thanks, but now form create always new image in database, and i want update
image.
how it can be done??
sunnuntaina 11. maaliskuuta 2012 13.37.36 UTC+2 Martin.Mulone kirjoitti:
>
> I don't know why you have two {{pass}} in your view. I think you can do:
>
> def jobs():
> images=db().select(db.ref.id,db.ref.file)
> form = SQLFORM(db.ref)
> if form.process().accepted:
> response.flash = "Image uploaded"
>
> return dict(form=form, images=images)
>
> in views:
> {{for img in images:}}
> <image><img width="85px"
> src="{{=URL('download', args=img.file)}}" /><br></image>
> <br><br><br><br><br>
> {{=img}}
> {{=img.id}}
> {{=form}}
> {{pass}}
>
> 2012/3/11 juvi1 <[email protected]>
>
>> Hello sorry my bad english and i hope that you understand my problem.
>>
>> I want show in one page all pictures which are saved in database.
>> In page i want that i can upload new picture and update it whit forms
>> (override old one)
>> My code are
>>
>> in controller:
>> def jobs():
>> images=db().select(db.ref.id,db.ref.file)
>> for img in images:
>> formname = "upload_f_%s"%img.id
>> form = FORM(INPUT(_type="file",_name=formname),
>> INPUT(_type='submit'))
>>
>> if form.accepts(request.vars, _name=formname):
>> response.flash = form.vars
>>
>> return dict(form=form, img=img, formname=formname, images=images)
>>
>> in views:
>> {{for img in images:}}
>> <image><img width="85px"
>> src="{{=URL('download', args=img.file)}}" /><br></image>
>> <br><br><br><br><br>
>> {{=img}}
>> {{formname=img.id}}
>> {{=formname}}
>> {{=form}}
>> {{pass}}{{pass}}
>>
>> please help!
>>
>
>
>
> --
> http://www.tecnodoc.com.ar
>
>
