Thx for your quick answer.
But i try to do my form without sqlform (i didn't show all my form) because
I have a form with a lot of fields, and they aren't all on the same page, i
mean it's with anchor on the same page, and i dunno how to do the same
thing with a factory. I can't upload a file with a simple form ?
And do you know how to change the name of the file when it's upload ? I
have several upload fields.
Le mercredi 16 décembre 2015 10:53:41 UTC+1, xmarx a écrit :
>
> in controller:
>
> def index():
> import os
> form=SQLFORM.factory(Field('name'),Field('file',
> 'upload',uploadfolder=os.path.join(request.folder,'uploads')))
> if form.process().accepted:
> request.flash='file uploaded!'
> return dict(form=form)
>
>
>
> in view index.html:
>
> {{extend 'layout.html'}}
>
> <h1>Upload File</h1>
> {{=form}}
>
>
>
>
> thats it.
>
> 16 Aralık 2015 Çarşamba 11:07:11 UTC+2 tarihinde Adrien yazdı:
>>
>>
>> Hi everyone,
>> Like i said in the title, i want to upload a file in a directory with a
>> form.
>> I saw another subject where he did what i want but with me, it doesn't
>> work and i don't know why.
>>
>> This is my controller default.py :
>> def test():
>> import shutil
>>
>> filename=request.vars.filename
>> file=request.vars.file
>> shutil.copyfileobj(file,open('path/'+filename,'wb'))
>> return dict()
>>
>> def index():
>> return dict()
>>
>> And the view index.html :
>> <form method="post" enctype="multipart/form-data" action="test">
>> <input name="upload" type="file" size="60" maxlength="100000">
>> <input type="Submit" value="Upload">
>> </form>
>>
>> Do someone knows how to fix this problem ? And sorry but I'm not english,
>> i hope i'm clear for you.
>>
>
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