Re: how to tell who the caller of a procedure is
- From: "sol beach" <sol.beach@xxxxxxxxx>
- To: oxnard@xxxxxxxxxxxxxxx
- Date: Wed, 30 Jan 2008 09:41:23 -0800
from asktom.oracle.com
create or replace procedure who_called_me( owner out varchar2,
name out varchar2,
lineno out number,
caller_t out varchar2 )
as
call_stack varchar2(4096) default dbms_utility.format_call_stack;
n number;
found_stack BOOLEAN default FALSE;
line varchar2(255);
cnt number := 0;
begin
--
loop
n := instr( call_stack, chr(10) );
exit when ( cnt = 3 or n is NULL or n = 0 );
--
line := substr( call_stack, 1, n-1 );
call_stack := substr( call_stack, n+1 );
--
if ( NOT found_stack ) then
if ( line like '%handle%number%name%' ) then
found_stack := TRUE;
end if;
else
cnt := cnt + 1;
-- cnt = 1 is ME
-- cnt = 2 is MY Caller
-- cnt = 3 is Their Caller
if ( cnt = 3 ) then
lineno := to_number(substr( line, 13, 6 ));
line := substr( line, 21 );
if ( line like 'pr%' ) then
n := length( 'procedure ' );
elsif ( line like 'fun%' ) then
n := length( 'function ' );
elsif ( line like 'package body%' ) then
n := length( 'package body ' );
elsif ( line like 'pack%' ) then
n := length( 'package ' );
elsif ( line like 'anonymous%' ) then
n := length( 'anonymous block ' );
else
n := null;
end if;
if ( n is not null ) then
caller_t := ltrim(rtrim(upper(substr( line, 1, n-1 ))));
else
caller_t := 'TRIGGER';
end if;
line := substr( line, nvl(n,1) );
n := instr( line, '.' );
owner := ltrim(rtrim(substr( line, 1, n-1 )));
name := ltrim(rtrim(substr( line, n+1 )));
end if;
end if;
end loop;
end;
/
create or replace function who_am_i return varchar2
is
l_owner varchar2(30);
l_name varchar2(30);
l_lineno number;
l_type varchar2(30);
begin
who_called_me( l_owner, l_name, l_lineno, l_type );
return l_owner || '.' || l_name;
end;
/
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