This is working for me. I have a similar requirement. Select a text file to
be processed and pass it to a separate routine to process the file:
form = SQLFORM.factory(
Field('brillXf1', 'upload',
uploadfolder=os.path.join(request.folder,'uploads'),
label='Brill XF1 File'))
if form.process().accepted:
fileName = form.vars.brillXf1
print fileName
from reports import brillToDalex as btd
btd.process('%s/%s' % (os.path.join(request.folder,'uploads'), fileName), 1
)
redirect(URL('nutrientConversion'))
response.flash('File has been uploaded and email will be sent with results'
)
The form field only returns the file name. You have to specify the path to
it if you are going to use it later as seen in this statement:
btd.process('%s/%s' % (os.path.join(request.folder,'uploads'), fileName), 1)
-Jim
On Thursday, November 1, 2012 3:33:50 AM UTC-5, praveen krishna wrote:
>
> Nol there is no such requirement.Actually I am able to upload and read a
> text file by zipping it but for my task I have upload a text file .
>
> On Thu, Nov 1, 2012 at 7:24 AM, Johann Spies <johann...@gmail.com<javascript:>
> > wrote:
>
>> On 31 October 2012 17:45, praveen krishna <praveenc...@gmail.com<javascript:>
>> > wrote:
>>
>>> Hi,
>>> For my task I have to upload a text file in form with out zipping it
>>> ,I have prepared the form as:
>>> form = FORM(TABLE(TR(TD('Upload File:', INPUT(_type='file',
>>> name='myfile', id='myfile',
>>> requires=IS_NOT_EMPTY()))),TR(TD(INPUT(_type='submit',_value='Submit')))))
>>>
>>> But I unable to read the file how could it be possible without using
>>> ZipFile.
>>>
>>
>> You refer to 'zip' twice. Why? Is there a requirement that it should
>> be zipped somewhere?
>>
>> Regards
>> Johann
>> --
>> Because experiencing your loyal love is better than life itself,
>> my lips will praise you. (Psalm 63:3)
>>
>> --
>>
>>
>>
>>
>
>
--
