Oracle SQL Developer Query on recommended password setup - sql

Question

Setting up a random password for user using
select
dbms_random.string('L',2) || dbms_random.string('X',6) || '1!' as deflvrpwd,
'${access_request_cri_acc_cas9}' as ACNTDN
from dual
New requirement
New Hire Details:
Name :John Doe
Region: America
WDID : 876214
WDID Reverse and split
Region in the middle with the letter A replaced with # symbol
Should read if we follow your formula.
= 412#meric#s678
Please suggest attribute are same as mentioned.
Thank You

Answer 1

Here's one option; read comments within code.
SQL> WITH
2 -- sample data
3 test (name, region, wdid)
4 AS
5 (SELECT 'John Doe', 'America', '876214' FROM DUAL),
6 temp
7 AS
8 -- reverse WDID; don't use undocumented REVERSE function
9 -- replace "A" (or "a") with "#" in REGION
10 ( SELECT name,
11 REPLACE (REPLACE (region, 'A', '#'), 'a', '#') new_region,
12 LISTAGG (letter, '') WITHIN GROUP (ORDER BY lvl DESC) new_wdid
13 FROM ( SELECT SUBSTR (wdid, LEVEL, 1) letter,
14 LEVEL lvl,
15 name,
16 region
17 FROM test
18 CONNECT BY LEVEL <= LENGTH (wdid))
19 GROUP BY name, region)
20 -- finally
21 SELECT SUBSTR (new_wdid, 1, 3) || new_region || SUBSTR (new_wdid, 4) AS result
22 FROM temp;
RESULT
--------------------------------------------------------------------------------
412#meric#678
SQL>
I don't know where s in your result comes from (this: 412#meric#s678).

Answer 2

There's a small cost in context switching between SQL and PL/SQL, but this doesn't sound like a high-volume or performance-critical thing, so you might find it cleaner to put the logic in a function:
create or replace function get_password (p_wdid varchar2, p_region varchar2)
return varchar2 as
l_split pls_integer;
l_password varchar2(30);
begin
-- split WDID halfway, but allow for odd lengths
l_split := floor(length(p_wdid)/2);
-- iterate over the WDID in reverse
for i in reverse 1..length(p_wdid) LOOP
-- when we reach the split point, append the modified region
if i = l_split then
l_password := l_password || translate(p_region, 'Aax', '##x');
end if;
-- append each WDID character, in reverse order
l_password := l_password || substr(p_wdid, i, 1);
end loop;
return l_password;
end get_password;
/
The WDID is reversed in a loop, and the modified region is included at the midway point, based on the length of the WDID value.
You can then do:
select get_password('876214', 'America') from dual;
GET_PASSWORD('876214','AMERICA')
--------------------------------
412#meric#678
This also doesn't have the unexplained 's' from the example in your question.
If you can't create a function but are on a recent version of Oracle then you can define an ad hoc function in a CTE:
with
function invert (p_input varchar2) return varchar2 as
l_output varchar2(30);
begin
for i in reverse 1..length(p_input) LOOP
l_output := l_output || substr(p_input, i, 1);
end loop;
return l_output;
end invert;
t (wdid, region) as (
select invert('876214'), translate('America', 'Aax', '##x')
from dual
)
select substr(wdid, 1, floor(length(wdid)/2))
|| region
|| substr(wdid, floor(length(wdid)/2) + 1)
from t;
which gets the same result. (I've called the function invert to avoid confusion with the undocumented reverse function.)
db<>fiddle showing both.