Can you email me a minimalist app and a sample file to upload to try
reproduce the problem?
On May 13, 2:57 pm, Hans <johann.scheibelho...@easytouch-edv.com>
wrote:
> web2py 1.61.4
> WinXP Pro SP3
> Python 2.5.1 (r251:54863, Apr 18 2007, 08:51:08) [MSC v.1310 32 bit
> (Intel)]
>
> If there is anything I can do to mail it down on my side then let me
> know!
>
> Thanks,
> Hans
>
> On May 13, 9:40 pm, mdipierro <mdipie...@cs.depaul.edu> wrote:
>
> > The problem is not so much to solve it but to reproduce it.
>
> > what version of web2py are you running? which os? which python
> > version?
>
> > Massimo
>
> > On May 13, 2:32 pm, Hans <johann.scheibelho...@easytouch-edv.com>
> > wrote:
>
> > > bumping this one since I still could not fix that. Has anyone managed
> > > to solve this yet?
>
> > > On May 9, 1:09 am, mdipierro <mdipie...@cs.depaul.edu> wrote:
>
> > > > My bad. This was correct:
> > > > shutil.copyfileobj(request.vars.filename.file,open(path,'w'))
>
> > > > I am surprised you still get a corrupted file. let me run some tests.
>
> > > > Massimo
>
> > > > On May 8, 5:05 pm, Hans <johann.scheibelho...@easytouch-edv.com>
> > > > wrote:
>
> > > > > > shutil.copyfileobj(request.vars.filename,open(path,'w'))
>
> > > > > raises exception: (<type 'exceptions.AttributeError'>, AttributeError
> > > > > ('read',), <traceback object at 0x03FC0CB0>)
>
> > > > > tried also
> > > > > shutil.copyfileobj(request.vars.filename.file,open(path,'w'))
> > > > > which creates a copy, but again a corrupt copy.
>
> > > > > This tests now were done just on windows xp.
>
> > > > > Any other ideas how to solve this?
>
> > > > > Hans
>
> > > > > On May 8, 10:29 pm, mdipierro <mdipie...@cs.depaul.edu> wrote:
>
> > > > > > It may be an issue with buffer size. Try
>
> > > > > > import shutil
> > > > > > shutil.copyfileobj(request.vars.filename,open(path,'w'))
>
> > > > > > Massimo
>
> > > > > > On May 8, 2:56 pm, Hans <johann.scheibelho...@easytouch-edv.com>
> > > > > > wrote:
>
> > > > > > > There is a problem with the following code line
>
> > > > > > > >
> > > > > > > > open(path,'w').write(request.vars.filename.file.read())
>
> > > > > > > It creates the file 'a_dummy_file_name.xls' but it is not the same
> > > > > > > size as the original file, for example:
> > > > > > > 6790KB original.xls
> > > > > > > 6862KB a_dummy_file_name.xls
>
> > > > > > > If I try to open the dummy file with excel or open office it says
> > > > > > > the
> > > > > > > file is corrupt.
> > > > > > > I've tried to copy the file manually right after this code line
> > > > > > > and
> > > > > > > continued execution from there onward...rest works nicely.
>
> > > > > > > I tried on windows and on ubuntu, seems the same problem.
>
> > > > > > > Any idea how to solve that issue with the corrupt file (=getting a
> > > > > > > exact copy) ?
>
> > > > > > > thanks,
> > > > > > > hans
>
> > > > > > > On May 7, 10:56 pm, mdipierro <mdipie...@cs.depaul.edu> wrote:
>
> > > > > > > > This should do
>
> > > > > > > > def excel_read():
> > > > > > > > result={}
> > > > > > > > if request.vars.filename != None:
> > > > > > > > try:
> > > > > > > > import xlrd
> > > > > > > > path=os.path.join
> > > > > > > > (request.folder,'private','a_dummy_file_name.xls')
> > > > > > > >
> > > > > > > > open(path,'w').write(request.vars.filename.file.read())
> > > > > > > > book = xlrd.open_workbook(path)
> > > > > > > > result['The number of worksheets is']=book.nsheets
> > > > > > > > result["Worksheet name(s)"]= book.sheet_names()
> > > > > > > > sh = book.sheet_by_index(0)
> > > > > > > > result['sh.name']=sh.name
> > > > > > > > result['sh.nrows']=sh.nrows
> > > > > > > > result['sh.ncols']=sh.ncols
> > > > > > > > result['Cell D30 is']=sh.cell_value(rowx=29, colx=3)
> > > > > > > > for rx in range(sh.nrows):
> > > > > > > > result['row ' + str(rx)]=sh.row(rx)
> > > > > > > > except:
> > > > > > > > response.flash = T('excel_read exception')
> > > > > > > > return dict(result=result)
>
> > > > > > > > for you request.vars.filename is a cgi.FieldStorage object. It
> > > > > > > > is not
> > > > > > > > a named file.
> > > > > > > > It containes
> > > > > > > > - request.vars.filename.file (a buffered input stream)
> > > > > > > > - request.vars.filename.filename (the original filename)
> > > > > > > > For you to access it you should read the stream and save it
> > > > > > > > with a
> > > > > > > > name. Do not use the original filename to avoid directory
> > > > > > > > traversal
> > > > > > > > attacks.
>
> > > > > > > > Massimo
>
> > > > > > > > On May 7, 2:58 pm, Hans <johann.scheibelho...@easytouch-edv.com>
> > > > > > > > wrote:
>
> > > > > > > > > I'm trying to import some content directly from an excel file
> > > > > > > > > which
> > > > > > > > > the user selects from its local files. I've taken and
> > > > > > > > > modified code
> > > > > > > > > from the admin controller for retrieving the path/filename
> > > > > > > > > thru a nice
> > > > > > > > > file selection window.
>
> > > > > > > > > The problem: I get only the filename of a locked temporary
> > > > > > > > > file, not
> > > > > > > > > the original filename. If I close the temporary file, to
> > > > > > > > > remove the
> > > > > > > > > lock, the file is automatically deleted...
>
> > > > > > > > > I would not mind to work with a temporary copy of the
> > > > > > > > > original file,
> > > > > > > > > but retrieving the original path/filename thru a similar file
> > > > > > > > > selection window would be prefered. How can this be achieved?
>
> > > > > > > > > Hans
> > > > > > > > > P.S. I'm using the xlrd module
> > > > > > > > > fromhttp://pypi.python.org/pypi/xlrd
>
> > > > > > > > > ### viewer ###
> > > > > > > > > {{extend 'layout.html'}}
> > > > > > > > > <h1>excel_read.html template</h1>
>
> > > > > > > > > <div class='frame'>
> > > > > > > > > <h2>{{=t2.T('Select File')}}</h2>
> > > > > > > > > {{=FORM('File: ',INPUT(_type='file',_name='filename'),INPUT
> > > > > > > > > (_type='submit', _value='submit'))}}
> > > > > > > > > </div>
>
> > > > > > > > > <div class='frame'>
> > > > > > > > > {{=BEAUTIFY(response._vars)}}
> > > > > > > > > </div>
>
> > > > > > > > > ### controller ###
> > > > > > > > > def excel_read():
> > > > > > > > > result={}
> > > > > > > > > if request.vars.filename != None:
> > > > > > > > > try:
> > > > > > > > > import xlrd
>
> > > > > > > > > # >>> can't locate the originally selected path/filename in
> > > > > > > > > request.vars.filename <<<
> > > > > > > > > book =
> > > > > > > > > xlrd.open_workbook(request.vars.filename.file.name)
> > > > > > > > > # >>> with a valid path/filename it works <<<
> > > > > > > > > book = xlrd.open_workbook('F:/Documents and
> > > > > > > > > Settings/Hans/
> > > > > > > > > Desktop/test.xls')
>
> > > > > > > > > result['The number of worksheets is']=book.nsheets
> > > > > > > > > result["Worksheet name(s)"]= book.sheet_names()
> > > > > > > > > sh = book.sheet_by_index(0)
> > > > > > > > > result['sh.name']=sh.name
> > > > > > > > > result['sh.nrows']=sh.nrows
> > > > > > > > > result['sh.ncols']=sh.ncols
> > > > > > > > > result['Cell D30 is']=sh.cell_value(rowx=29,
> > > > > > > > > colx=3)
> > > > > > > > > for rx in range(sh.nrows):
> > > > > > > > > result['row ' + str(rx)]=sh.row(rx)
> > > > > > > > > except:
> > > > > > > > > response.flash = T('excel_read exception')
> > > > > > > > > return dict(result=result)
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