How to convert comma separated values to rows in oracle?

Question

Here is the DDL --

create table tbl1 (
   id number,
   value varchar2(50)
);

insert into tbl1 values (1, 'AA, UT, BT, SK, SX');
insert into tbl1 values (2, 'AA, UT, SX');
insert into tbl1 values (3, 'UT, SK, SX, ZF');

Notice, here value is comma separated string.

But, we need result like following-

ID VALUE
-------------
1  AA
1  UT
1  BT
1  SK
1  SX
2  AA
2  UT
2  SX
3  UT
3  SK
3  SX
3  ZF

How do we write SQL for this?

Answer 1

I agree that this is a really bad design. Try this if you can't change that design:

select distinct id, trim(regexp_substr(value,'[^,]+', 1, level) ) value, level
  from tbl1
   connect by regexp_substr(value, '[^,]+', 1, level) is not null
   order by id, level;

OUPUT

id value level
1   AA  1
1   UT  2
1   BT  3
1   SK  4
1   SX  5
2   AA  1
2   UT  2
2   SX  3
3   UT  1
3   SK  2
3   SX  3
3   ZF  4

Credits to this

To remove duplicates in a more elegant and efficient way (credits to @mathguy)

select id, trim(regexp_substr(value,'[^,]+', 1, level) ) value, level
  from tbl1
   connect by regexp_substr(value, '[^,]+', 1, level) is not null
      and PRIOR id =  id 
      and PRIOR SYS_GUID() is not null  
   order by id, level;

If you want an "ANSIer" approach go with a CTE:

with t (id,res,val,lev) as (
           select id, trim(regexp_substr(value,'[^,]+', 1, 1 )) res, value as val, 1 as lev
             from tbl1
            where regexp_substr(value, '[^,]+', 1, 1) is not null
            union all           
            select id, trim(regexp_substr(val,'[^,]+', 1, lev+1) ) res, val, lev+1 as lev
              from t
              where regexp_substr(val, '[^,]+', 1, lev+1) is not null
              )
select id, res,lev
  from t
order by id, lev;

OUTPUT

id  val lev
1   AA  1
1   UT  2
1   BT  3
1   SK  4
1   SX  5
2   AA  1
2   UT  2
2   SX  3
3   UT  1
3   SK  2
3   SX  3
3   ZF  4

Another recursive approach by MT0 but without regex:

WITH t ( id, value, start_pos, end_pos ) AS
  ( SELECT id, value, 1, INSTR( value, ',' ) FROM tbl1
  UNION ALL
  SELECT id,
    value,
    end_pos                    + 1,
    INSTR( value, ',', end_pos + 1 )
  FROM t
  WHERE end_pos > 0
  )
SELECT id,
  SUBSTR( value, start_pos, DECODE( end_pos, 0, LENGTH( value ) + 1, end_pos ) - start_pos ) AS value
FROM t
ORDER BY id,
  start_pos;

I've tried 3 approaches with a 30000 rows dataset and 118104 rows returned and got the following average results:

• My recursive approach: 5 seconds
• MT0 approach: 4 seconds
• Mathguy approach: 16 seconds
• MT0 recursive approach no-regex: 3.45 seconds

@Mathguy has also tested with a bigger dataset:

In all cases the recursive query (I only tested the one with regular substr and instr) does better, by a factor of 2 to 5. Here are the combinations of # of strings / tokens per string and CTAS execution times for hierarchical vs. recursive, hierarchical first. All times in seconds

• 30,000 x 4: 5 / 1.
• 30,000 x 10: 15 / 3.
• 30,000 x 25: 56 / 37.
• 5,000 x 50: 33 / 14.
• 5,000 x 100: 160 / 81.
• 10,000 x 200: 1,924 / 772

Answer 2

This will get the values without requiring you to remove duplicates or having to use a hack of including SYS_GUID() or DBMS_RANDOM.VALUE() in the CONNECT BY:

SELECT t.id,
       v.COLUMN_VALUE AS value
FROM   TBL1 t,
       TABLE(
         CAST(
           MULTISET(
             SELECT TRIM( REGEXP_SUBSTR( t.value, '[^,]+', 1, LEVEL ) )
             FROM   DUAL
             CONNECT BY LEVEL <= REGEXP_COUNT( t.value, '[^,]+' )
           )
           AS SYS.ODCIVARCHAR2LIST
         )
       ) v

Update:

Returning the index of the element in the list:

Option 1 - Return a UDT:

CREATE TYPE string_pair IS OBJECT( lvl INT, value VARCHAR2(4000) );
/

CREATE TYPE string_pair_table IS TABLE OF string_pair;
/

SELECT t.id,
       v.*
FROM   TBL1 t,
       TABLE(
         CAST(
           MULTISET(
             SELECT string_pair( level, TRIM( REGEXP_SUBSTR( t.value, '[^,]+', 1, LEVEL ) ) )
             FROM   DUAL
             CONNECT BY LEVEL <= REGEXP_COUNT( t.value, '[^,]+' )
           )
           AS string_pair_table
         )
       ) v;

Option 2 - Use ROW_NUMBER():

SELECT t.id,
       v.COLUMN_VALUE AS value,
       ROW_NUMBER() OVER ( PARTITION BY id ORDER BY ROWNUM ) AS lvl
FROM   TBL1 t,
       TABLE(
         CAST(
           MULTISET(
             SELECT TRIM( REGEXP_SUBSTR( t.value, '[^,]+', 1, LEVEL ) )
             FROM   DUAL
             CONNECT BY LEVEL <= REGEXP_COUNT( t.value, '[^,]+' )
           )
           AS SYS.ODCIVARCHAR2LIST
         )
       ) v;

Answer 3

Vercelli posted a correct answer. However, with more than one string to split, connect by will generate an exponentially-growing number of rows, with many, many duplicates. (Just try the query without distinct.) This will destroy performance on data of non-trivial size.

One common way to overcome this problem is to use a prior condition and an additional check to avoid cycles in the hierarchy. Like so:

select id, trim(regexp_substr(value,'[^,]+', 1, level) ) value, level
  from tbl1
   connect by regexp_substr(value, '[^,]+', 1, level) is not null
          and prior id = id
          and prior sys_guid() is not null
   order by id, level;

See, for example, this discussion on OTN: https://community.oracle.com/thread/2526535

Answer 4

An alternate method is to define a simple PL/SQL function:

CREATE OR REPLACE FUNCTION split_String(
  i_str    IN  VARCHAR2,
  i_delim  IN  VARCHAR2 DEFAULT ','
) RETURN SYS.ODCIVARCHAR2LIST DETERMINISTIC
AS
  p_result       SYS.ODCIVARCHAR2LIST := SYS.ODCIVARCHAR2LIST();
  p_start        NUMBER(5) := 1;
  p_end          NUMBER(5);
  c_len CONSTANT NUMBER(5) := LENGTH( i_str );
  c_ld  CONSTANT NUMBER(5) := LENGTH( i_delim );
BEGIN
  IF c_len > 0 THEN
    p_end := INSTR( i_str, i_delim, p_start );
    WHILE p_end > 0 LOOP
      p_result.EXTEND;
      p_result( p_result.COUNT ) := SUBSTR( i_str, p_start, p_end - p_start );
      p_start := p_end + c_ld;
      p_end := INSTR( i_str, i_delim, p_start );
    END LOOP;
    IF p_start <= c_len + 1 THEN
      p_result.EXTEND;
      p_result( p_result.COUNT ) := SUBSTR( i_str, p_start, c_len - p_start + 1 );
    END IF;
  END IF;
  RETURN p_result;
END;
/

Then the SQL becomes very simple:

SELECT t.id,
       v.column_value AS value
FROM   TBL1 t,
       TABLE( split_String( t.value ) ) v

Answer 5

--converting row of data into comma sepaerated string
SELECT
    department_id,
    LISTAGG(first_name, ',') WITHIN GROUP(
        ORDER BY
            first_name
    ) comma_separted_data
FROM
    hr.employees
GROUP BY
    department_id;

--comma-separated string into row of data

CREATE TABLE t (
    deptno          NUMBER,
    employee_name   VARCHAR2(255)
);

INSERT INTO t VALUES (
    10,
    'mohan,sam,john'
);

INSERT INTO t VALUES (
    20,
    'manideeep,ashok,uma'
);

INSERT INTO t VALUES (
    30,
    'gopal,gopi,manoj'
);

SELECT
    deptno,
    employee_name,
    regexp_count(employee_name, ',') + 1,
    regexp_substr(employee_name, '\w+', 1, 1)
FROM
    t,
    LATERAL (
        SELECT
            level l
        FROM
            dual
        CONNECT BY
            level < regexp_count(employee_name, ',') + 1
    );

DROP TABLE t;

Answer 6

SELECT  COL1,   COL2
FROM    (   SELECT INDX, MY_STR1, MY_STR2, COL1_ELEMENTS, COL1, COL2_ELEMENTS, COL2
            FROM    (   SELECT 0 "INDX", COL1 "MY_STR1", COL1_ELEMENTS, COL1, '' "MY_STR2", COL2_ELEMENTS, COL2
                        FROM(
                                SELECT
                                    REPLACE(COL1, ', ', ',') "COL1",    -- In case there is a space after comma
                                    Trim(Length(Replace(COL1, ' ', ''))) - Trim(Length(Translate(REPLACE(COL1, ', ', ','), 'A,', 'A'))) + 1 "COL1_ELEMENTS",    -- Number of elements
                                    Replace(COL2, ', ', ',') "COL2",    -- In case there is a space after comma
                                    Trim(Length(Replace(COL2, ' ', ''))) - Trim(Length(Translate(REPLACE(COL2, ', ', ','), 'A,', 'A'))) + 1 "COL2_ELEMENTS"     -- Number of elements
                                FROM
                                    (SELECT 'aaa,bbb,ccc' "COL1", 'qq, ww, ee' "COL2" FROM DUAL)        -- Your example data
                            )
                    )
                MODEL       -- Modeling --> INDX = 0    COL1='aaa,bbb,ccc'      COL2='qq,ww,ee'
                    DIMENSION BY(0 as INDX)
                    MEASURES(COL1, COL1_ELEMENTS, COL2, CAST('a' as VarChar2(4000)) as MY_STR1, CAST('a' as VarChar2(4000)) as MY_STR2)
                    RULES ITERATE (10)      --UNTIL (ITERATION_NUMBER <= COL1_ELEMENTS[ITERATION_NUMBER + 1]) -- If you don't know the number of elements this should be bigger then you aproximation. Othewrwise it will split given number of elements
                    (
                        COL1_ELEMENTS[ITERATION_NUMBER + 1] = COL1_ELEMENTS[0],
                        MY_STR1[0] = COL1[CV()],
                        MY_STR1[ITERATION_NUMBER + 1] = SubStr(MY_STR1[ITERATION_NUMBER], InStr(MY_STR1[ITERATION_NUMBER], ',', 1) + 1),
                        COL1[ITERATION_NUMBER + 1] = SubStr(MY_STR1[ITERATION_NUMBER], 1, CASE WHEN InStr(MY_STR1[ITERATION_NUMBER], ',') <> 0 THEN InStr(MY_STR1[ITERATION_NUMBER], ',')-1 ELSE Length(MY_STR1[ITERATION_NUMBER]) END),
                        MY_STR2[0] = COL2[CV()],
                        MY_STR2[ITERATION_NUMBER + 1] = SubStr(MY_STR2[ITERATION_NUMBER], InStr(MY_STR2[ITERATION_NUMBER], ',', 1) + 1),
                        COL2[ITERATION_NUMBER + 1] = SubStr(MY_STR2[ITERATION_NUMBER], 1, CASE WHEN InStr(MY_STR2[ITERATION_NUMBER], ',') <> 0 THEN InStr(MY_STR2[ITERATION_NUMBER], ',')-1 ELSE Length(MY_STR2[ITERATION_NUMBER]) END)
                    )
        )
WHERE INDX > 0 And INDX <= COL1_ELEMENTS    -- INDX 0 contains starting strings
--
--  COL1  COL2
--  ----  ----
--  aaa   qq
--  bbb   ww
--  ccc   ee