Oracle SQL Developer Query on recommended password setup - sql

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

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).

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.

Related

PL/SQL LOOP - Return a row with mixed capital letters

I know this question probably has an easy answer, but I can't get my head around it.
I'm trying to, inside a loop, return a string (in the SQL output) with mixed capital and non-capital letters.
Example: If a name in the row is John Doe, the output will print JoHn DoE, or MiXeD CaPiTaL.
This is my code (which I know is poor written but I need to use the cursor!):
declare
aa_ VARCHAR2(2000);
bb_ NUMBER:=0;
cc_ NUMBER:=0;
CURSOR cur_ IS
SELECT first_name namn, last_name efternamn FROM person_info
;
begin
FOR rec_ IN cur_ LOOP
dbms_output.put_line(rec_.namn);
FOR bb_ IN 1.. LENGTH(rec_.namn) LOOP
dbms_output.put(UPPER(SUBSTR(rec_.namn,bb_,1)));
cc_ := MOD(bb_,2);
IF cc_ = 0 THEN
dbms_output.put(UPPER(SUBSTR(rec_.namn,cc_,1)));
ELSE
dbms_output.put(LOWER(SUBSTR(rec_.namn,2)));
END IF;
end loop;
dbms_output.new_line;
end loop;
end;
Again, I know the code is really bad but yeah, trying to learn!
Thanks in advance :)
You may use plain SQL for this purpose, without any loop:
Split input text by pairs separated with some special character (that doesn't appear in the text).
Use initcap SQL function to turn each first letter to upper case.
Remove the special separator.
with a as (
select 'John Doe' as a
from dual
union all
select 'mixed capital and non-capital letters'
from dual
)
select
replace(
initcap(
/*Convert case*/
regexp_replace(a, '([a-zA-Z]{2})',
/*Add ASCII nul after each two letters*/
'\1' || chr(0)
)
),
/*Remove ASCII nul to revert changes*/
chr(0)
) as mixed_case
from a
| MIXED_CASE |
| :------------------------------------ |
| JoHn DoE |
| MiXeD CaPiTaL AnD NoN-CaPiTaL LeTtErS |
db<>fiddle here
I'd put the text transformation into a function, rather than including all the logic in the body of the loop.
declare
cursor c_people is
select 'John' as first_name, 'Doe' as last_name from dual union all
select 'Mixed', 'Capitals'
from dual;
function mixCaps(inText varchar2) return varchar2
is
letter varchar2(1);
outText varchar2(4000);
begin
for i in 1..length(inText) loop
letter := substr(inText,i,1);
outText := outText ||
case mod(i,2)
when 0 then lower(letter)
else upper(letter)
end;
end loop;
return outText;
end mixCaps;
begin
for person in c_people loop
dbms_output.put_line(mixCaps(person.first_name|| ' ' || person.last_name));
end loop;
end;
If performance was critical and you had large numbers of values, you might consider inlining the function using pragma inline (but then you wouldn't be using dbms_output anyway).
For learning purpose you can use code below (it is not efficient it is for learning of oracle features)
Steps :
split word on letters using connect by level
get Nth (level) occurence of one letter ('.?') from word using reg exp
convert to upper case every 2nd letter
concatenate back using list agg and sorting by letter number
used here function in with so you can apply it to any sql table
with
function mixed(iv_name varchar2) return varchar2 as
l_result varchar2(4000);
begin
with src_letters as
(select REGEXP_SUBSTR(iv_name, '.?', level) as letter
,level lvl
from dual
connect by level <= length(iv_name)),
mixed_letters as
(select case
when mod(lvl, 2) = 0 then
letter
else
upper(letter)
end as letter
,lvl
from src_letters
order by lvl)
select listagg(letter) within group(order by lvl)
into l_result
from mixed_letters;
return l_result;
end;
select mixed('text') from dual

How to skip comma from csv using double quotes

I am getting a string in the below format after reading data from a csv file
v_lastline = '29218368,8062115," Benedict Canyon Equities, Inc",CLS,,FAX';
I just want to convert it into an array while will contain 6 values, the comma before the , Inc needs to be escaped.
Can any one please suggest whats the best way to do it in PL/SQL?
This is similar to this question, but you have empty elements in your list; and a simple translation of one of the patterns I tried there skips those:
var v_lastline varchar2(50);
exec :v_lastline := '29218368,8062115," Benedict Canyon Equities, Inc",CLS,,FAX';
select level as lvl,
regexp_substr(:v_lastline, '("[^"]*"|[^,]+)', 1, level) as element
from dual
connect by level <= regexp_count(:v_lastline, '("[^"]*"|[^,]+)');
LVL ELEMENT
---------- ----------------------------------------
1 29218368
2 8062115
3 " Benedict Canyon Equities, Inc"
4 CLS
5 FAX
If you can identify a special character that will never appear in the data then you can work around that by putting that into the empty elements by changing every comma to comma+character, and then removing it after the split:
select level as lvl,
replace(regexp_substr(replace(:v_lastline, ',', ',§'),
'(§"[^"]*"|[^,]+)', 1, level), '§', null) as element
from dual
connect by regexp_substr(replace(:v_lastline, ',', ',§'),
'(§"[^"]*"|[^,]+)', 1, level) is not null;
LVL ELEMENT
---------- ----------------------------------------
1 29218368
2 8062115
3 " Benedict Canyon Equities, Inc"
4 CLS
5
6 FAX
It's an extension of a common method to split delimited strings, which is explained in detail here.
replace(:v_lastline, ',', ',§') changes ...,CLS,,FAX to ...,§CLS,§,§FAX, where § is a character you'll never see.
regexp_substr(..., '(§"[^"]*"|[^,]+)', 1, level) tokenises the updated value with a regex that looks for any double-quote-enclosed value (now preceded by the special character too) or a non-comma; the order of the evaluation means commas inside the quoted part are ignored.
the level is part of the hierarchical query syntax, where:
connect by regexp_substr(<same value and pattern>) is not null just figured out how many tokens there are.
and finally replace(regexp_substr(...), , '§', null) removes the special character used in the first step.
You can then remove the double-quotes too with a further level of replace(), and trim whitespace, if you want/need to.
You have't said quite what you mean by an array, but you can run that query in PL/SQL and bulk-collect into a collection if that's what you intend to work with. For example, using the built-in ODCIVARCHAR2LIST collection type:
set serveroutput on
declare
v_lastline varchar2(50);
v_array sys.odcivarchar2list;
begin
v_lastline := '29218368,8062115," Benedict Canyon Equities, Inc",CLS,,FAX';
select trim(replace(replace(
regexp_substr(replace(:v_lastline, ',', ',§'),
'(§"[^"]*"|[^,]+)', 1, level), '§', null), '"', null))
bulk collect into v_array
from dual
connect by regexp_substr(replace(:v_lastline, ',', ',§'),
'(§"[^"]*"|[^,]+)', 1, level) is not null;
dbms_output.put_line('Number of elements: ' || v_array.count);
for i in 1..v_array.count loop
dbms_output.put_line('Index ' || i || ' has: ' || v_array(i));
end loop;
end;
/
Number of elements: 6
Index 1 has: 29218368
Index 2 has: 8062115
Index 3 has: Benedict Canyon Equities, Inc
Index 4 has: CLS
Index 5 has:
Index 6 has: FAX
With multiple empty elements this also (now) works:
exec :v_lastline := '29218368,8062115," Benedict Canyon Equities, Inc",,,,,,,CLS,,,,,FAX,,,,,,,,,,,,,,,,,,INVOICE';
select level as lvl,
replace(regexp_substr(replace(:v_lastline, ',', ',§'),
'(§"[^"]*"|[^,]+)', 1, level), '§', null) as element
from dual
connect by regexp_substr(replace(:v_lastline, ',', ',§'),
'(§"[^"]*"|[^,]+)', 1, level) is not null;
LVL ELEMENT
---------- ----------------------------------------
1 29218368
2 8062115
3 " Benedict Canyon Equities, Inc"
4
...
9
10 CLS
11
...
14
15 FAX
16
...
32
33 INVOICE
If the structure of your CSV if fixed, you can try with something like this:
with text(text) as ( select '29218368,8062115," Benedict Canyon Equities, Inc",CLS,,FAX' from dual)
select level,
trim(',' from
case
when level in (1,2) then
regexp_substr(text, '(.*??)\,', 1, level)
when level = 3 then
regexp_substr(text, '"(.*??)"', 1, 1)
when level in (4,5) then
regexp_substr(text, '(.*??)\,', instr(text, '"', 1, 2), level -2)
when level = 6 then
regexp_substr(text, '\,([^\,]*)', instr(text, '"', 1, 2), 3)
end
)
from text
connect by level <= 6
This makes strong assumptions on the structure of CSV, by treating each part in a different way, but it seems to me difficult to find a really generic solution to the problem.
Here is a solution without regular expressions, first of all create two helper functions
/* CAR select car('hello,world,bla') from dual --> hello */
create or replace function car(PI_STR in varchar2,
PI_SEPARATOR in varchar2 default ',')
return varchar2 is l_pos number;
begin
l_pos := instr(PI_STR, PI_SEPARATOR);
if l_pos > 0 then
return substr(PI_STR, 1, l_pos - 1);
end if;
return PI_STR;
end;
/* CDR select cdr('hello,world,bla') from dual --> world,bla */
create or replace function cdr(PI_STR in varchar2,
PI_SEPARATOR in varchar2 default ',')
return varchar2 is l_pos number;
begin
l_pos := instr(PI_STR, PI_SEPARATOR);
if l_pos > 0 then
return substr(PI_STR, l_pos + length(PI_SEPARATOR));
end if;
return '';
end;
now: extract by ',' and for each result concat with next entry if escape character is found upto next escape character:
create or replace type csv_col is table of varchar2(4000);
create or replace function get_columns(PI_STR in varchar2,
PI_SEPARATOR in varchar2,
PI_ESC_CHAR in varchar2)
return csv_col pipelined is l_car varchar2(4000);
l_cdr varchar2(4000);
l_car_esc varchar2(4000);
begin
l_car := car(PI_STR, PI_SEPARATOR);
l_cdr := cdr(PI_STR, PI_SEPARATOR);
-- check for escape char
l_car_esc := cdr(l_car, PI_ESC_CHAR);
if l_car_esc is not null then
l_car := l_car_esc || PI_SEPARATOR || car(l_cdr, PI_ESC_CHAR);
l_cdr := cdr(cdr(l_cdr, PI_ESC_CHAR), PI_SEPARATOR);
end if;
loop
if l_car is null and l_cdr is null then
exit;
end if;
pipe row(l_car);
l_car := car(l_cdr, PI_SEPARATOR);
l_cdr := cdr(l_cdr, PI_SEPARATOR);
l_car_esc := cdr(l_car, PI_ESC_CHAR);
if l_car_esc is not null then
l_car := l_car_esc || PI_SEPARATOR || car(l_cdr, PI_ESC_CHAR);
l_cdr := cdr(cdr(l_cdr, PI_ESC_CHAR), PI_SEPARATOR);
dbms_output.put_line(l_car);
dbms_output.put_line(l_cdr);
end if;
end loop;
end;
call it like this:
select *
from table(get_columns('29218368,8062115," Benedict Canyon Equities, Inc",CLS,,FAX',
',',
'"'));
--> result
29218368
8062115
Benedict Canyon Equities, Inc
CLS
FAX

separate values from a string using SQL

I have this requirement where we need to separate values from a string the format of the is like
{Feature1=Value1} | {Feature2=Value2} | .. | {FeatureN=ValueN}
{12345=Gold}|{12346=Silver}
so need to separate features and values from the given srting..
To separate PIPE separated values i am using..
select *
from xmltable('r/c' passing xmltype('<r><c>' || replace('{12345=Gold}|{12346=Silver}','|','</c><c>') || '</c></r>')
columns new_str varchar2(30) path '.');
NEW_STR
------------------------------
{12345=Gold}
{12346=Silver}
I am writing a PLSQL block which iterate through each pipe separate values using above query.
I can store these values in PLSQL variable.
Now another task here is to get features and values from two above strings for this i write below SQL
select substr ('{12345=Gold}',2, instr('{12345=Gold}', '=')-2) features from dual;
FEATURES
----------------------
12345
SELECT SUBSTR('{12345=Gold}', instr('{12345=Gold}', '=')+1, LENGTH(substr ('{12345=Gold}', instr('{12345=Gold}', '=')+1, INSTR('{12345=Gold}', '}', 2)))-1) value FROM DUAL;
VALUE
--------------
Gold
So here i am able to get the features and values from a string......
I am looking for another or alternate SQL for my SQL's specially for the last one i find it complex function use so if you have any better idea for the above scenario then please Post !
Please ask if the scenario is not clear
MY DB is --
Oracle Database 12c Enterprise Edition Release 12.1.0.2.0 - 64bit Production
with line as (
select '{12345=Gold}|{12346=Silver}|{12399=Copper}' str from dual)
select substr (parse, 2, instr(parse,'=')-2) as feature
,substr (parse, instr(parse,'=')+1, length(parse)-instr(parse,'=')-1 ) as value
from
(select distinct regexp_substr(str, '[^|]+', 1, level) as parse
from line
connect by regexp_substr(str, '[^|]+', 1, level) is not null)
Here is a way using types and a handy built-in package called apex_util:
create type keyval_t is object (key varchar2(10), value varchar2(100));
/
create type keyval_tab_t is table of keyval_t;
/
create or replace package test_pkg is
function keyval_tab (p_keyval_string varchar2) return keyval_tab_t;
end;
/
create or replace package body test_pkg is
function keyval_tab (p_keyval_string varchar2) return keyval_tab_t
is
l_tab apex_application_global.vc_arr2;
l_tab2 apex_application_global.vc_arr2;
l_keyval_tab keyval_tab_t := keyval_tab_t();
l_str long;
begin
-- Split string at pipe delimiters
l_tab := apex_util.string_to_table (p_keyval_string, '|');
-- For each {key=value} pair
for i in 1..l_tab.count loop
l_str := l_tab(i);
-- Remove the {}
l_str := ltrim (l_str, '{ ');
l_str := rtrim (l_str, '} ');
-- Split string into key and value
l_tab2 := apex_util.string_to_table (l_str, '=');
if l_tab2.count = 2 then
l_keyval_tab.extend;
l_keyval_tab(i) := keyval_t (l_tab2(1), l_tab2(2));
else
-- ?? invalid string
null;
end if;
end loop;
return l_keyval_tab;
end;
end;
/
Now you can query:
SQL> select value from table(test_pkg.keyval_tab ('{12345=Gold}|{12346=Silver}'))
2* where key='12346';
VALUE
--------------------------------------------------------------------------------
Silver
SQL> select key from table(test_pkg.keyval_tab ('{12345=Gold}|{12346=Silver}'))
2 where value='Gold';
KEY
----------
12345
Use this query to get your expected output. Sorry if there are too many replace functions. But this is quite easier.
select EXTRACTVALUE (COLUMN_VALUE, '/row/Code') code,
EXTRACTVALUE (COLUMN_VALUE, '/row/Value') Value
from TABLE(XMLSEQUENCE(EXTRACT(XMLTYPE('<rowset><row>'||replace(replace(replace(replace('{12345=Gold}|{12346=Silver}','}|{','</Value></row><row><Code>'),'{','<Code>'),'=','</Code><Value>'),'}','</Value>')||'</row></rowset>'),'/rowset/row')));
If you already have separated columns into rows the work to do is fairly simple with REGEXP_REPLACE function.
Considering that your cols now is:
NEW_STR
--------------
{12345=Gold}
{12346=Silver}
You can do this sql to transform it into two different columns:
select regexp_replace( col, '\{(\d+)=\w+\}', '\1' ) as feature,
regexp_replace( col, '\{\d+=(\w+)\}', '\1' ) as value
from testTable
Transform that into a view and then just select with columns as you like:
create or replace view testView as
select regexp_replace( col, '\{(\d+)=\w+\}', '\1' ) as feature,
regexp_replace( col, '\{\d+=(\w+)\}', '\1' ) as value
from testTable
Then just do:
select * from testView where feature = '12345'
Or
Select * from testView where value = 'Gold'
If you like to transform the feature value as a number just use the to_number function on that column as:
to_number(regexp_replace( col, '\{(\d+)=\w+\}', '\1' ))
Remember that in order to do this you must be absolute sure that it is only numbers on that, otherwise you will have conversion errors
You can also use a pivot table and REGEXP_SUBSTR
with MyStrings as
(select '{Feature1=Value1}|{Feature2=Value2}|{FeatureN=ValueN}' Str from dual
union all
select '{12345=Gold}|{12346=Silver}' from dual
)
,pivot as (
Select Rownum Pnum
From dual
Connect By Rownum <= 100
)
SELECT rownum rn
,REGEXP_SUBSTR (ms.Str,'[^|]+',1,pv.pnum) TXT
FROM MyStrings ms
,pivot pv
where REGEXP_SUBSTR (ms.Str,'[^|]+',1,pv.pnum) is not null

How do I expand a string with wildcards in PL/SQL using string functions

I have a column, which stores a 4 character long string with 4 or less wild characters (for eg. ????, ??01', 0??1 etc). For each such string like 0??1 I have to insert into another table values 0001 to 0991; for the string ??01, values will be be 0001 to 9901; for string ???? values will be 0000 to 9999 and so on.
How could I accomplish this using PL/SQL and string functions?
EDIT
The current code is:
declare
v_rule varchar2(50) := '????52132';
v_cc varchar2(50);
v_nat varchar2(50);
v_wild number;
n number;
begin
v_cc := substr(v_rule,1,4);
v_nat := substr(v_rule,5);
dbms_output.put_line (v_cc || ' '|| v_nat);
if instr(v_cc, '????') <> 0 then
v_wild := 4;
end if;
n := power(10,v_wild);
for i in 0 .. n - 1 loop
dbms_output.put_line(substr(lpad(to_char(i),v_wild,'0' ),0,4));
end loop;
end;
/
Would something like the following help?
BEGIN
FOR source_row IN (SELECT rule FROM some_table)
LOOP
INSERT INTO some_other_table (rule_match)
WITH numbers AS (SELECT LPAD(LEVEL - 1, 4, '0') AS num FROM DUAL CONNECT BY LEVEL <= 10000)
SELECT num FROM numbers WHERE num LIKE REPLACE(source_row.rule, '?', '_');
END LOOP;
END;
/
This assumes you have a table called some_table with a column rule, which contains text such as ??01, 0??1 and ????. It inserts into some_other_table all numbers from 0000 to 9999 that match these wild-carded patterns.
The subquery
SELECT LPAD(LEVEL - 1, 4, '0') AS num FROM DUAL CONNECT BY LEVEL <= 10000)
generates all numbers in the range 0000 to 9999. We then filter out from this list of numbers any that match this pattern, using LIKE. Note that _ is the single-character wildcard when using LIKE, not ?.
I set this up with the following data:
CREATE TABLE some_table (rule VARCHAR2(4));
INSERT INTO some_table (rule) VALUES ('??01');
INSERT INTO some_table (rule) VALUES ('0??1');
INSERT INTO some_table (rule) VALUES ('????');
COMMIT;
CREATE TABLE some_other_table (rule_match VARCHAR2(4));
After running the above PL/SQL block, the table some_other_table had 10200 rows in it, all the numbers that matched all three of the patterns given.
Replace * to %, ? to _ and use LIKE clause with resulting values.
To expand on #Oleg Dok's answer, which uses the little known fact that an underscore means the same as % but only for a single character and using PL\SQL I think the following is the simplest way to do it. A good description of how to use connect by is here.
declare
cursor c_min_max( Crule varchar2 ) is
select to_number(min(numb)) as min_n, to_number(max(numb)) as max_n
from ( select '0000' as numb
from dual
union
select lpad(level, 4, '0') as numb
from dual
connect by level <= 9999 )
where to_char(numb) like replace(Crule, '?', '_');
t_mm c_min_max%rowtype;
l_rule varchar2(4) := '?091';
begin
open c_min_max(l_rule);
fetch c_min_max
into t_mm;
close c_min_max;
for i in t_mm.min_n .. t_mm.max_n loop
dbms_output.put_line(lpad(i, 4, '0'));
end loop;
end;
/

How to reverse a string after tokenizing it in SQL

I need to tokenize a string and reverse it in SQL. For example if the string is, 'L3:L2:L1:L0', i need to reverse it as 'L0:L1:L2:L3'. The tokenizing could be done using a delimiter ':' and then reverse it. Please suggest a Function in SQL for the same.
Thanks in advance,
Geetha
If possible, the best solution would be to change your data so that each value is stored in a different row.
If that doesn't work, you can create a PL/SQL function.
If you want a purely SQL solution, typically you'll have to split each value into multiple rows (cross join with an object table, or connect by level <= max number of items), and then re-aggregate the data using one of a dozen different methods (listagg, collect, stragg, xml, sys_connect_by_path, etc.)
Another SQL-only way is to use regular expressions. This is probably the fastest, but it only works with up to 9 items because Oracle only supports 9 back references:
--Get everything except the extra ':' at the end.
select substr(string, 1, length(string) - 1) string from
(
select regexp_replace(
--Add a delimter to the end so all items are the same
'L3:L2:L1:L0'||':'
--Non-greedy search for anything up to a : (I bet there's a better way to do this)
,'(.*?:)?(.*?:)?(.*?:)?(.*?:)?(.*?:)?(.*?:)?(.*?:)?(.*?:)?(.*?:)?(.*?:)?'
--Reverse the back-references
,'\9\8\7\6\5\4\3\2\1') string
from dual
);
Something like :
SELECT
REGEXP_REPLACE('L1:L2:L3',
'([[:alnum:]]{1,}):([[:alnum:]]{1,}):([[:alnum:]]{1,})',
'\3 \2 \1') "REGEXP_REPLACE"
from dual
But you might need to detail what constitutes a token.
Here is a solution using a PL/SQL pipelined function to split the elements:
create type t_str_array as table of varchar2(4000);
create or replace function split_str (p_str in varchar2,
p_separator in varchar2 := ':') return t_str_array pipelined
as
l_str varchar2(32000) := p_str || p_separator;
l_pos pls_integer;
begin
loop
l_pos := instr(l_str, p_separator);
exit when (nvl(l_pos,0) = 0);
pipe row (ltrim(rtrim(substr(l_str,1,l_pos-1))));
l_str := substr(l_str, l_pos+1);
end loop;
return;
end split_str;
Then you would use normal SQL to order the elements:
select * from table(split_str('L3:L2:L1:L0')) order by column_value
declare
s varchar2(1000) := 'L 1 0:L9:L8:L7:L6:L5:L4:L3:L2:L1:L0';
j number := length(s);
begin
for i in reverse 1..length(s) loop
if substr(s, i, 1) = ':' then
dbms_output.put(substr(s, i + 1, j - i) || ':');
j := i - 1;
end if;
end loop;
dbms_output.put_line(substr(s, 1, j));
end;
Convert elements in a CSV string into records, suppressing all NULLs:
SELECT REGEXP_SUBSTR( :csv,'[^,]+', 1, LEVEL ) AS element
FROM dual
CONNECT BY REGEXP_SUBSTR( :csv, '[^,]+', 1, LEVEL ) IS NOT NULL ;
Convert elements in a CSV string into records, preserving NULLs (but not order):
SELECT REGEXP_SUBSTR( :csv,'[^,]+', 1, LEVEL ) AS element
FROM dual
CONNECT BY LEVEL <= LENGTH( :csv ) - LENGTH( REPLACE( :CSV, ',' ) ) + 1 ;
Improving upon Kevan's answer, here is what I tried:
select listagg(TOKEN, ':') WITHIN GROUP (ORDER BY TOKEN_LEVEL DESC)
from
(SELECT REGEXP_SUBSTR( myStr,'[^:]+', 1, LEVEL ) AS TOKEN, LEVEL TOKEN_LEVEL
FROM dual
CONNECT BY REGEXP_SUBSTR( myStr, '[^:]+', 1, LEVEL ) IS NOT NULL);
Since you use Oracle it would be easy to generate a java stored procedure passing the string and then
split sting into array
loop array backwards and concate the resulting string
return the resulting string
this will be a small java code and not slower then pl/sql. but if you want to use pl/sql you can possibly also use DBMS_UTILITY.table_to_comma/.comma_to_table. But as the function name let assume -> you have to use "," as token.