정규 표현 식 regexpsubstr

     regexp_substr     
 
①select regexp_substr(string, '[^,]+', 1, level)
  from dual
connect by level <= regexp_count(string, '[^,]+') 
   and rowid = prior (rowid)
         and (prior dbms_random.value) is not null;
②select regexp_substr(string, '[^,]+', 1, level)
  from dual
connect by level <= length(string) - LENGTH(REPLACE(string, ',', '')) + 1
 and rowid = prior (rowid)
         and (prior dbms_random.value) is not null;
		
REGEXP_SUBSTR              ,
（  ）     ，          ，level，connect    ，
         -      =      +1

Test1:  ：         N1 N2      ，         N1       ，     N2       
N1：
    ：
SQL> select regexp_substr('i,am,a,pretty,girl', '[^,]+', 1, level) substring
  2    from dual
  3  connect by level <= regexp_count('i,am,a,pretty,girl', '[^,]+')
  4  ;
SUBSTRING
------------------------------------------------------------------------
i
am
a
pretty
girl

       ，       
create table ta as 
select 'i,am,a,pretty,girl' substring
  from dual
union all
select '1,2,3,4,5,6'
  from dual
union all
select ' , , , '
  from dual;
SQL> select count(regexp_substr(substring, '[^,]+', 1, level)) cnt
  2    from ta
  3  connect by level <= regexp_count(substring, '[^,]+')
  4  ;
       CNT
----------
       444  
	   
N2：（  N1      dbms_random.value）
       ：
SQL> select regexp_substr('i,am,a,pretty,girl', '[^,]+', 1, level)substring1
  2    from dual
  3  connect by level <= regexp_count('i,am,a,pretty,girl', '[^,]+')
  4         and rowid = prior (rowid)
  5         and (prior dbms_random.value) is not null;
SUBSTRING1
------------------------------------------------------------------------
i
am
a
pretty
girl

       
SQL> select count(regexp_substr(substring, '[^,]+', 1, level)) cnt
  2    from ta
  3  connect by level <= regexp_count(substring, '[^,]+')
  4   and rowid = prior (rowid)
  5           and (prior dbms_random.value) is not null;
       CNT
----------
        15

Test2:         

SQL> SELECT REGEXP_SUBSTR('1 meeting you,2  waiting for you.7',
  2                       '[0-9]+',
  3                       1,
  4                       level)
  5    FROM dual
  6  connect by level <=
  7             regexp_count('1 meeting you,2  waiting for you.7', '[0-9]+');
REGEXP_SUBSTR('1MEETINGYOU,2WA
--------------------------------------------------------------------------------
1
2
7

Test3:  REGEP_SUBSTR                        

SQL> SELECT REGEXP_SUBSTR('u are best best ', '([[:alnum:]]+)([[:space:]]+)\1') AS substr
  2    FROM dual;
SUBSTR
---------
best best

Test4:IP     0

select listagg(tring1) within group(order by rn) tring
  from (select rn, lpad(tring, 3, 0) tring1
          from (select level rn,
                       regexp_substr('12.34.6.110', '[^.]+', 1, level) tring
                  from dual
                connect by level <= 4));

TRING
--------------------------------------------------------------------------------
012034006110