34

Need Split function which will take two parameters, string to split and delimiter to split the string and return a table with columns Id and Data.And how to call Split function which will return a table with columns Id and Data. Id column will contain sequence and data column will contain data of the string. Eg.

SELECT*FROM Split('A,B,C,D',',')

Result Should be in below format:

|Id | Data
 --   ----
|1  | A  |
|2  | B  |
|3  | C  |
|4  | D  |
Lalit Kumar B
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AKBAR ALI
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  • Above link fixed, please see [**Split single comma delimited string into rows in Oracle**](https://lalitkumarb.wordpress.com/2014/12/02/split-comma-delimited-string-into-rows-in-oracle/) – Lalit Kumar B Aug 13 '17 at 08:06

10 Answers10

30

Here is how you could create such a table:

 SELECT LEVEL AS id, REGEXP_SUBSTR('A,B,C,D', '[^,]+', 1, LEVEL) AS data
   FROM dual
CONNECT BY REGEXP_SUBSTR('A,B,C,D', '[^,]+', 1, LEVEL) IS NOT NULL;

With a little bit of tweaking (i.e., replacing the , in [^,] with a variable) you could write such a function to return a table.

David Faber
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25

There are multiple options. See Split single comma delimited string into rows in Oracle

You just need to add LEVEL in the select list as a column, to get the sequence number to each row returned. Or, ROWNUM would also suffice.

Using any of the below SQLs, you could include them into a FUNCTION.

INSTR in CONNECT BY clause:

SQL> WITH DATA AS
  2    ( SELECT 'word1, word2, word3, word4, word5, word6' str FROM dual
  3    )
  4  SELECT trim(regexp_substr(str, '[^,]+', 1, LEVEL)) str
  5  FROM DATA
  6  CONNECT BY instr(str, ',', 1, LEVEL - 1) > 0
  7  /

STR
----------------------------------------
word1
word2
word3
word4
word5
word6

6 rows selected.

SQL>

REGEXP_SUBSTR in CONNECT BY clause:

SQL> WITH DATA AS
  2    ( SELECT 'word1, word2, word3, word4, word5, word6' str FROM dual
  3    )
  4  SELECT trim(regexp_substr(str, '[^,]+', 1, LEVEL)) str
  5  FROM DATA
  6  CONNECT BY regexp_substr(str , '[^,]+', 1, LEVEL) IS NOT NULL
  7  /

STR
----------------------------------------
word1
word2
word3
word4
word5
word6

6 rows selected.

SQL>

REGEXP_COUNT in CONNECT BY clause:

SQL> WITH DATA AS
  2        ( SELECT 'word1, word2, word3, word4, word5, word6' str FROM dual
  3        )
  4      SELECT trim(regexp_substr(str, '[^,]+', 1, LEVEL)) str
  5      FROM DATA
  6      CONNECT BY LEVEL 

Using XMLTABLE

SQL> WITH DATA AS
  2    ( SELECT 'word1, word2, word3, word4, word5, word6' str FROM dual
  3    )
  4  SELECT trim(COLUMN_VALUE) str
  5    FROM DATA, xmltable(('"' || REPLACE(str, ',', '","') || '"'))
  6  /
STR
------------------------------------------------------------------------
word1
word2
word3
word4
word5
word6

6 rows selected.

SQL>

Using MODEL clause:

SQL> WITH t AS
  2  (
  3         SELECT 'word1, word2, word3, word4, word5, word6' str
  4         FROM   dual ) ,
  5  model_param AS
  6  (
  7         SELECT str AS orig_str ,
  8                ','
  9                       || str
 10                       || ','                                 AS mod_str ,
 11                1                                             AS start_pos ,
 12                Length(str)                                   AS end_pos ,
 13                (Length(str) - Length(Replace(str, ','))) + 1 AS element_count ,
 14                0                                             AS element_no ,
 15                ROWNUM                                        AS rn
 16         FROM   t )
 17  SELECT   trim(Substr(mod_str, start_pos, end_pos-start_pos)) str
 18  FROM     (
 19                  SELECT *
 20                  FROM   model_param MODEL PARTITION BY (rn, orig_str, mod_str)
 21                  DIMENSION BY (element_no)
 22                  MEASURES (start_pos, end_pos, element_count)
 23                  RULES ITERATE (2000)
 24                  UNTIL (ITERATION_NUMBER+1 = element_count[0])
 25                  ( start_pos[ITERATION_NUMBER+1] = instr(cv(mod_str), ',', 1, cv(element_no)) + 1,
 26                  end_pos[iteration_number+1] = instr(cv(mod_str), ',', 1, cv(element_no) + 1) ) )
 27  WHERE    element_no != 0
 28  ORDER BY mod_str ,
 29           element_no
 30  /

STR
------------------------------------------
word1
word2
word3
word4
word5
word6

6 rows selected.

SQL>

You could also use DBMS_UTILITY package provided by Oracle. It provides various utility subprograms. One such useful utility is COMMA_TO_TABLE procedure, which converts a comma-delimited list of names into a PL/SQL table of names.

Read DBMS_UTILITY.COMMA_TO_TABLE

Lalit Kumar B
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7

Oracle Setup:

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;
/

Query

SELECT ROWNUM AS ID,
       COLUMN_VALUE AS Data
FROM   TABLE( split_String( 'A,B,C,D' ) );

Output:

ID DATA
-- ----
 1 A
 2 B
 3 C
 4 D
MT0
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4

If you need a function try this.
First we'll create a type:

CREATE OR REPLACE TYPE T_TABLE IS OBJECT
(
    Field1 int
    , Field2 VARCHAR(25)
);
CREATE TYPE T_TABLE_COLL IS TABLE OF T_TABLE;
/

Then we'll create the function:

CREATE OR REPLACE FUNCTION TEST_RETURN_TABLE
RETURN T_TABLE_COLL
    IS
      l_res_coll T_TABLE_COLL;
      l_index number;
    BEGIN
      l_res_coll := T_TABLE_COLL();
      FOR i IN (
        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)
      LOOP
        IF i.ID = 1001 THEN
          l_res_coll.extend;
          l_index := l_res_coll.count;
          l_res_coll(l_index):= T_TABLE(i.ID, i.name);
        END IF;
      END LOOP;
      RETURN l_res_coll;
    END;
    /

Now we can select from it:

select * from table(TEST_RETURN_TABLE()); 

Output:

SQL> select * from table(TEST_RETURN_TABLE());

    FIELD1 FIELD2
---------- -------------------------
      1001 A
      1001 B
      1001 C
      1001 D
      1001 E
      1001 F

6 rows selected.

Obviously you'd need to replace the WITH TAB AS... bit with where you would be getting your actual data from. Credit Credit

Community
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mmmmmpie
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2

Use this 'Split' function:

CREATE OR REPLACE FUNCTION Split (p_str varchar2) return sys_refcursor is
v_res sys_refcursor;

begin
  open v_res for 
  WITH TAB AS 
  (SELECT p_str STR FROM DUAL)
  select substr(STR, instr(STR, ',', 1, lvl) + 1, instr(STR, ',', 1, lvl + 1) - instr(STR, ',', 1, lvl) - 1) name 
  from
    ( select ',' || STR || ',' as STR from TAB ),
    ( select level as lvl from dual connect by level <= 100 )
    where lvl <= length(STR) - length(replace(STR, ',')) - 1;

     return v_res;
   end;

You can't use this function in select statement like you described in question, but I hope you will find it still useful.

EDIT: Here are steps you need to do. 1. Create Object: create or replace type empy_type as object(value varchar2(512)) 2. Create Type: create or replace type t_empty_type as table of empy_type 3. Create Function:

CREATE OR REPLACE FUNCTION Split (p_str varchar2) return sms.t_empty_type is
v_emptype t_empty_type := t_empty_type();
v_cnt     number := 0;
v_res sys_refcursor;
v_value nvarchar2(128);
begin
  open v_res for
  WITH TAB AS
  (SELECT p_str STR FROM DUAL)
  select substr(STR, instr(STR, ',', 1, lvl) + 1, instr(STR, ',', 1, lvl +     1) - instr(STR, ',', 1, lvl) - 1) name
  from
    ( select ',' || STR || ',' as STR from TAB ),
    ( select level as lvl from dual connect by level <= 100 )
    where lvl <= length(STR) - length(replace(STR, ',')) - 1;


  loop
     fetch v_res into v_value;
      exit when v_res%NOTFOUND;
      v_emptype.extend;
      v_cnt := v_cnt + 1;
     v_emptype(v_cnt) := empty_type(v_value);
    end loop;
    close v_res;

    return v_emptype;
end;

Then just call like this:

SELECT * FROM (TABLE(split('a,b,c,d,g'))) 
1

This function returns the nth part of input string MYSTRING. Second input parameter is separator ie., SEPARATOR_OF_SUBSTR and the third parameter is Nth Part which is required.

Note: MYSTRING should end with the separator.

create or replace FUNCTION PK_GET_NTH_PART(MYSTRING VARCHAR2,SEPARATOR_OF_SUBSTR VARCHAR2,NTH_PART NUMBER)
RETURN VARCHAR2
IS
NTH_SUBSTR VARCHAR2(500);
POS1 NUMBER(4);
POS2 NUMBER(4);
BEGIN
IF NTH_PART=1 THEN
SELECT REGEXP_INSTR(MYSTRING,SEPARATOR_OF_SUBSTR, 1, 1)  INTO POS1 FROM DUAL; 
SELECT SUBSTR(MYSTRING,0,POS1-1) INTO NTH_SUBSTR FROM DUAL;
ELSE
SELECT REGEXP_INSTR(MYSTRING,SEPARATOR_OF_SUBSTR, 1, NTH_PART-1) INTO  POS1 FROM DUAL; 
SELECT REGEXP_INSTR(MYSTRING,SEPARATOR_OF_SUBSTR, 1, NTH_PART)  INTO POS2 FROM DUAL; 
SELECT SUBSTR(MYSTRING,POS1+1,(POS2-POS1-1)) INTO NTH_SUBSTR FROM DUAL;
END IF;
RETURN NTH_SUBSTR;
END;

Hope this helps some body, you can use this function like this in a loop to get all the values separated:

SELECT REGEXP_COUNT(MYSTRING, '~', 1, 'i') INTO NO_OF_RECORDS FROM DUAL;
WHILE NO_OF_RECORDS>0
LOOP
    PK_RECORD    :=PK_GET_NTH_PART(MYSTRING,'~',NO_OF_RECORDS);
    -- do some thing
    NO_OF_RECORDS  :=NO_OF_RECORDS-1;
END LOOP;

Here NO_OF_RECORDS,PK_RECORD are temp variables.

Hope this helps.

0

Best Query For comma separated in This Query we Convert Rows To Column ...

SELECT listagg(BL_PRODUCT_DESC, ', ') within
   group(   order by BL_PRODUCT_DESC) PROD
  FROM GET_PRODUCT
--  WHERE BL_PRODUCT_DESC LIKE ('%WASH%')
  WHERE Get_Product_Type_Id = 6000000000007
MartenCatcher
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0

Created PL/SQL function that can split string by specified delimiter and return result as VARRAY.

CREATE OR REPLACE FUNCTION split(p_parameters VARCHAR2, p_delimiter VARCHAR2) RETURN string_varray AS
    v_delimiter_position NUMBER := 0;
    v_read_position NUMBER :=1;
    v_list string_varray := string_varray();
    v_substring VARCHAR2(4000);
    
    FUNCTION normalize(v_substring VARCHAR2, p_delimiter VARCHAR2) RETURN VARCHAR2 AS
    BEGIN
        RETURN trim(TRAILING p_delimiter FROM trim(BOTH ' ' FROM v_substring));
    END normalize;
BEGIN
    LOOP
         v_delimiter_position := instr(p_parameters, p_delimiter, v_read_position);
         IF v_delimiter_position = 0 THEN
            v_delimiter_position := LENGTH(p_parameters);
         END IF;
         v_substring := substr(p_parameters, v_read_position, v_delimiter_position-v_read_position+1);
         v_list.EXTEND;
         v_list(v_list.LAST) := normalize(v_substring, p_delimiter);
         v_read_position := v_delimiter_position+1;
         IF v_delimiter_position = LENGTH(p_parameters) THEN
            EXIT;
         END IF; 
    END LOOP;  

    RETURN v_list;
END split;

string_varray is VARRAY of VARCHAR2(4000) type. Function also removes whitespaces and the start and end of each value. Invocation example:

select * from table(split('zaa, dddd,ccc', ','));

Will produce three rows in output: zaa dddd ccc

enter image description here

Maksim Sirotkin
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begin for rec in (select * from table(split('shfgjsdfg,242535', ','))) loop dbms_output.put_line(rec.COLUMN_VALUE);

end loop; end;

-- Output shfgjsdfg 242535

  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Sep 29 '22 at 02:17
-4

Try like below

select 
    split.field(column_name,1,',','"') name1,
    split.field(column_name,2,',','"') name2
from table_name
techspider
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