Oracle SQL display distinct none "standard alphanumeric caracters - sql

Question

I need to find the way to list all the characters used in the column in order to narrow down the "Approved" values within the insert template we are creating...
the idea is to allow all letters (only standard) without any dialect / country specific ones.
trying something like this... but need to have a list of the characters left over... like "$%()* etc.
SELECT * FROM mytable WHERE REGEXP_LIKE(column_1,^[a-zA-Z0-9-]+$)

Answer 1

To find the other characters, you could remove the ones you do expect and then see what is left:
SELECT REGEXP_REPLACE( column1, '[a-zA-Z0-9-]' ) AS other_characters
FROM mytable
WHERE REGEXP_REPLACE( column1, '[a-zA-Z0-9-]' ) IS NOT NULL
If you want to concatenate and remove duplicate characters:
WITH replace_expected ( str ) AS (
SELECT REGEXP_REPLACE( column1, '[a-zA-Z0-9-]' )
FROM mytable
WHERE REGEXP_REPLACE( column1, '[a-zA-Z0-9-]' ) IS NOT NULL
),
split_strings ( str, pos, ch ) AS (
SELECT str, 1, SUBSTR(str, 1, 1)
FROM replace_expected
UNION ALL
SELECT str, pos + 1, SUBSTR(str, pos + 1, 1)
FROM split_strings
WHERE pos < LENGTH(str)
)
SELECT LISTAGG(DISTINCT ch) WITHIN GROUP (ORDER BY ch) AS other_characters
FROM split_strings;
fiddle

Answer 2

Two steps:
Extract the special characters from all strings in the column and concatenate them into one long string (with possibly many characters appearing multifold in the string).
Use a recursive query to loop through the characters and return a string with distinct characters.
The query:
with one_row (str) as
(
select listagg(regexp_replace(column_1, '[a-zA-Z0-9\-]'))
from mytable
where regexp_like(column_1, '[^a-zA-Z0-9\-]')
)
select listagg(distinct substr(str, level, 1)) as c
from one_row
connect by level <= length(str);

Answer 3

create table dummy_data as
select rownum as id, dbms_random.string(opt, len) str
from (
select case round(dbms_random.value(0, 3))
when 0 then 'a'
when 1 then 'x'
else 'p'
end opt,
round(dbms_random.value(5, 60)) len
from dual connect by level <= 30 );
SELECT * FROM DUMMY_DATA;
ID STR
1 UMUUJ0R5VM1T3X10TDCNIWC3MQ5ELOB041YMNEJSLT
2 _t8 }LeZhjiMB"8/a'/~a
3 BLSE6XX6SL3M7W0DG3HH28SCHPSAT11ZH2E5DOSKEV3KW9
4 1]Mh58(l<Wa}{
5 :_QiWUkwp}V$}O
6 NC911A4SRN35CNXT2EU5H2GZ67IQQLKH
7 e"8,z$=Yvy5egvEH2KUkNoVjkitd9IMm0ZktsB i(bk4uU]c3;E
8 MgbpIsLZpWEcAghOUKOISA
9 7H02ASKO3CZRN4D5FUNPEU6YUZD
10 KbJ+QrI\l.th%>^f!Io%wshsVA%
11 PO9A47VU7AXI17XYD5VMSWW8E
12 1ILWL4V
13 FgubwibYBytNvmJHxUfG
14 ?[ngH?0!k.onN>mF(nrkO
15 86G0HP3
16 WXDBV3OBMVSDKQ59YT73G0II3U94
17 GP375CFIQPPN6216I5A7L54O
18 i\L<K,"d'ye 6s~_MB0O1 aC$q;T"EaqpZ^s\gIiYu&:%OnhVj]<a]CmOgqM
19 WxUEtr\II(97i7PQ-Z]yqd#&`#CQB0M"c0;{.by9qo#HT
20 IF5OP7KS9AXW91
21 HNcKwxXozXjTVwKeFZDLdmNOzFKKKq
22 4D8CINXIVT244RDDRZ5TSDQ4CRF4
23 3)oxevW-(~=+#cP[^g)##|1.TL-_N9O-Zdgj"cwJC'*NR; FtK)K
24 AndzeLIEPklDuTWWEBrKrdNKdXwMGeLauJkRzKpKHEGAsxlEXliwBTHdK
25 dlEX1tGFuU5\5+{5`R
26 /W0.{B&)ax&lWEE#OSw
27 CBKOVLKDFKC3EVR
28 :V#Lc.Z"8[O-)cAWUpMjc?j\Kj?xV#%`Yp [VkEV1
29 P9P047
30 W)S<fB`F;N_brMP
with
h (ok_chars) as (
select '0123456789' || 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' ||
'abcdefghijklmnopqrstuvwxyz'
from dual
)
select c, count(*) as cnt
from dummy_data cross join lateral (
select substr(str, level, 1) as c
from h
where instr(ok_chars, substr(str, level, 1)) = 0
connect by level <= length(str)
)
group by c;
C CNT
} 4
- 5
' 3
< 4
, 2
= 2
_ 5
( 5
^ 3
[ 4
" 7
/ 3
: 3
. 6
% 5
! 2
? 4
` 4
8
; 5
+ 3
> 2
) 6
| 1
* 1
~ 3
\ 6
& 4
# 5
] 5
{ 4
# 4
$ 3