[mod_python] how do I uplaod a file - code snippets sought

Hi I use the following snippet to upload my files. Additionally a
foldername is given, but this is only my usecase:

def uploadFile(req, myFile=None, folderName=None):
    """
    Uploads the file and then the file will be written in the given folder.
    """
    req.content_type = "text/html"
    folderPath = _util.getUploadFolderPath( \
                                        uploadRootPath=config.upload_root_path,
                                        folder=folderName)
    fileName = myFile.filename.replace('\\','/').split('/')[-1]
    req.log_error("httpUpload.uploadFile  file: %s/%s" % (folderName,fileName),
                            apache.APLOG_INFO)
    filePath = os.path.join(folderPath,  fileName)
    fileHandle = open(filePath, 'wb')
    size = 0
    while True:
        data = myFile.file.read(8192)
        if not data:
            break
        fileHandle.write(data)
        size += len(data)
        #~ req.log_error("%s - write chunk" % myFile.filename,
apache.APLOG_INFO)
    fileHandle.close()
    os.chmod(filePath, stat.S_IREAD | stat.S_IWRITE | stat.S_IRGRP |
stat.S_IWGRP)

    template = psp.PSP(req, filename='_httpupload/psp/uploadFinished.psp')
    template.run({
                'size':size,
                'fileName':fileName,
                'folderName':folderName,
                'type':myFile.type})


HTH Alex.

On 10/14/06, Dave Britton <dave at davebritton.com> wrote:
>
>
> What is the way to use mod_python.utils to get a file object create and
> uploaded from an html form? I can't quite get my head around it and I can't
> find any useful snippets of code to copy. I'd appreciate any help,
> especially an example. Thanks!
>
> Say I have an html form like this:
>  """<form action="upload" method="POST" enctype="multipart/form-data">
>  File name: <input name="file" type="file"><br>
>  <input name="submit" type="submit" value = "Send this file to the web
> server">
>  </form>"""
>
> What does upload.py have to do to process the file whose name is submitted?
>
> the only cookbook version I have found is for  regular python in cgi mode:
>
> ==================
>  form = cgi.FieldStorage()
>     if not form.has_key(form_field):
>         print HTML_TEMPLATE2% {'SCRIPT_NAME':os.environ['SCRIPT_NAME']}
>         return
>     fileitem = form[form_field]
>     if not fileitem.file:
>         return
>     fname=os.path.basename(fileitem.filename)
>     print "fileitem.filename=%s,<br> uploadddir=%s,<br> fout=%s" \
>        %(fileitem.filename, upload_dir, os.path.join(upload_dir, fname) )
>
>     fout = file (os.path.join(upload_dir, fname), 'wb')
>     while 1:
>         chunk = fileitem.file.read(100000)
>         if not chunk: break
>         fout.write (chunk)
>     fout.close()
> ===============
> How do I adapt this to use mod_python FieldStorage ? Especially for early
> mod_python as the installation I need this for has version 2.7 on Apache
> 1.3, debian linux
>
>
> Dave Britton
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>