Database users loop with password generator - oracle

Question

I have created some kind of code block till now, and I'm stuck. I'm selecting from table called FND_ORACLE_USERID for EBS database user schemas. Idea is to have those accounts printed in format like this (password is placeholder):
FNDCPASS user/password 0 Y system/test11 ALLORACLE password
Till now I have done this:
set serveroutput on;
DECLARE
c1 SYS_REFCURSOR;
l_pass varchar(16);
l_count number(3);
BEGIN
select count (*) into l_count from FND_ORACLE_USERID
where READ_ONLY_FLAG='A';
OPEN c1 for
select xmlagg(xmlelement("r", ch)).extract('//text()').getstringval() ch
from
(
select distinct first_value(ch) over (partition by lower(ch)) as ch
from (
select substr('abcd$efghijklmn#pqrstuvw#xyzABC$DEFGHIJK$LMNPQR!STUVWXYZ1!23456789',
level, 1) as ch
from dual
connect by level <= 59
order by dbms_random.value
)
where rownum <= dbms_random.value(17,17)
);
LOOP
FETCH c1
INTO l_pass;
EXIT WHEN c1%NOTFOUND;
DBMS_OUTPUT.PUT_LINE('FNDCPASS user/'||l_pass||' 0 Y system/test11 ALLORACLE '||l_pass||'');
END LOOP;
CLOSE c1;
END;
/
So I have started to count schemas, and I would like to stop printing new lines from above, when it reach end of count variable.
I'm not so good with plsql so please could you give some advice?
Thanks

Answer 1

Ok, i have created password function which returns string value for my password generator code now, like this :
create or replace function generic_passwd
return varchar2 is passwd VARCHAR2(17);
begin
select xmlagg(xmlelement("r", ch)).extract('//text()').getstringval() ch into passwd
from
(
select distinct first_value(ch) over (partition by lower(ch)) as ch
from (
select substr('abcd$efghijklmn#pqrstuvw#xyzABC$DEFGHIJK$LMNPQR!STUVWXYZ1!23456789',
level, 1) as ch
from dual
connect by level <= 59
order by dbms_random.value
)
where rownum <= dbms_random.value(17,17)
);
return passwd;
end;
/
Then i called this function from my code block and it is fine to me, like this:
set serveroutput on size 1000000
DECLARE
c1 SYS_REFCURSOR;
l_user varchar(10);
l_passd varchar(17);
BEGIN
OPEN c1 for
select oracle_username, generic_passwd() from FND_ORACLE_USERID where READ_ONLY_FLAG='A';
LOOP
FETCH c1
INTO l_user,l_passd;
EXIT WHEN c1%NOTFOUND ;
DBMS_OUTPUT.PUT_LINE('FNDCPASS '||l_user||'/'||l_passd||' 0 Y system/oracle11 ALLORACLE '||l_passd||'');
END LOOP;
CLOSE c1;
END;
/
Thank you for your time Jeff.

Answer 2

Found your own solution. Good on you. There are perhaps a couple inconsistencies, which may actually be intentional. But I think at least one or two are not:
The digits 3-9 can never appear in the output. You sub string
individual characters up to the 59th in the base string, however
that string is 66 characters in length. Perhaps the original base
string changed but the hard coded value did not. Intentional?
The output cannot have a repeated character, seems intentional, but
neither can it have a letter and the corresponding alternate case of
that letter. I.E cannot have both "a" and "A" in the output. Intentional?
The Expression: dbms_random.value(17,17) will always return 17, at least in a piratical since as rownum is always as integer.
So there is no reason to pay the function call overhead. Just use
the equivalent rownum <= 17;
Just a side note: This seems to be a limited use function. If that
is not the case then creating a standalone function is the way to
go. If, however this limited and it is version 12.1 or higher then
you may want to build the function directly into query using WITH
FUNCTION .... Doing so eliminate context switch between the
PL/SQL and SQL thus gaining performance.
The following addresses all these: See Demo But i reiterate. Except for #1 and #2 above what you have will work well.
with function generic_passwd --<<< define function directly as part of the query
return varchar2
is
l_generic_password varchar2(17);
begin
with base (stg) as
( select 'abcd$efghijklmn#pqrstuvw#xyzABC$DEFGHIJK$LMNPQR!STUVWXYZ1!23456789' from dual)
select xmlagg(xmlelement("r", ch)).extract('//text()').getstringval()
into l_generic_password
from
(
select distinct ch as ch --<< do not allow duplicate letters, but allow lower and upper of same letter
from (
select substr(stg,level, 1) as ch
from base
connect by level <= length(stg) --<< determine stop point from base string itself.
order by dbms_random.value
)
where rownum <= 17 --<< avoid subroutine overhead that always returns same value.
);
return l_generic_password;
end;
select username, generic_passwd() password
from fnd_oracle_userid;