {
> // TODO Auto-generated catch block
> e.printStackTrace();
> } catch(NumberFormatException e){
> e.printStackTrace();
> }
> }
>
> }
>
>
> however for some reason FileItem is being read first so I can not
> store this file at desired location it is (/usr/sim/0/0/0/filename),
> can some one please suggest how can I make sure that I read form item
> first then file items.
>
> I do appreciate all the help and time. Thanks
>
> -sim
>
--
View this message in context: http://www.nabble.com/File-Upload-and-request-parameters-tp20238511p20239931.html
Sent from the Commons - User mailing list archive at Nabble.com.
---------------------------------------------------------------------
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