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Posted to user@commons.apache.org by Adam Pelletier <ad...@moebius-solutions.com> on 2004/10/07 19:59:10 UTC
FileUpload Parameter Handling
I'm using commons-fileupload-1.0.jar.
I'm trying to upload a file using enctype="multipart/form-data".
However, I'm also passing a hidden form input element with an ID stuck in it. This ID is critical for the state of the servlet. Oddly, when I use the enctype="multipart/form-data", my HttpServletRequest no longer seems to contain the parameters usually accessible by req.getParameter("ID");. In the debugger I see that the parameter hashtable is empty. As a test I remove enctype="multipart/form-data" from the form. The parameters then show up properly in the parameter map, but now the file is not accessible.
Is there a way to utilize both the file uploading capability and still have access to the parameter list as it comes in as a HttpServletRequest?
Thanks in advance.
Adam
Re: FileUpload Parameter Handling
Posted by Ben Souther <bs...@fwdco.com>.
Regular form params show up in in the DiskFileUpload.
When you iterate through the values, you can test to see
if they are form fields with "isFormField"
Example:......................................
DiskFileUpload upload = new DiskFileUpload();
List files = upload.parseRequest(request);
Iterator it = files.iterator();
while(it.hasNext()){
FileItem item = (FileItem)it.next();
if(item.isFormField()){....
On Thu, 2004-10-07 at 13:59, Adam Pelletier wrote:
> I'm using commons-fileupload-1.0.jar.
>
> I'm trying to upload a file using enctype="multipart/form-data".
>
> However, I'm also passing a hidden form input element with an ID stuck in it. This ID is critical for the state of the servlet. Oddly, when I use the enctype="multipart/form-data", my HttpServletRequest no longer seems to contain the parameters usually accessible by req.getParameter("ID");. In the debugger I see that the parameter hashtable is empty. As a test I remove enctype="multipart/form-data" from the form. The parameters then show up properly in the parameter map, but now the file is not accessible.
>
> Is there a way to utilize both the file uploading capability and still have access to the parameter list as it comes in as a HttpServletRequest?
>
> Thanks in advance.
> Adam
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Re: FileUpload Parameter Handling
Posted by Michael McGrady <mi...@michaelmcgrady.com>.
What are you doing with commons upload? You should be able to get
whatever values are present as values of parameter keys in the form.
There is nothing about a multipart upload that precludes regular
parameters. For example, I parse the DiskFileUpload as follows:
DiskFileUpload dfu = new DiskFileUpload();
dfu.setSizeMax(fileSizeLimit);
dfu.setSizeThreshold(UploadConstant.MEMORY_BUFFER_SIZE);
dfu.setRepositoryPath(tmpFolder);
if(encoding != null) {
dfu.setHeaderEncoding(encoding);
}
List list = null;
try {
list = dfu.parseRequest(req);
StdOut.log("log.development","UploadParser list = " + list);
} catch(FileUploadException fue) {
throw new IOException(fue.getMessage());
}
And my log test (StdOut) that you see here results in
UploadParser list =
[org.apache.commons.fileupload.DefaultFileItem@1d56bbe,
org.apache.commons.fileupload.DefaultFileItem@94a28e,
org.apache.commons.fileupload.DefaultFileItem@3c40f0,
org.apache.commons.fileupload.DefaultFileItem@1cb4cae,
org.apache.commons.fileupload.DefaultFileItem@176ee6,
org.apache.commons.fileupload.DefaultFileItem@71d29a,
org.apache.commons.fileupload.DefaultFileItem@b98a06,
org.apache.commons.fileupload.DefaultFileItem@114a3c6,
org.apache.commons.fileupload.DefaultFileItem@c4cee,
org.apache.commons.fileupload.DefaultFileItem@18ec029,
org.apache.commons.fileupload.DefaultFileItem@e9c592,
org.apache.commons.fileupload.DefaultFileItem@1a9fcea,
org.apache.commons.fileupload.DefaultFileItem@11b99c4,
org.apache.commons.fileupload.DefaultFileItem@10c29fe,
org.apache.commons.fileupload.DefaultFileItem@1991ba7]
UploadParser fieldName = uploadedFile1
UploadParser fieldName = uploadedFile2
UploadParser fieldName = uploadedFile3
UploadParser fieldName = uploadedFile4
UploadParser fieldName = uploadedFile5
UploadParser fieldName = uploadedFile6
UploadParser fieldName = uploadedFile7
UploadParser fieldName = uploadedFile8
UploadParser fieldName = uploadedFile9
UploadParser fieldName = uploadedFile10
UploadParser fieldName = uploadedFile11
UploadParser fieldName = uploadedFile12
UploadParser fieldName = submit.dispatch.x
UploadParser fieldName = submit.dispatch.y
UploadParser fieldName = fileOp
From a page that has twelve <input type='file'> tags. Hope this helps.
Michael McGrady
Adam Pelletier wrote:
> No, this is just a plain servlet. Another guy wrote that regular form
> params show up in in the DiskFileUpload.
>
> When you iterate through the values, you can test to see
> if they are form fields with "isFormField"
>
>
> Example:......................................
> DiskFileUpload upload = new DiskFileUpload();
> List files = upload.parseRequest(request);
> Iterator it = files.iterator();
>
> while(it.hasNext()){
> FileItem item = (FileItem)it.next();
> if(item.isFormField()){....
>
>
> But that still breaks my architecture - although I can get to that
> architecture if I have to.
>
> Adam
>
> ----- Original Message ----- From: "Michael McGrady"
> <mi...@michaelmcgrady.com>
> To: "Jakarta Commons Users List" <co...@jakarta.apache.org>
> Sent: Thursday, October 07, 2004 11:45 AM
> Subject: Re: FileUpload Parameter Handling
>
>
>> Are you using Struts, Adam?
>>
>> Michael McGrady
>>
>> Adam Pelletier wrote:
>>
>>> I'm using commons-fileupload-1.0.jar.
>>>
>>> I'm trying to upload a file using enctype="multipart/form-data".
>>>
>>> However, I'm also passing a hidden form input element with an ID
>>> stuck in it. This ID is critical for the state of the servlet.
>>> Oddly, when I use the enctype="multipart/form-data", my
>>> HttpServletRequest no longer seems to contain the parameters usually
>>> accessible by req.getParameter("ID");. In the debugger I see that
>>> the parameter hashtable is empty. As a test I remove
>>> enctype="multipart/form-data" from the form. The parameters then
>>> show up properly in the parameter map, but now the file is not
>>> accessible.
>>> Is there a way to utilize both the file uploading capability and
>>> still have access to the parameter list as it comes in as a
>>> HttpServletRequest?
>>>
>>> Thanks in advance.
>>> Adam
>>>
>>
>>
>>
>> ---------------------------------------------------------------------
>> To unsubscribe, e-mail: commons-user-unsubscribe@jakarta.apache.org
>> For additional commands, e-mail: commons-user-help@jakarta.apache.org
>>
>>
>
>
>
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Re: FileUpload Parameter Handling
Posted by Michael McGrady <mi...@michaelmcgrady.com>.
My pleasure, Adam. I don't know who wrote the commons upload stuff, but
I think they did one hell of a job. You have to study it a bit,
however, because it is not entirely intuitive at first. I will tell you
this, however, it really provides a wonderful base for working with
multipart issues. I have a hugely efficient file upload application
built on top of this functionality. Anyway, glad you are on track.
These things can bedevil us.
Michael McGrady
Adam Pelletier wrote:
> I fixed my problem. My problem is that I had a Servlet that would
> serve up any number of Pages. The way the navigation was wired was
> through keys/values pairs in the parameter list (i.e.
> req.getParameter("ID)). So the HttpServletRequest itself would get
> passed down to a couple of different delgation layers. Well in
> DiskFileUpload the parameters are stored differently in the Parts list
> and then have to be iterated over. So all my code depended upon:
>
> // block 1
> req.getParameter("foo");
>
> as opposed to
>
> // block 2
> Iterator iter = m_req_parts.iterator();
> while (iter.hasNext()) {
> FileItem fi = (FileItem) iter.next();
> if (fi.isFormField() && fi.getFieldName().equals("foo")) {
> return fi.getString();
> }
> }
>
> So what I did is write an HttpServletRequestWrapper so when my code
> calls black 1 above, what really happens is block 2 above executes.
>
> Thanks all for the help,
> Adam
>
> ----- Original Message ----- From: "Michael McGrady"
> <mi...@michaelmcgrady.com>
> To: "Jakarta Commons Users List" <co...@jakarta.apache.org>
> Sent: Thursday, October 07, 2004 3:29 PM
> Subject: Re: FileUpload Parameter Handling
>
>
>> I cannot see how a servlet stepping through parameter values could
>> break an architecture. What do you mean by that? If the parameter
>> is there, then the servlet can read it. How that could affect
>> architecture is not clear to me. Can you guys explain what you mean?
>>
>> Michael McGrady
>>
>> Paul DeCoursey wrote:
>>
>>> I know what you are saying about it breaking your architecture... i
>>> use a
>>> servlet that calls scripts based on a parameter, I found that using the
>>> multipart-encoding broke that. I later discovered that if I post the
>>> multipart but have my parameter on the querystring it works fine.
>>>
>>> Paul
>>>
>>>
>>>
>>>> No, this is just a plain servlet. Another guy wrote that regular form
>>>> params show up in in the DiskFileUpload.
>>>>
>>>> When you iterate through the values, you can test to see
>>>> if they are form fields with "isFormField"
>>>>
>>>>
>>>> Example:......................................
>>>> DiskFileUpload upload = new DiskFileUpload();
>>>> List files = upload.parseRequest(request);
>>>> Iterator it = files.iterator();
>>>>
>>>> while(it.hasNext()){
>>>> FileItem item = (FileItem)it.next();
>>>> if(item.isFormField()){....
>>>>
>>>>
>>>> But that still breaks my architecture - although I can get to that
>>>> architecture if I have to.
>>>>
>>>> Adam
>>>>
>>>>
>>>>
>>>
>>>
>>> ---------------------------------------------------------------------
>>> To unsubscribe, e-mail: commons-user-unsubscribe@jakarta.apache.org
>>> For additional commands, e-mail: commons-user-help@jakarta.apache.org
>>>
>>>
>>>
>>>
>>>
>>
>>
>>
>> ---------------------------------------------------------------------
>> To unsubscribe, e-mail: commons-user-unsubscribe@jakarta.apache.org
>> For additional commands, e-mail: commons-user-help@jakarta.apache.org
>>
>>
>
>
>
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>
>
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Re: FileUpload Parameter Handling
Posted by Adam Pelletier <ad...@moebius-solutions.com>.
I fixed my problem. My problem is that I had a Servlet that would serve up
any number of Pages. The way the navigation was wired was through
keys/values pairs in the parameter list (i.e. req.getParameter("ID)). So
the HttpServletRequest itself would get passed down to a couple of different
delgation layers. Well in DiskFileUpload the parameters are stored
differently in the Parts list and then have to be iterated over. So all my
code depended upon:
// block 1
req.getParameter("foo");
as opposed to
// block 2
Iterator iter = m_req_parts.iterator();
while (iter.hasNext()) {
FileItem fi = (FileItem) iter.next();
if (fi.isFormField() && fi.getFieldName().equals("foo")) {
return fi.getString();
}
}
So what I did is write an HttpServletRequestWrapper so when my code calls
black 1 above, what really happens is block 2 above executes.
Thanks all for the help,
Adam
----- Original Message -----
From: "Michael McGrady" <mi...@michaelmcgrady.com>
To: "Jakarta Commons Users List" <co...@jakarta.apache.org>
Sent: Thursday, October 07, 2004 3:29 PM
Subject: Re: FileUpload Parameter Handling
>I cannot see how a servlet stepping through parameter values could break an
>architecture. What do you mean by that? If the parameter is there, then
>the servlet can read it. How that could affect architecture is not clear
>to me. Can you guys explain what you mean?
>
> Michael McGrady
>
> Paul DeCoursey wrote:
>
>>I know what you are saying about it breaking your architecture... i use a
>>servlet that calls scripts based on a parameter, I found that using the
>>multipart-encoding broke that. I later discovered that if I post the
>>multipart but have my parameter on the querystring it works fine.
>>
>>Paul
>>
>>
>>
>>>No, this is just a plain servlet. Another guy wrote that regular form
>>>params show up in in the DiskFileUpload.
>>>
>>>When you iterate through the values, you can test to see
>>>if they are form fields with "isFormField"
>>>
>>>
>>>Example:......................................
>>>DiskFileUpload upload = new DiskFileUpload();
>>>List files = upload.parseRequest(request);
>>>Iterator it = files.iterator();
>>>
>>>while(it.hasNext()){
>>> FileItem item = (FileItem)it.next();
>>> if(item.isFormField()){....
>>>
>>>
>>>But that still breaks my architecture - although I can get to that
>>>architecture if I have to.
>>>
>>>Adam
>>>
>>>
>>>
>>
>>
>>---------------------------------------------------------------------
>>To unsubscribe, e-mail: commons-user-unsubscribe@jakarta.apache.org
>>For additional commands, e-mail: commons-user-help@jakarta.apache.org
>>
>>
>>
>>
>>
>
>
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: commons-user-unsubscribe@jakarta.apache.org
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>
>
---------------------------------------------------------------------
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Re: FileUpload Parameter Handling
Posted by Michael McGrady <mi...@michaelmcgrady.com>.
I cannot see how a servlet stepping through parameter values could break
an architecture. What do you mean by that? If the parameter is there,
then the servlet can read it. How that could affect architecture is not
clear to me. Can you guys explain what you mean?
Michael McGrady
Paul DeCoursey wrote:
>I know what you are saying about it breaking your architecture... i use a
>servlet that calls scripts based on a parameter, I found that using the
>multipart-encoding broke that. I later discovered that if I post the
>multipart but have my parameter on the querystring it works fine.
>
>Paul
>
>
>
>
>>No, this is just a plain servlet. Another guy wrote that regular form
>>params show up in in the DiskFileUpload.
>>
>>When you iterate through the values, you can test to see
>>if they are form fields with "isFormField"
>>
>>
>>Example:......................................
>>DiskFileUpload upload = new DiskFileUpload();
>>List files = upload.parseRequest(request);
>>Iterator it = files.iterator();
>>
>>while(it.hasNext()){
>> FileItem item = (FileItem)it.next();
>> if(item.isFormField()){....
>>
>>
>>But that still breaks my architecture - although I can get to that
>>architecture if I have to.
>>
>>Adam
>>
>>
>>
>>
>
>
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>
>
>
>
>
>
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Re: FileUpload Parameter Handling
Posted by Paul DeCoursey <pa...@decoursey.net>.
I know what you are saying about it breaking your architecture... i use a
servlet that calls scripts based on a parameter, I found that using the
multipart-encoding broke that. I later discovered that if I post the
multipart but have my parameter on the querystring it works fine.
Paul
> No, this is just a plain servlet. Another guy wrote that regular form
> params show up in in the DiskFileUpload.
>
> When you iterate through the values, you can test to see
> if they are form fields with "isFormField"
>
>
> Example:......................................
> DiskFileUpload upload = new DiskFileUpload();
> List files = upload.parseRequest(request);
> Iterator it = files.iterator();
>
> while(it.hasNext()){
> FileItem item = (FileItem)it.next();
> if(item.isFormField()){....
>
>
> But that still breaks my architecture - although I can get to that
> architecture if I have to.
>
> Adam
>
>
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Re: FileUpload Parameter Handling
Posted by Adam Pelletier <ad...@moebius-solutions.com>.
No, this is just a plain servlet. Another guy wrote that regular form
params show up in in the DiskFileUpload.
When you iterate through the values, you can test to see
if they are form fields with "isFormField"
Example:......................................
DiskFileUpload upload = new DiskFileUpload();
List files = upload.parseRequest(request);
Iterator it = files.iterator();
while(it.hasNext()){
FileItem item = (FileItem)it.next();
if(item.isFormField()){....
But that still breaks my architecture - although I can get to that
architecture if I have to.
Adam
----- Original Message -----
From: "Michael McGrady" <mi...@michaelmcgrady.com>
To: "Jakarta Commons Users List" <co...@jakarta.apache.org>
Sent: Thursday, October 07, 2004 11:45 AM
Subject: Re: FileUpload Parameter Handling
> Are you using Struts, Adam?
>
> Michael McGrady
>
> Adam Pelletier wrote:
>
>>I'm using commons-fileupload-1.0.jar.
>>
>>I'm trying to upload a file using enctype="multipart/form-data".
>>
>>However, I'm also passing a hidden form input element with an ID stuck in
>>it. This ID is critical for the state of the servlet. Oddly, when I use
>>the enctype="multipart/form-data", my HttpServletRequest no longer seems
>>to contain the parameters usually accessible by req.getParameter("ID");.
>>In the debugger I see that the parameter hashtable is empty. As a test I
>>remove enctype="multipart/form-data" from the form. The parameters then
>>show up properly in the parameter map, but now the file is not accessible.
>>Is there a way to utilize both the file uploading capability and still
>>have access to the parameter list as it comes in as a HttpServletRequest?
>>
>>Thanks in advance.
>>Adam
>>
>
>
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: commons-user-unsubscribe@jakarta.apache.org
> For additional commands, e-mail: commons-user-help@jakarta.apache.org
>
>
---------------------------------------------------------------------
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Re: FileUpload Parameter Handling
Posted by Michael McGrady <mi...@michaelmcgrady.com>.
Are you using Struts, Adam?
Michael McGrady
Adam Pelletier wrote:
>I'm using commons-fileupload-1.0.jar.
>
>I'm trying to upload a file using enctype="multipart/form-data".
>
>However, I'm also passing a hidden form input element with an ID stuck in it. This ID is critical for the state of the servlet. Oddly, when I use the enctype="multipart/form-data", my HttpServletRequest no longer seems to contain the parameters usually accessible by req.getParameter("ID");. In the debugger I see that the parameter hashtable is empty. As a test I remove enctype="multipart/form-data" from the form. The parameters then show up properly in the parameter map, but now the file is not accessible.
>
>Is there a way to utilize both the file uploading capability and still have access to the parameter list as it comes in as a HttpServletRequest?
>
>Thanks in advance.
>Adam
>
>
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