Split comma separated values to columns in Oracle

Question

I have values being returned with 255 comma separated values. Is there an easy way to split those into columns without having 255 substr?

ROW  | VAL
--------         


        

Answer 1

If you only have one row, and time to create your

• create your own built-in cto_table function to split a string on any separator, then you can use PIVOT + LISTAGG to do it like follows:

---

select * from (
  select rownum r , collection.*  
    from TABLE(cto_table(',','1.25, 3.87, 2, 19,, 1, 9, ')) collection
)
PIVOT ( 
  LISTAGG(column_value) within group (order by 1) as val 
  for r in (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
)

---

FYI: here is how to create the cto_table function:

CREATE OR REPLACE TYPE t_my_list AS TABLE OF VARCHAR2(100);
CREATE OR REPLACE
FUNCTION cto_table(p_sep in Varchar2, p_list IN VARCHAR2)
  RETURN t_my_list
AS
  l_string VARCHAR2(32767) := p_list || p_sep;
  l_sep_index PLS_INTEGER;
  l_index PLS_INTEGER := 1;
  l_tab t_my_list     := t_my_list();
BEGIN
  LOOP
    l_sep_index := INSTR(l_string, p_sep, l_index);
    EXIT
  WHEN l_sep_index = 0;
    l_tab.EXTEND;
    l_tab(l_tab.COUNT) := TRIM(SUBSTR(l_string,l_index,l_sep_index - l_index));
    l_index            := l_sep_index + 1;
  END LOOP;
  RETURN l_tab;
END cto_table;
/

Answer 2

hierarchical query could be used. pivoting can be done with case and group by.

with value_t as
( 
  select row_t,row_number() OVER (partition by row_t order by rownum )rn,
  regexp_substr(val, '[^,]+', 1, LEVEL) val from Table1
CONNECT BY LEVEL <= regexp_count(val, '[^,]+') 
AND prior row_t = row_t 
AND prior sys_guid() is not null
  ) select row_t, max( case when rn = 1 THEN val end ) val_1,
  max( case when rn = 2 THEN val end ) val_2,
  max( case when rn = 3 THEN val end ) val_3
  from value_t
  group by row_t;

Answer 3

You can use regexp_substr():

select regexp_substr(val, '[^,]+', 1, 1) as val1, 
       regexp_substr(val, '[^,]+', 1, 2) as val2, 
       regexp_substr(val, '[^,]+', 1, 3) as val3, 
       . . .

I would suggest that you generate a column of 255 numbers in Excel (or another spreadsheet), and use the spreadsheet to generate the SQL code.

Answer 4

Beware! The regexp_substr expression of the format '[^,]+' will not return the expected value if there is a null element in the list and you want that item or one after it. Consider this example where the 4th element is NULL and I want the 5th element and thus expect the '5' to be returned:

SQL> select regexp_substr('1,2,3,,5,6', '[^,]+', 1, 5) from dual;

R
-
6

Surprise! It returns the 5th NON-NULL element, not the actual 5th element! Incorrect data returned and you may not even catch it. Try this instead:

SQL> select regexp_substr('1,2,3,,5,6', '(.*?)(,|$)', 1, 5, NULL, 1) from dual;

R
-
5

So, the above corrected REGEXP_SUBSTR says to look for the 5th occurrence of 0 or more comma-delimited characters followed by a comma or the end of the line (allows for the next separator, be it a comma or the end of the line) and when found return the 1st subgroup (the data NOT including the comma or end of the line).

The search match pattern '(.*?)(,|$)' explained:

(             = Start a group
.             = match any character
*             = 0 or more matches of the preceding character
?             = Match 0 or 1 occurrences of the preceding pattern
)             = End the 1st group
(             = Start a new group (also used for logical OR)
,             = comma
|             = OR
$             = End of the line
)             = End the 2nd group

EDIT: More info added and simplified the regex.

See this post for more info and a suggestion to encapsulate this in a function for easy reuse: REGEX to select nth value from a list, allowing for nulls It's the post where I discovered the format '[^,]+' has the problem. Unfortunately it's the regex format you will most commonly see as the answer for questions regarding how to parse a list. I shudder to think of all the incorrect data being returned by '[^,]+'!