Обсуждение: How can I find the schema that a table belongs to?
Hi, I am trying to tidy up my perl script that runs as a cgi and allows remote users (aka me) to interact with my Pg database. I primarily want to use this tool as a way for my iPad to browse my data base. It seems to be working quite well but one certainly does not want to 'select * from ginormous_table'. I am trying to build a 'describe' function. ie if the user types "describe tablename" in the sql box I want to display some interesting attributes for that table, mainly names and data types for each of the columns. I have done this in other programs but I forced the user to specify a schema qualified name ( and mimiced what psql -E did). So I guess the question is: Given a bare table name, how can I recover the schema qualified name with whatever the current search path happens to be? This task has to be done using simple sql from the perl dbi. Thanks, Jerry
one of the way to find out schema name is like below
select * from pg_tables where tablename like '%xyz%';
schemaname | tablename | tableowner | tablespace | hasindexes | hasrules | hastriggers
------------+-----------+------------+------------+------------+----------+-------------
(0 rows)
~Deepak
select * from pg_tables where tablename like '%xyz%';
schemaname | tablename | tableowner | tablespace | hasindexes | hasrules | hastriggers
------------+-----------+------------+------------+------------+----------+-------------
(0 rows)
~Deepak
On Wed, Jan 19, 2011 at 8:26 AM, Jerry LeVan <jerry.levan@mac.com> wrote:
Hi,
I am trying to tidy up my perl script that runs as a cgi and allows
remote users (aka me) to interact with my Pg database.
I primarily want to use this tool as a way for my iPad to
browse my data base. It seems to be working quite well but
one certainly does not want to 'select * from ginormous_table'.
I am trying to build a 'describe' function. ie if the user
types "describe tablename" in the sql box I want to display
some interesting attributes for that table, mainly names and
data types for each of the columns.
I have done this in other programs but I forced the user
to specify a schema qualified name ( and mimiced what psql -E
did).
So I guess the question is:
Given a bare table name, how can I recover the schema
qualified name with whatever the current search path happens
to be?
This task has to be done using simple sql from the perl dbi.
Thanks,
Jerry
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If your looking for Views then you could use pg_views ==> select * from pg_views limit 1;
Here is one more, there was a recent post same as your request, Please see the below email, hope this helps you
Tom Lane =================>
Thomas Kellerer <spam_eater@gmx.net> writes:
> Jerry LeVan, 19.01.2011 17:35:
>> So I guess the question is:
>> Given a bare table name, how can I recover the schema
>> qualified name with whatever the current search path happens
>> to be?
> SELECT table_schema
> FROM information_schema.tables
> WHERE table_name = 'your_table'
> ;
That's not going to work, at least not in the interesting case where youHere is one more, there was a recent post same as your request, Please see the below email, hope this helps you
Tom Lane =================>
Thomas Kellerer <spam_eater@gmx.net> writes:
> Jerry LeVan, 19.01.2011 17:35:
>> So I guess the question is:
>> Given a bare table name, how can I recover the schema
>> qualified name with whatever the current search path happens
>> to be?
> SELECT table_schema
> FROM information_schema.tables
> WHERE table_name = 'your_table'
> ;
have more than one candidate table --- that SELECT will list all of 'em.
In most cases the answer to this type of problem is "use regclass",
but regclass doesn't quite solve Jerry's problem because it won't
schema-qualify the name if the table is visible in the search path.
The best solution I can think of is
select nspname from pg_namespace n join pg_class c on n.oid = c.relnamespace
where c.oid = 'my_table_name'::regclass;
which works but seems a bit brute-force.
regards, tom lane
- Hide quoted text -
==========================================>
On Wed, Jan 19, 2011 at 12:58 PM, Jerry LeVan <jerry.levan@mac.com> wrote:
That does not work if the user entered a 'view' in the sql box.
On Jan 19, 2011, at 3:12 PM, DM wrote:
> one of the way to find out schema name is like below
>
> select * from pg_tables where tablename like '%xyz%';
> schemaname | tablename | tableowner | tablespace | hasindexes | hasrules | hastriggers
> ------------+-----------+------------+------------+------------+----------+-------------
> (0 rows)
>
> ~Deepak
>
Also a table/view could be in several schemas...
Jerry
or you could use the below query
********* QUERY **********
SELECT n.nspname as "Schema",
c.relname as "Name",
CASE c.relkind WHEN 'r' THEN 'table' WHEN 'v' THEN 'view' WHEN 'i' THEN 'index' WHEN 'S' THEN 'sequence' WHEN 's' THEN 'special' END as "Type",
pg_catalog.pg_get_userbyid(c.relowner) as "Owner"
FROM pg_catalog.pg_class c
LEFT JOIN pg_catalog.pg_namespace n ON n.oid = c.relnamespace
WHERE c.relkind IN ('r','v','S','')
AND n.nspname <> 'pg_catalog'
AND n.nspname <> 'information_schema'
AND n.nspname !~ '^pg_toast'
AND pg_catalog.pg_table_is_visible(c.oid)
AND c.relname like '%my_table_name%'
ORDER BY 1,2;
**************************
********* QUERY **********
SELECT n.nspname as "Schema",
c.relname as "Name",
CASE c.relkind WHEN 'r' THEN 'table' WHEN 'v' THEN 'view' WHEN 'i' THEN 'index' WHEN 'S' THEN 'sequence' WHEN 's' THEN 'special' END as "Type",
pg_catalog.pg_get_userbyid(c.relowner) as "Owner"
FROM pg_catalog.pg_class c
LEFT JOIN pg_catalog.pg_namespace n ON n.oid = c.relnamespace
WHERE c.relkind IN ('r','v','S','')
AND n.nspname <> 'pg_catalog'
AND n.nspname <> 'information_schema'
AND n.nspname !~ '^pg_toast'
AND pg_catalog.pg_table_is_visible(c.oid)
AND c.relname like '%my_table_name%'
ORDER BY 1,2;
**************************
On Wed, Jan 19, 2011 at 1:54 PM, DM <dm.aeqa@gmail.com> wrote:
If your looking for Views then you could use pg_views ==> select * from pg_views limit 1;That's not going to work, at least not in the interesting case where you
Here is one more, there was a recent post same as your request, Please see the below email, hope this helps you
Tom Lane =================>
Thomas Kellerer <spam_eater@gmx.net> writes:
> Jerry LeVan, 19.01.2011 17:35:> SELECT table_schema
>> So I guess the question is:
>> Given a bare table name, how can I recover the schema
>> qualified name with whatever the current search path happens
>> to be?
> FROM information_schema.tables
> WHERE table_name = 'your_table'
> ;
have more than one candidate table --- that SELECT will list all of 'em.
In most cases the answer to this type of problem is "use regclass",
but regclass doesn't quite solve Jerry's problem because it won't
schema-qualify the name if the table is visible in the search path.
The best solution I can think of is
select nspname from pg_namespace n join pg_class c on n.oid = c.relnamespace
where c.oid = 'my_table_name'::regclass;
which works but seems a bit brute-force.
regards, tom lane- Hide quoted text -
==========================================>On Wed, Jan 19, 2011 at 12:58 PM, Jerry LeVan <jerry.levan@mac.com> wrote:That does not work if the user entered a 'view' in the sql box.
On Jan 19, 2011, at 3:12 PM, DM wrote:
> one of the way to find out schema name is like below
>
> select * from pg_tables where tablename like '%xyz%';
> schemaname | tablename | tableowner | tablespace | hasindexes | hasrules | hastriggers
> ------------+-----------+------------+------------+------------+----------+-------------
> (0 rows)
>
> ~Deepak
>
Also a table/view could be in several schemas...
Jerry
On Jan 19, 2011, at 3:12 PM, DM wrote: > one of the way to find out schema name is like below > > select * from pg_tables where tablename like '%xyz%'; > schemaname | tablename | tableowner | tablespace | hasindexes | hasrules | hastriggers > ------------+-----------+------------+------------+------------+----------+------------- > (0 rows) > > ~Deepak > That does not work if the user entered a 'view' in the sql box. Also a table/view could be in several schemas... Jerry