ORACLE - Substring - sql

I found a problem in ORACLE-CERTIFICATE about substring (I guess) that I don't have any idea how to do that. Does anyone understand this??
Create table Bonus and do DML insert to Bonus table for the following script:
Hint: you should be able to ran this script against your table without any error and generate the output "Here is the result".
select (substr(
substr(
substr(
l.object,
instr(l.object,',')+1
),
instr(
substr(
l.object,
instr(l.object,',')+1
),
','
)+1
),
instr(
substr(
substr(l.object,instr(l.object,',')+1),
instr(
substr(
l.object,
instr(l.object,',')+1
),
','
)+1
),
','
)+1
)
) ref
from Bonus l
where l.l_index = 't234'
Here is the output:
REF
------------------------------------------------------------
Here is the result


Yes, it is looking for the remainder of the string that follows the 3rd comma:
Oracle Setup:
CREATE TABLE Bonus ( l_index, object ) AS
SELECT 't234', ',,,Here is the result' FROM DUAL;
Query:
select (substr(
substr(
substr(
l.object,
instr(l.object,',')+1
),
instr(
substr(
l.object,
instr(l.object,',')+1
),
','
)+1
),
instr(
substr(
substr(
l.object,
instr(l.object,',')+1
),
instr(
substr(
l.object,
instr(l.object,',')+1
),
','
)+1
),
','
)+1
)
) ref
from Bonus l
where l.l_index = 't234'
Output:
REF
---------------------
Here is the result
A simpler version of this query would be:
SELECT SUBSTR( object, INSTR( object, ',', 1, 3 ) + 1 ) AS ref
FROM Bonus
WHERE l_index = 't234';

Related

transpose multiple strings Oracle [duplicate]

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?

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

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;

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

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

--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;

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

Find space after nth characters and split into new row

I have a large string stored in table as a single line. I need a select query to split the large string to rows after every 100 characters and it should not split in middle of the word. Basically, the query should find a space after 100 characters and split into new line.
I have used this query, it is splitting after 100 lines, but it is breaking in the middle of words.
SELECT REGEXP_REPLACE ( col_large_string , '(.{100})' , '\1' || CHR (10) ) AS split_to_rows
FROM tab_large_string where string_id = 1;

You do not need (slow) regular expressions and can do it with simple (quicker) string functions.
If you want to replace spaces with newlines then:
WITH bounds ( str, end_pos ) AS (
SELECT col_large_string,
INSTR(col_large_string, ' ', 101)
FROM tab_large_string
UNION ALL
SELECT SUBSTR(str, 1, end_pos - 1)
|| CHR(10)
|| SUBSTR(str, end_pos + 1),
INSTR(str, ' ', end_pos + 101)
FROM bounds
WHERE end_pos > 0
)
SELECT str AS split_to_lines
FROM bounds
WHERE end_pos = 0;
and if you want to have each line in a new row then:
WITH bounds ( str, start_pos, end_pos ) AS (
SELECT col_large_string,
1,
INSTR(col_large_string, ' ', 101)
FROM tab_large_string
UNION ALL
SELECT str,
end_pos + 1,
INSTR(str, ' ', end_pos + 101)
FROM bounds
WHERE end_pos > 0
)
SELECT CASE end_pos
WHEN 0
THEN SUBSTR(str, start_pos)
ELSE SUBSTR(str, start_pos, end_pos - start_pos)
END AS split_to_rows
FROM bounds;
If you do want to use regular expressions then:
SELECT REGEXP_REPLACE(
col_large_string,
'(.{100,}?) ',
'\1' || CHR (10)
) AS split_to_lines
FROM tab_large_string
WHERE string_id = 1;
db<>fiddle here

You can use this regular expression:
SELECT REGEXP_REPLACE ( col_large_string , '((\w+\s+){100})' , '\1' || CHR (10) ) AS split_to_rows
FROM tab_large_string where string_id = 1;
\w+ matches one or more occurrence of word character.
\s+ matches one or more occurrence of space character.
(\w+\s+) matches a word followed by space
(\w+\s+){100} then matches (a word followed by space) x100.

How to convert comma separated values to rows in oracle?

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?

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

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;

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

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

--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;

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

change the order of a string in pl/sql

I want to change the order of a string like :
name/surname by surname/name OR name/ surname by surname/ name (taking account on spaces after/)
below my request.but the second part is wrong :
select SUBSTR ('name/surname' , INSTR ('name/surname','/')+1) ||SUBSTR ('name/surname' , 1,INSTR('name/surname','/')-1)
from dual

Using a regular expression:
SELECT REGEXP_REPLACE( 'name/surname', '^(.*?)/(.*)$', '\2/\1' ) FROM DUAL;
or trimming all whitespaces:
SELECT REGEXP_REPLACE( ' name / surname ', '^\s*(.*?)\s*/\s*(.*?)\s*$', '\2/\1' ) FROM DUAL;
or preserving a whitespace before and after the slash (and trimming leading/training whitespace):
SELECT REGEXP_REPLACE( ' name / surname ', '^\s*(.*?)(\s?/\s?)(.*?)\s*$', '\3\2\1' ) FROM DUAL;
Using string functions:
WITH names ( text ) AS (
SELECT 'name/surname' FROM DUAL
)
SELECT SUBSTR( text, INSTR( text, '/' ) + 1 ) || '/' || SUBSTR( text, 1, INSTR( text, '/' ) - 1 )
FROM names;
or trimming whitespace:
WITH names ( text ) AS (
SELECT ' name / surname ' FROM DUAL
)
SELECT TRIM( SUBSTR( text, INSTR( text, '/' ) + 1 ) ) || '/' || TRIM( SUBSTR( text, 1, INSTR( text, '/' ) - 1 ) )
FROM names;

select substr('firstname/lastname',instr('firstname/lastname','/')+1)||'/'
|| substr('firstname/lastname',1,instr('firstname/lastname','/')-1) from dual;

SQL Concatenate strings across multiple columns with corresponding values

I'm looking for a way to achieve this in a SELECT statement.
FROM
Column1 Column2 Column3
A,B,C 1,2,3 x,y,z
TO
Result
A|1|x,B|2|y,C|3|z
The delimiters don't matter. I'm just trying to to get all the data in one single column. Ideally I am looking to do this in DB2. But I'd like to know if there's an easier way to get this done in Oracle.
Thanks

You can do it like this using INSTR and SUBSTR:
select
substr(column1,1,instr(column1,',',1)-1) || '|' ||
substr(column2,1,instr(column2,',',1)-1) || '|' ||
substr(column3,1,instr(column3,',',1)-1) || '|' ||
',' ||
substr(column1 ,instr(column1 ,',',1,1)+1,instr(column1 ,',',1,2) - instr(column1 ,',',1)-1) || '|' ||
substr(column2 ,instr(column2 ,',',1,1)+1,instr(column2 ,',',1,2) - instr(column2 ,',',1)-1) || '|' ||
substr(column3 ,instr(column3 ,',',1,1)+1,instr(column3 ,',',1,2) - instr(column3 ,',',1)-1) || '|' ||
',' ||
substr(column1 ,instr(column1 ,',',1,2)+1) || '|' ||
substr(column2 ,instr(column2 ,',',1,2)+1) || '|' ||
substr(column3 ,instr(column3 ,',',1,2)+1)
from yourtable

i tried some thing. just look into link
first i created a table called t_ask_test and inserted the data based on the above question. Achieved the result by using the string functions
sample table
create table t_ask_test(column1 varchar(10), column2 varchar(10),column3 varchar(10));
inserted a row
insert into T_ASK_TEST values ('A,B,C','1,2,3','x,y,z');
the following query will be in dynamic way
select substr(column1,1,instr(column1,',',1,1)-1)||'|'||substr(column2,1,instr(column1,',',1,1)-1)||'|'||substr(column3,1,instr(column1,',',1,1)-1) ||','||
substr(column1,instr(column1,',',1,1)+1,instr(column1,',',1,2)-instr(column1,',',1,1)-1)||'|'||substr(column2,instr(column2,',',1,1)+1,instr(column2,',',1,2)-instr(column2,',',1,1)-1)||'|'||substr(column3,instr(column3,',',1,1)+1,instr(column3,',',1,2)-instr(column3,',',1,1)-1) ||','||
substr(column1,instr(column1,',',1,2)+1,length(column1)-instr(column1,',',1,2))||'|'||substr(column2,instr(column2,',',1,2)+1,length(column2)-instr(column2,',',1,2))||'|'||substr(column3,instr(column3,',',1,2)+1,length(column3)-instr(column3,',',1,2)) as test from t_ask_test;
output will be as follows
TEST
---------------
A|1|x,B|2|y,C|3|z

If you have a dynamic number of entries for each row then:
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE TEST ( Column1, Column2, Column3 ) AS
SELECT 'A,B,C', '1,2,3', 'x,y,z' FROM DUAL
UNION ALL SELECT 'D,E', '4,5', 'v,w' FROM DUAL;
Query 1:
WITH ids AS (
SELECT t.*, ROWNUM AS id
FROM TEST t
)
SELECT LISTAGG(
REGEXP_SUBSTR( i.Column1, '[^,]+', 1, n.COLUMN_VALUE )
|| '|' || REGEXP_SUBSTR( i.Column2, '[^,]+', 1, n.COLUMN_VALUE )
|| '|' || REGEXP_SUBSTR( i.Column3, '[^,]+', 1, n.COLUMN_VALUE )
, ','
) WITHIN GROUP ( ORDER BY n.COLUMN_VALUE ) AS value
FROM ids i,
TABLE(
CAST(
MULTISET(
SELECT LEVEL
FROM DUAL
CONNECT BY LEVEL <= GREATEST(
REGEXP_COUNT( i.COLUMN1, '[^,]+' ),
REGEXP_COUNT( i.COLUMN2, '[^,]+' ),
REGEXP_COUNT( i.COLUMN3, '[^,]+' )
)
)
AS SYS.ODCINUMBERLIST
)
) n
GROUP BY i.ID
Results:
| VALUE |
|-------------------|
| A|1|x,B|2|y,C|3|z |
| D|4|v,E|5|w |

You need to use:
SUBSTR
INSTR
|| concatenation operator
It would be easy if you break your output, and then understand how it works.
SQL> WITH t AS
2 ( SELECT 'A,B,C' Column1, '1,2,3' Column2, 'x,y,z' Column3 FROM dual
3 )
4 SELECT SUBSTR(column1, 1, instr(column1, ',', 1) -1)
5 ||'|'
6 || SUBSTR(column2, 1, instr(column2, ',', 1) -1)
7 ||'|'
8 || SUBSTR(column3, 1, instr(column1, ',', 1) -1)
9 ||','
10 || SUBSTR(column1, instr(column1, ',', 1, 2) +1 - instr(column1, ',', 1),
11 instr(column1, ',', 1) -1)
12 ||'|'
13 || SUBSTR(column2, instr(column2, ',', 1, 2) +1 - instr(column2, ',', 1),
14 instr(column2, ',', 1) -1)
15 ||'|'
16 || SUBSTR(column3, instr(column3, ',', 1, 2) +1 - instr(column3, ',', 1),
17 instr(column3, ',', 1) -1)
18 ||','
19 || SUBSTR(column1, instr(column1, ',', 1, 3) +1 - instr(column1, ',', 1),
20 instr(column1, ',', 2) -1)
21 as "new_column"
22 FROM t;
new_column
-------------
A|1|x,B|2|y,C
On a side note, you should avoid storing delimited values in a single column. Consider normalizing the data.
From Oracle 11g and above, you could create a VIRTUAL COLUMN using the above expression and use it instead of executing the SQL frequently.

Its very simple in oracle. just use the concatenation operatort ||.
In the below solution, I have used underscore as the delimiter
select Column1 ||'_'||Column2||'_'||Column3 from table_name;