On 12/22/20 4:39 PM, Tom Lane wrote:
Adrian Klaver <adrian.klaver@xxxxxxxxxxx> writes:
So how does one go about using a table name from
information_schema.tables in pg_table_size()?
You want something like
select pg_table_size(quote_ident(table_schema)||'.'||quote_ident(table_name))
from information_schema.tables;
I imagine that the failures you got are a consequence of having
some table names that aren't valid unless quoted (ie contain
spaces, funny characters, etc). In a general-purpose query,
you can't ignore the schema name either.
I might be more excited about v12's failure to provide an implicit
cast to regclass if there were any prospect of queries like this
working in a bulletproof way without accounting for schema names
and funny characters. But there isn't, so the query shown in SO
is a house of cards to start with. When you do it right, with
quote_ident() or format(), no special casting is needed.
Thanks, that pushed me in right direction.
I see now the previous query worked because the alias table_name and the
column table_name where the same and the column previously was a
varchar. This meant the pg_table_size() was actually working on the
column value not the concatenated value.
So the query can be simplified to:
SELECT
pg_size_pretty(pg_table_size(quote_ident(table_name))),
pg_size_pretty(pg_indexes_size(quote_ident(table_name))) AS
indexes_size,
pg_size_pretty(pg_total_relation_size(quote_ident(table_name))) AS
total_size
FROM
information_schema.tables
WHERE
table_schema = 'public'
;
regards, tom lane
--
Adrian Klaver
adrian.klaver@xxxxxxxxxxx
