On Apr 27, 3:14 pm, Gary <gary.k.ma...@gmail.com> wrote:
> Like myapp/default/display_manual_form() in the book, right?
>
> Two questions.  I assume that I need a:
>
> def data(): return dict(form=crud())
>
> for each table that uses the crud controller to properly define the
> URL.  If that's true, it's the part that I missed.

No. This controller will work for every table in the database. In
fact, as you do below, you specify the table in the URL

> 1)http://hostname/myapp/default/data/update/person/2

You create a CRUD action per table

def mycreate():
    return dict(form=crud.create(db.mytable))

or an update action

def myupdate():
    id=request.args[0]
    return dict(form=crud.update(db.mytable,id))



> or create a controller like:
>
> 2) def update1():
>       redirect(URL(r=request,f='data/update/person/2'))

Do do not do actions that just redirect.

> In either case, what should the view be named?

the view is the controller/action.html

> Since one might want
> to have a different view for read,update,delete,etc., it seems that it
> should be:
>
> default/data/update.html, but my testing seems to work at default/
> data.html.

you can do

def data():
    response.view="%s/%s/%s.html" % (request.controller,
request.function, request.args[0])
    return dict(form=crud())

> That seems to imply that using method #2 above with a response.view()
> would be the better solution.  Would you agree with that conclusion or
> is there a better way to handle this?
>
> Thank you.
> Gary
>
> On Apr 26, 9:22 pm, mdipierro <mdipie...@cs.depaul.edu> wrote:
>
> > Yes, you can make the form in html as usual for SQLFORM.
>
> > On 26 Apr, 18:37, Gary <gary.k.ma...@gmail.com> wrote:
>
> > > Is there a simple way to replace the automatically generated SQLFORM
> > > in the various crud controllers with a custom SQLFORM?
>
> > > Thanks.
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