Re: ToDo: show size of partitioned table

ne 7. 4. 2019 v 18:07 odesílatel Alvaro Herrera <alvherre(at)2ndquadrant(dot)com>
napsal:

> On 2019-Apr-07, Pavel Stehule wrote:
>
> > ne 7. 4. 2019 v 17:27 odesílatel Justin Pryzby <pryzby(at)telsasoft(dot)com>
> > napsal:
>
> > > I think there's an issue with showing indices. You said that \dP
> should be
> > > same as \dPti, no? Right now, indices are not shown in \dP, unless a
> > > pattern is given. I see you add that behavior in the regression
> > > tests; is that really what's intended ? Also, right now adding a
> > > pattern affects how sizes are computed, I don't see why that's
> > > desirable or, if so, how to resolve that inconsistency, or how to
> > > document it.
> >
> > That depends. If there are not pattern, then \dP show only tables, but
> with
> > total relation size (so size of indexes are nested). It is different than
> > \dPti, but I think so it is useful - when you don't specify object type,
> > then usually you would to see a tables, but with total size.
> >
> > I don't see a benefit from \dP == \dPti. When there are a pattern (that
> can
> > choose some index, then, indexes are displayed and \dP == \dPti.
>
> Well, I think Justin has it right --- \dP should be just like \df, which
> means to list "everything". If you add the "t" or the "i", that means
> to list only those kinds of things (just like adding one of a, n, p, t,
> w does for \df). You can add both, and then it list both kinds, just
> like \dfanptw list the same things that \df does.
>
> That's also what I changed the docs to say, but I failed to update the
> code correctly, and didn't verify the expected output closely either.
> So I'm due to resubmit this ...
>

ok

Pavel

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