Do not sure crud. It is old, to be deprected, less flexible than SQLFORM.
Do this:
form = SQLFORM(db.table, record_id)
form.process(dbio=False)
if form.accepted:
if form.deleted: # to be deleted because dbio=False
count = db(db.table3.field2t3 == form.record_id).count()
if count:
session.flash = T('The record you try to delete is still
referenced by other records and can\'t be deleted')
redirect(URL(c='default', f='create_update',
args=request.args(0)))
else:
form.record.delete()
else:
form.record.update_record(**form.post_vars)
On Wednesday, 12 December 2012 11:37:26 UTC-6, Richard wrote:
>
> Ok, I understand, but I don't want the record to be deleted, so SET TO
> NULL not a solution.
>
> Now I try what I was trying to avoid :
>
> def ondelete_func2(form):
> count = 0
> count += db(db.table3.field2t3 == request.args(1)).count()
> if count > 0:
> pass
> else:
> session.flash = T('The record you try to delete is still
> referenced by other records and can\'t be deleted')
> redirect(URL(c='default', f='create_update', args=request.args(0)))
>
> I want to look into table3 if a id of table2 is still referenced.
>
> I know that I don't need count+= it is like that because in my own app
> there is not only one table3 there is 30+ tables that are referencing id of
> table2 and that why I would avoid using query to be informed about that.
>
> Now what I don't know how to do is to prevent the crud.update() deletion
> to occur in case of count > 0...
>
> Help is appreciate.
>
> Richard
>
> On Wed, Dec 12, 2012 at 12:29 PM, Massimo Di Pierro
> <[email protected]<javascript:>
> > wrote:
>
>> add it because web2py does not do migration when ondelete changes. You
>> have to delete your database, you .tables and start again.
>
>
>
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