>
> equipment is the table.
>
Yes, but in your code, it simply said args=[equipment.slug]. If your db
object has a table named "equipment", you would refer to it as
db.equipment, but in any case, that would not be relevant here.
> the equipment details page is 'show' or 'details'. In reality, I thought I
> could get away with the default 'view' from sqlform.grid, using a
> slug...but it appears I cannot, so I have to create a custom 'show' or
> 'details' page. If I do
> links = [lambda row: A('Details',_href=URL('show','show',
> args=[row.slug]))]
> i get a key error.
>
What is the key error? Is your grid selecting from the equipment table
(with the slug field)?
Why is the URL now URL('show', 'show')? Is the controller named show.py,
with a function called show()? In the URL() function, the first argument
should be the controller and the second the function.
Anthony
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