Here after default/delformforscreen.load accepts the form
default/screen.html should load. How do I do that? It ain't working the way
I thought. So I marked comment signs infront of that (see below) .
def screen():
row=db(db.info.info_id==auth.user_id).select().first()
if row:
grouprow=db(db.joiners.joiner_id==auth.user_id).select()
return locals()
def delformforscreen():
row=db(db.info.info_id==auth.user_id).select().first()
form=0
if row:
form=SQLFORM(db.info,row.id,fields=['first_name','last_name','dob','sex','hometown',
'highschool', 'university', 'oneself'],showid=False)
#if form.process().accepted:
# redirect(URL('default','screen'))
else:
form=SQLFORM(db.info,fields=['first_name','last_name','dob','sex','hometown',
'highschool', 'university', 'oneself'])
return locals()
{{extend 'layout.html'}}
{{if row:}}
Hi
{{=row.firstname}}
{{pass}}
<br>
<br>
<h4>
Update/Edit your Info
</h4>
<br>
{{=LOAD('default', 'delformforscreen.load', ajax=True)}}
--
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