You may need to db.commit() because the manual insert may not commit
normally maybe??
if form.process().accepted:
for i in range(0, int(request.vars.row)):
db.company.insert(company=form.vars['company_%s' % i])
db.commit()
But I am not sure I understand you onkeyup ajax call what's the intent of
it???
Richard
On Sun, Mar 17, 2013 at 9:48 AM, 黄祥 <[email protected]> wrote:
> @anthony : thank you for correcting me anthony
> @richard : i'm learning to use 2 input form too but not work as expected
> (no error occured), the data is not insert after i submit it. any idea
> about this?
> thank you
>
> *default.py*
> def test():
> inputs = []
> for i in range(0, int(request.vars.row)):
> inputs.append(db.company.company.clone(name='%s_%s' % (
> db.company.company.name, i)))
> form = SQLFORM.factory(*inputs)
> if form.process().accepted:
> for i in range(0, int(request.vars.row)):
> db.company.insert(company=form.vars['company_%s' % i])
> elif form.errors:
> response.flash = 'form has errors'
> return dict(form=form)
>
> def index():
> return dict()
>
> *index.html*
> {{extend 'layout.html'}}
>
> <form>
> <input name="row" onkeyup="ajax('test', ['row'], 'target')" />
> </form>
>
> <div id='target'></div>
>
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