How to split string words with regexp_substr in Oracle SQL?-CodePudding

select  
regexp_substr('a-b--->d--->e f','[^--->] ',1,1) col1
,regexp_substr('a-b--->d--->e f','[^--->] ',1,2) col2
,regexp_substr('a-b--->d--->e f','[^--->] ',1,3) col3
,regexp_substr('a-b--->d--->e f','[^--->] ',1,4) col4
 from dual

output

col1	col2	col3	col4
a	b	d	e f

Required output

col1	col2	col3	col4
a-b	d	e f

CodePudding user response：

You can use:

select regexp_substr(value,'(.*?)(- >|$)',1,1, NULL, 1) AS col1
,      regexp_substr(value,'(.*?)(- >|$)',1,2, NULL, 1) AS col2
,      regexp_substr(value,'(.*?)(- >|$)',1,3, NULL, 1) AS col3
,      regexp_substr(value,'(.*?)(- >|$)',1,4, NULL, 1) AS col4
 from  table_name

or, if you are looking for the fixed delimiter string ---> then you can use simple string functions (which are more to type but are likely going to be much faster):

SELECT CASE
       WHEN pos1 = 0 THEN value
                     ELSE SUBSTR(value, 1, pos1 - 1)
       END AS col1,
       CASE
       WHEN pos1 = 0 THEN NULL
       WHEN pos2 = 0 THEN SUBSTR(value, pos1   4)
                     ELSE SUBSTR(value, pos1   4, pos2 - pos1 - 4)
       END AS col2,
       CASE
       WHEN pos2 = 0 THEN NULL
       WHEN pos3 = 0 THEN SUBSTR(value, pos2   4)
                     ELSE SUBSTR(value, pos3   4, pos3 - pos2 - 4)
       END AS col3,
       CASE
       WHEN pos3 = 0 THEN NULL
                     ELSE SUBSTR(value, pos3   4)
       END AS col4
FROM   (
  SELECT value,
         INSTR(value, '--->', 1, 1) AS pos1,
         INSTR(value, '--->', 1, 2) AS pos2,
         INSTR(value, '--->', 1, 3) AS pos3
  FROM   table_name
)

Which, for the sample data:

CREATE TABLE table_name (value) AS
SELECT 'a-b--->d--->e f' FROM DUAL;

Both output:

COL1	COL2	COL3	COL4
a-b	d	e f	null

fiddle

CodePudding user response：

select  
 regexp_substr('a-b--->d--->e f','[a-z] ([ \-][a-z] ){0,1}',1,1) col1
,regexp_substr('a-b--->d--->e f','[a-z] ([ \-][a-z] ){0,1}',1,2) col2
,regexp_substr('a-b--->d--->e f','[a-z] ([ \-][a-z] ){0,1}',1,3) col3
,regexp_substr('a-b--->d--->e f','[a-z] ([ \-][a-z] ){0,1}',1,4) col4
 from dual
 ;

a-b d   e f (null)

CodePudding user response：

[^--->] is a character set, so there is no point in specifying twice the same character in it.

A working regex is:

(-*[^->])

That matches any character that is not an hyphen or a > with any number of possible trailing hyphens, and the pattern is repeated in a greedy way.

See a demo here.