hi,
i would do the follow:
1) in model use compute to create image form base64 string:
db.define_table('user_images',
Field('my_image', 'upload', label=T('My Image'), compute=lambda r:
create_image_from_base64_str(r.my_image_b64)),
Field('my_image_b64', 'text')
)
def create_image_from_base65_str(r):
...
return filename
2) in controller:
form = SQLFORM(db.user_images)
if form.process().accepted:
#form processed with no errors
#image is created
#make redirect
redirect(URL('xyz', args=[form.vars.id]), client_side=True)
...
3) in view i would use form.custom to create form with hidden fields and
fake button. with js i would save the base64 string in the hidden field and
do form.submit()
Am Montag, 7. April 2014 00:46:11 UTC+2 schrieb Ari Lion BR Sp:
>
> Any answer? I have the very same need.
>
> Thanks
> Ari
>
> Em segunda-feira, 2 de dezembro de 2013 14h02min30s UTC-2, Michael Hall
> escreveu:
>>
>> If I have the following model:
>>
>> db.define_table('user_images',
>> Field('my_image', 'upload',
>> label=T('My Image')),
>> )
>>
>> and In my rendered html view I have a canvas, the canvas has produced an
>> image saved in the variable
>> document.getElementById('canvas').src = dataURL;
>>
>> Underneath the image is an upload button, I wish to POST the canvas image
>> to user_images.myimage when the button is pressed and then move to a new
>> view and controller which displays the uploaded image and has some more
>> editing options.
>>
>> What is the best way to achieve this? Is it possible to render a sqlform
>> that only has a button which when pressed uploads dataURL and redirects to
>> the next controller/view?
>>
>>
>>
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