Thread: PHP PDO getting data from pgSQL stored function
Hi guys, i need to do a ugly select which i dont want to place in my php code. I use PDO all the time. I want to return a structure which is the same as a table i have created. I want to write it in pgSQL. Now my question, how can i return that and access it with PHP PDO? Should I use a cursor or shoul I use the return of the function (if it is possible)? Can someone provide a piece of example code? Thanks for help
On Sat, Nov 2, 2013 at 11:54 AM, Michael Schmidt <css.liquid@gmail.com> wrote:
--
Jonah H. Harris
Blog: http://www.oracle-internals.com/
i need to do a ugly select which i dont want to place in my php code.
I use PDO all the time.
All database access from our application is functionalized using either PL/pgSQL or SQL functions.
I want to return a structure which is the same as a table i have created.
CREATE FUNCTION blah (IN param 1, ..., IN param n, OUT param 1, ... OUT param n) RETURNS SETOF RECORD
Now my question, how can i return that and access it with PHP PDO?
Should I use a cursor or shoul I use the return of the function (if it is possible)?
Just do a SELECT * FROM blah(...); as a normal PDO query.
Jonah H. Harris
Blog: http://www.oracle-internals.com/
Hi
basically create a function returning a table, then select * from function() can be called from php.
below is a complete sql language function i wrote returning a table.
create or replace function
show_privilege(p_grantee name)
returns
table(grantee name
,role_name name
,grantor name
,table_catalog name
,table_name name
,privilege_type varchar)
as $$
select
AR.grantee::name
,AR.role_name::name
,RTG.grantor::name
,RTG.table_catalog::name
,RTG.table_name::name
,privilege_type
from
information_schema.applicable_roles AR
left outer join
information_schema.role_table_grants RTG on (AR.role_name = RTG.grantee)
where
AR.grantee = p_grantee;
$$ language sql;
you'll notice the returns table defines the rows in the return.
on one of my databases, if i run:
select * from show_privilege('wuggly_ump_admin');
i get
grantee | role_name | grantor | table_catalog | table_name | privilege_type
------------------+-----------+---------+---------------+------------+----------------
wuggly_ump_admin | sys_user | | | |
(1 row)
i hope that helps.
reiner
On 2 nov 2013, at 16:54, Michael Schmidt <css.liquid@gmail.com> wrote:
> Hi guys,
>
> i need to do a ugly select which i dont want to place in my php code.
> I use PDO all the time.
>
> I want to return a structure which is the same as a table i have created.
>
> I want to write it in pgSQL.
>
> Now my question, how can i return that and access it with PHP PDO?
> Should I use a cursor or shoul I use the return of the function (if it is possible)?
>
> Can someone provide a piece of example code?
>
> Thanks for help
>
>
> --
> Sent via pgsql-php mailing list (pgsql-php@postgresql.org)
> To make changes to your subscription:
> http://www.postgresql.org/mailpref/pgsql-php
Hi Reiner,
On Mon, Nov 4, 2013 at 3:35 AM, reiner peterke <zedaardv@drizzle.com> wrote:
basically create a function returning a table, then select * from function() can be called from php.
below is a complete sql language function i wrote returning a table.
create or replace function
show_privilege(p_grantee name)
returns
table(grantee name
,role_name name
,grantor name
,table_catalog name
,table_name name
,privilege_type varchar)
as $
select
AR.grantee::name
,AR.role_name::name
,RTG.grantor::name
,RTG.table_catalog::name
,RTG.table_name::name
,privilege_type
from
information_schema.applicable_roles AR
left outer join
information_schema.role_table_grants RTG on (AR.role_name = RTG.grantee)
where
AR.grantee = p_grantee;
$ language sql;
you'll notice the returns table defines the rows in the return.
on one of my databases, if i run:
select * from show_privilege('wuggly_ump_admin');
i get
grantee | role_name | grantor | table_catalog | table_name | privilege_type
------------------+-----------+---------+---------------+------------+----------------
wuggly_ump_admin | sys_user | | | |
(1 row)
i hope that helps.
I'm not sure what is your problem, but it seems all you need is VIEW if you have
problem with your stored procedure.
Anyway, I suppose PostgreSQL returns result set resource. Are you saying
PDO pgsql wouldn't? or missing columns? How about pgsql module? Does it work?
I guess you have been tested, but the "select" in procedure works as expected
Hi,
sorry for the late response.
I did it with the solution posted by Jonah.
It is not necessary for me to return a table(...) because i want to return the structure which is already defined in my database so i can use SETOF RECORD as return type. So i use the real table to define the return of my function.
PDO select works just like a normal select on a table.
Thank you.
Am 03.11.2013 17:06, schrieb Jonah H. Harris:
sorry for the late response.
I did it with the solution posted by Jonah.
It is not necessary for me to return a table(...) because i want to return the structure which is already defined in my database so i can use SETOF RECORD as return type. So i use the real table to define the return of my function.
PDO select works just like a normal select on a table.
Thank you.
Am 03.11.2013 17:06, schrieb Jonah H. Harris:
On Sat, Nov 2, 2013 at 11:54 AM, Michael Schmidt <css.liquid@gmail.com> wrote:--i need to do a ugly select which i dont want to place in my php code.
I use PDO all the time.All database access from our application is functionalized using either PL/pgSQL or SQL functions.I want to return a structure which is the same as a table i have created.CREATE FUNCTION blah (IN param 1, ..., IN param n, OUT param 1, ... OUT param n) RETURNS SETOF RECORDNow my question, how can i return that and access it with PHP PDO?
Should I use a cursor or shoul I use the return of the function (if it is possible)?Just do a SELECT * FROM blah(...); as a normal PDO query.
Jonah H. Harris
Blog: http://www.oracle-internals.com/