CODEWITHSUNDEEP
sqloracle-database
Rajeena Safeer
Rajeena Safeer

Reputation: 91

Oracle query to find all occurrences of a charcter in a string

I have to write an Oracle query in toad to find all the occurrences of a character in a string. For example if I'm searching for R in the string SSSRNNSRSSR, it should return positions 4, 8 and 11.

I am new to Oracle and tried this.

select instr(mtr_ctrl_flags, 'R', pos + 1, 1) as pos1 
  from mer_trans_reject  
 where pos in ( select instr(mtr_ctrl_flags, 'R', 1, 1) as pos 
                  from mer_trans_reject
                       );

where mtr_ctrl_flags is the column name. I'm getting an error indicating that pos is an invalid identifier.

Upvotes: 7

Views: 22220

Answers (3)

Ben
Ben

Reputation: 52903

Extending GolezTrol's answer you can use regular expressions to significantly reduce the number of recursive queries you do:

 select instr('SSSRNNSRSSR','R', 1, level)
   from dual
connect by level <= regexp_count('SSSRNNSRSSR', 'R')

REGEXP_COUNT() returns the number of times the pattern matches, in this case the number of times R exists in SSSRNNSRSSR. This limits the level of recursion to the exact number you need to.

INSTR() simply searches for the index of R in your string. level is the depth of the recursion but in this case it's also the level th occurrence of the string as we restricted to the number of recurses required.

If the string you're wanting to pick out is more complicated you could go for regular expressions ans REGEXP_INSTR() as opposed to INSTR() but it will be slower (not by much) and it's unnecessary unless required.

Simple benchmark as requested:

The two CONNECT BY solutions would indicate that using REGEXP_COUNT is 20% quicker on a string of this size.

SQL> set timing on
SQL>
SQL> -- CONNECT BY with REGEX
SQL> declare
  2     type t__num is table of number index by binary_integer;
  3     t_num t__num;
  4  begin
  5    for i in 1 .. 100000 loop
  6       select instr('SSSRNNSRSSR','R', 1, level)
  7         bulk collect into t_num
  8         from dual
  9      connect by level <= regexp_count('SSSRNNSRSSR', 'R')
 10              ;
 11     end loop;
 12  end;
 13  /

PL/SQL procedure successfully completed.

Elapsed: 00:00:03.94
SQL>
SQL> -- CONNECT BY with filter
SQL> declare
  2     type t__num is table of number index by binary_integer;
  3     t_num t__num;
  4  begin
  5    for i in 1 .. 100000 loop
  6       select pos
  7         bulk collect into t_num
  8         from ( select substr('SSSRNNSRSSR', level, 1) as character
  9                     , level as pos
 10                  from dual t
 11               connect by level <= length('SSSRNNSRSSR') )
 12        where character = 'R'
 13              ;
 14     end loop;
 15  end;
 16  /

PL/SQL procedure successfully completed.

Elapsed: 00:00:04.80

The pipelined table function is a fair bit slower, though it would be interesting to see how it performs over large strings with lots of matches.

SQL> -- PIPELINED TABLE FUNCTION
SQL> declare
  2     type t__num is table of number index by binary_integer;
  3     t_num t__num;
  4  begin
  5    for i in 1 .. 100000 loop
  6       select *
  7         bulk collect into t_num
  8         from table(string_indexes('SSSRNNSRSSR','R'))
  9              ;
 10     end loop;
 11  end;
 12  /

PL/SQL procedure successfully completed.

Elapsed: 00:00:06.54

Upvotes: 16

Ben
Ben

Reputation: 52903

To take up a_horse_with_no_name's challenge here is another answer with a pipelined table function.

A pipelined function returns an array, which you can query normally. I would expect that over strings with large numbers of matches this will perform better than the recursive query but as with everything test yourself first.

create type num_array as table of number
/


create function string_indexes (
    PSource_String in varchar2
  , PSearch_String in varchar2
    ) return num_array pipelined is

begin

   for i in 1 .. length(PSource_String) loop

      if substr(PSource_String, i, 1) = PSearch_String then
         pipe row(i);
      end if;

   end loop;

   return;

end;
/

Then in order to access it:

select *
  from table(string_indexes('SSSRNNSRSSR','R'))

SQL Fiddle

Upvotes: 4

GolezTrol
GolezTrol

Reputation: 116160

This is a solution:

select
  pos
from
  (select
    substr('SSSRNNSRSSR', level, 1) as character,
    level as pos
  from
    dual
  connect by
    level <= length(t.text))
where
  character = 'R'

dual is a built in table that just returns a single row. Very convenient!

connect by lets you build recursive queries. This is often used to generate lists from tree-like data (parent/child relations). It allows you to more or less repeat the query in front of it. And you've got special fields, like level that allows you to check how deeply the recursion went.

In this case, I use it to split the string to characters and return a row for each character. Using level, I can repeat the query and get a character until the end of the string is reached.

Then it is just a matter of returning the pos for all rows containing the character 'R'

Upvotes: 9

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