this code work but i need create form in controller but how i send the
img_id variable in controller (form)?
{{for img in images:}}
<image><img width="85px"
src="{{=URL('download', args=img.file)}}" /><br></image>
<br><br><br><br><br>
{{=img}}
{{=img.id}}
{{row_id=img.id}}
{{form = SQLFORM(db.ref, row_id)
if form.process().accepted:
response.flash = "Image uploaded"+str(form.vars)
else:
response.flash = form.vars.id}}
{{=form}}
{{pass}}{{pass}}
maanantaina 12. maaliskuuta 2012 14.02.10 UTC+2 juvi1 kirjoitti:
>
>
> <https://lh4.googleusercontent.com/-C5e04mE4s8I/T13jl7HXizI/AAAAAAAAAAU/ngyK-9pCkAA/s1600/Screenshot%2520at%25202012-03-12%252013%253A51%253A57.png>
> Sry. I dont understand. now all forms update same image (image_id1)
> can i send img_id (in view) to form (in controller)?
> I also try this code, but that insert always image_id variable = last
> image id in database.
> def jobs():
> images=db().select(db.ref.id,db.ref.file)
> for img in images:
> form = SQLFORM(db.ref, img.id)
>
> if form.process().accepted:
> response.flash = "Image uploaded"+str(form.vars)
> else:
> response.flash = form.vars.id
>
> return dict(form=form, images=images)
>
> maanantaina 12. maaliskuuta 2012 12.23.36 UTC+2 Martin.Mulone kirjoitti:
>>
>> To edit an image you need to pass the id of the row to SQLFORM.
>>
>> Ex.:
>> row_id = 1
>> form = SQLFORM(db.ref, row_id)
>>
>> 2012/3/11 juvi1 <juha.wilh...@gmail.com>
>>
>>> 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 <juha.wilh...@gmail.com>
>>>>
>>>>> 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
>>>>
>>>>
>>
>>
>> --
>> http://www.tecnodoc.com.ar
>>
>>
