How to reverse a string after tokenizing it in SQL - sql

Question

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

Answer 1

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
);

Answer 2

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.

Answer 3

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

Answer 4

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;

Answer 5

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 ;

Answer 6

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);

Answer 7

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.