9

I have a table like below:

-------------
ID   | NAME
-------------
1001 | A,B,C
1002 | D,E,F
1003 | C,E,G
-------------

I want these values to be displayed as:

-------------
ID   | NAME
-------------
1001 | A
1001 | B
1001 | C
1002 | D
1002 | E
1002 | F
1003 | C
1003 | E
1003 | G
-------------

I tried doing:

select split('A,B,C,D,E,F', ',') from dual; -- WILL RETURN COLLECTION

select column_value
from table (select split('A,B,C,D,E,F', ',') from dual); -- RETURN COLUMN_VALUE
VLAZ
  • 26,331
  • 9
  • 49
  • 67
vipin.huddar
  • 91
  • 1
  • 1
  • 6

7 Answers7

7

Try using below query:

 WITH T AS (SELECT 'A,B,C,D,E,F' STR  FROM DUAL)   SELECT    
 REGEXP_SUBSTR (STR, '[^,]+', 1, LEVEL) SPLIT_VALUES  FROM T 
 CONNECT BY LEVEL <= (SELECT LENGTH (REPLACE (STR, ',', NULL)) FROM T)

Below Query with ID:

WITH TAB AS 
(SELECT '1001' ID, 'A,B,C,D,E,F' STR FROM DUAL
)
SELECT    ID, 
REGEXP_SUBSTR (STR, '[^,]+', 1, LEVEL) SPLIT_VALUES  FROM TAB 
CONNECT BY LEVEL <= (SELECT LENGTH (REPLACE (STR, ',', NULL)) FROM TAB);

EDIT: Try using below query for multiple IDs and multiple separation:

WITH TAB AS 
(SELECT '1001' ID, 'A,B,C,D,E,F' STR FROM DUAL
UNION
SELECT '1002' ID, 'D,E,F' STR FROM DUAL
UNION
SELECT '1003' ID, 'C,E,G' STR FROM DUAL
)
select id, substr(STR, instr(STR, ',', 1, lvl) + 1, instr(STR, ',', 1, lvl + 1) - instr(STR, ',', 1, lvl) - 1) name 
from
( select ',' || STR || ',' as STR, id from TAB ),
( select level as lvl from dual connect by level <= 100 )
where lvl <= length(STR) - length(replace(STR, ',')) - 1

order by ID, NAME
ngrashia
  • 9,869
  • 5
  • 43
  • 58
  • I want value from 'ID' column as well. – vipin.huddar May 14 '14 at 09:03
  • @user2531405: What is the maximum number of commas expected in your name column? – ngrashia May 14 '14 at 09:08
  • Not fixed. There can be any number of comma separated values, for a particular ID. – vipin.huddar May 14 '14 at 09:09
  • @Nisha This won't work for multiple IDs in the input data (single-row subquery returns more than one row) - I think you need to add a WHERE clause for the ID in your subquery for the LENGTH. – Frank Schmitt May 14 '14 at 09:18
  • Thanks for the help, I got some direction. But value of ID is hard coded, which will not be the case with me. Ya, and as Frank said, it will not work for multiple IDs. – vipin.huddar May 14 '14 at 09:20
  • You can just replace the with clause with your table name. IDs need not be hardcoded. Check the latest edit. Its generic. – ngrashia May 14 '14 at 09:21
7

There are multiple options. See Split comma delimited strings in a table in Oracle.

Using REGEXP_SUBSTR:

SQL> WITH sample_data AS(
  2  SELECT 10001 ID, 'A,B,C' str FROM dual UNION ALL
  3  SELECT 10002 ID, 'D,E,F' str FROM dual UNION ALL
  4  SELECT 10003 ID, 'C,E,G' str FROM dual
  5  )
  6  -- end of sample_data mimicking real table
  7  SELECT distinct id, trim(regexp_substr(str, '[^,]+', 1, LEVEL)) str
  8  FROM sample_data
  9  CONNECT BY LEVEL <= regexp_count(str, ',')+1
 10  ORDER BY ID
 11  /

        ID STR
---------- -----
     10001 A
     10001 B
     10001 C
     10002 D
     10002 E
     10002 F
     10003 C
     10003 E
     10003 G

9 rows selected.

SQL>

Using XMLTABLE:

SQL> WITH sample_data AS(
  2  SELECT 10001 ID, 'A,B,C' str FROM dual UNION ALL
  3  SELECT 10002 ID, 'D,E,F' str FROM dual UNION ALL
  4  SELECT 10003 ID, 'C,E,G' str FROM dual
  5  )
  6  -- end of sample_data mimicking real table
  7  SELECT id,
  8        trim(COLUMN_VALUE) str
  9  FROM sample_data,
 10       xmltable(('"'
 11          || REPLACE(str, ',', '","')
 12          || '"'))
 13  /

        ID STR
---------- ---
     10001 A
     10001 B
     10001 C
     10002 D
     10002 E
     10002 F
     10003 C
     10003 E
     10003 G

9 rows selected.
Lalit Kumar B
  • 47,486
  • 13
  • 97
  • 124
2

i solved similar problem this way...

    select YT.ID,
           REPLACE(REGEXP_SUBSTR(','||YT.STR||',',',.*?,',1,lvl.lvl),',','') AS STR
    from YOURTABLE YT
    join (select level as lvl 
          from dual 
          connect by level <= (select max(regexp_count(STR,',')+1) from YOURTABLE)
         ) lvl on lvl.lvl <= regexp_count(YT.STR,',')+1
KARLOS
  • 21
  • 1
1

You may try something like this:

CREATE OR REPLACE TYPE "STR_TABLE"
as table of varchar2


create or replace function GetCollection( iStr varchar2, iSplit char default ',' ) return STR_TABLE as
pStr varchar2(4000) := trim(iStr);
rpart varchar(255);
pColl STR_TABLE := STR_TABLE();
begin
   while nvl(length(pStr),0) > 0 loop
         pos := inStr(pStr, iSplit );
         if pos > 0 then
            rpart := substr(pStr,1, pos-1);
            pStr  := substr(pStr,pos+1,length(pStr));
         else
            rpart := pStr;
            pStr := null;
         end if;
         if rpart is not null then
           pColl.Extend;
           pColl(pColl.Count) := rpart;
         end if;
   end loop;
   return pColl;
end;
i100
  • 4,529
  • 1
  • 22
  • 20
  • Can't I do it through a single query, as I want to avoid adding a new function or something like that on my required DB environment. – vipin.huddar May 14 '14 at 09:00
1

Do not use CONNECT BY or REGEXP which results in a Cartesian product on a complex query. Furthermore the above solutions expect you know the possible results (A,B,C,D,E,F) rather than a list of combinations

Use XMLTable:

SELECT c.fname, c.lname,
trim(COLUMN_VALUE) EMAIL_ADDRESS
 FROM 
  CONTACTS c, CONTACT_STATUS s,
  xmltable(('"'
  || REPLACE(EMAIL_ADDRESS, ';', '","')
  || '"'))
where  c.status = s.id

The COLUMN_VALUE is a pseudocolumn that belongs to xmltable. This is quick and correct and allows you to reference a column w/o know its values.

This takes the column and makes a table of values "item","item2","item3" and automatically joins to its source table (CONTACTS). This was tested on thousands of rows

Note the ';' in the xmltable is the separator in the column field.

Adam
  • 51
  • 1
  • 5
1

I tried the solution of Lalit Kumar B and it worked so far. But with more data I ran into an performance issue (> 60 Rows, >7 Level). Therefore I used a more static variation, I would like to share as alternative.

WITH T AS (
      SELECT 1001 AS ID, 'A,B,C' AS NAME FROM DUAL
UNION SELECT 1002 AS ID, 'D,E,F' AS NAME FROM DUAL
UNION SELECT 1003 AS ID, 'C,E,G' AS NAME FROM DUAL
     )   --SELECT * FROM T
SELECT ID as ID,
       distinct_column AS NAME
  FROM ( SELECT t.ID,       
                trim(regexp_substr(t.NAME, '[^,]+', 1,1)) AS c1,
                trim(regexp_substr(t.NAME, '[^,]+', 1,2)) AS c2,
                trim(regexp_substr(t.NAME, '[^,]+', 1,3)) AS c3,
                trim(regexp_substr(t.NAME, '[^,]+', 1,4)) AS c4 -- etc.
           FROM T )
UNPIVOT ( distinct_column FOR cn IN ( c1, c2, c3, c4 ) )    


    ID NAME               
------ ------
  1001 A                    
  1001 B                    
  1001 C                    
  1002 D                    
  1002 E                    
  1002 F                    
  1003 C                    
  1003 E                    
  1003 G                    

9 Zeilen gewählt
Benjamin
  • 139
  • 6
  • Fantastic. I was stuck with the performance of the `CONNECT BY` approach, and this way works perfectly as long as you know the max. number of elements in the string -- which you can check by something like: `select t.name, regexp_count(t.name, ',') from t group by t.name having regexp_count(t.name, ',') > 2;` – Andrew Spencer Apr 08 '19 at 15:32
0

this version works also with strings longer than one char:

select regexp_substr('A,B,C,Karl-Heinz,D','[^,]+', 1, level) from dual
  connect by regexp_substr('A,B,C,Karl-Heinz,D', '[^,]+', 1, level) is not null;

see How to split comma separated string and pass to IN clause of select statement

stackunderflow
  • 387
  • 4
  • 9