I have table :
id | name
1 | a,b,c
2 | b
i want output like this :
id | name
1 | a
1 | b
1 | c
2 | b
If you can create a numbers table, that contains numbers from 1 to the maximum fields to split, you could use a solution like this:
select
tablename.id,
SUBSTRING_INDEX(SUBSTRING_INDEX(tablename.name, ',', numbers.n), ',', -1) name
from
numbers inner join tablename
on CHAR_LENGTH(tablename.name)
-CHAR_LENGTH(REPLACE(tablename.name, ',', ''))>=numbers.n-1
order by
id, n
Please see fiddle here.
If you cannot create a table, then a solution can be this:
select
tablename.id,
SUBSTRING_INDEX(SUBSTRING_INDEX(tablename.name, ',', numbers.n), ',', -1) name
from
(select 1 n union all
select 2 union all select 3 union all
select 4 union all select 5) numbers INNER JOIN tablename
on CHAR_LENGTH(tablename.name)
-CHAR_LENGTH(REPLACE(tablename.name, ',', ''))>=numbers.n-1
order by
id, n
an example fiddle is here.
If the name column were a JSON array (like '["a","b","c"]'), then you could extract/unpack it with JSON_TABLE() (available since MySQL 8.0.4):
select t.id, j.name
from mytable t
join json_table(
t.name,
'$[*]' columns (name varchar(50) path '$')
) j;
Result:
| id | name |
| --- | ---- |
| 1 | a |
| 1 | b |
| 1 | c |
| 2 | b |
View on DB Fiddle
If you store the values in a simple CSV format, then you would first need to convert it to JSON:
select t.id, j.name
from mytable t
join json_table(
replace(json_array(t.name), ',', '","'),
'$[*]' columns (name varchar(50) path '$')
) j
Result:
| id | name |
| --- | ---- |
| 1 | a |
| 1 | b |
| 1 | c |
| 2 | b |
View on DB Fiddle
I have take the reference from here with changed column name.
DELIMITER $$
CREATE FUNCTION strSplit(x VARCHAR(65000), delim VARCHAR(12), pos INTEGER)
RETURNS VARCHAR(65000)
BEGIN
DECLARE output VARCHAR(65000);
SET output = REPLACE(SUBSTRING(SUBSTRING_INDEX(x, delim, pos)
, LENGTH(SUBSTRING_INDEX(x, delim, pos - 1)) + 1)
, delim
, '');
IF output = '' THEN SET output = null; END IF;
RETURN output;
END $$
CREATE PROCEDURE BadTableToGoodTable()
BEGIN
DECLARE i INTEGER;
SET i = 1;
REPEAT
INSERT INTO GoodTable (id, name)
SELECT id, strSplit(name, ',', i) FROM BadTable
WHERE strSplit(name, ',', i) IS NOT NULL;
SET i = i + 1;
UNTIL ROW_COUNT() = 0
END REPEAT;
END $$
DELIMITER ;
Here is my attempt:
The first select presents the csv field to the split.
Using recursive CTE, we can create a list of numbers that are limited to the number of terms in the csv field.
The number of terms is just the difference in the length of the csv field and itself with all the delimiters removed.
Then joining with this numbers, substring_index extracts that term.
with recursive
T as ( select 'a,b,c,d,e,f' as items),
N as ( select 1 as n union select n + 1 from N, T
where n <= length(items) - length(replace(items, ',', '')))
select distinct substring_index(substring_index(items, ',', n), ',', -1)
group_name from N, T
My variant: stored procedure that takes table name, field names and delimiter as arguments. Inspired by post http://www.marcogoncalves.com/2011/03/mysql-split-column-string-into-rows/
delimiter $$
DROP PROCEDURE IF EXISTS split_value_into_multiple_rows $$
CREATE PROCEDURE split_value_into_multiple_rows(tablename VARCHAR(20),
id_column VARCHAR(20), value_column VARCHAR(20), delim CHAR(1))
BEGIN
DECLARE id INT DEFAULT 0;
DECLARE value VARCHAR(255);
DECLARE occurrences INT DEFAULT 0;
DECLARE i INT DEFAULT 0;
DECLARE splitted_value VARCHAR(255);
DECLARE done INT DEFAULT 0;
DECLARE cur CURSOR FOR SELECT tmp_table1.id, tmp_table1.value FROM
tmp_table1 WHERE tmp_table1.value IS NOT NULL AND tmp_table1.value != '';
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1;
SET #expr = CONCAT('CREATE TEMPORARY TABLE tmp_table1 (id INT NOT NULL, value VARCHAR(255)) ENGINE=Memory SELECT ',
id_column,' id, ', value_column,' value FROM ',tablename);
PREPARE stmt FROM #expr;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;
DROP TEMPORARY TABLE IF EXISTS tmp_table2;
CREATE TEMPORARY TABLE tmp_table2 (id INT NOT NULL, value VARCHAR(255) NOT NULL) ENGINE=Memory;
OPEN cur;
read_loop: LOOP
FETCH cur INTO id, value;
IF done THEN
LEAVE read_loop;
END IF;
SET occurrences = (SELECT CHAR_LENGTH(value) -
CHAR_LENGTH(REPLACE(value, delim, '')) + 1);
SET i=1;
WHILE i <= occurrences DO
SET splitted_value = (SELECT TRIM(SUBSTRING_INDEX(
SUBSTRING_INDEX(value, delim, i), delim, -1)));
INSERT INTO tmp_table2 VALUES (id, splitted_value);
SET i = i + 1;
END WHILE;
END LOOP;
SELECT * FROM tmp_table2;
CLOSE cur;
DROP TEMPORARY TABLE tmp_table1;
END; $$
delimiter ;
Usage example (normalization):
CALL split_value_into_multiple_rows('my_contacts', 'contact_id', 'interests', ',');
CREATE TABLE interests (
interest_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
interest VARCHAR(30) NOT NULL
) SELECT DISTINCT value interest FROM tmp_table2;
CREATE TABLE contact_interest (
contact_id INT NOT NULL,
interest_id INT NOT NULL,
CONSTRAINT fk_contact_interest_my_contacts_contact_id FOREIGN KEY (contact_id) REFERENCES my_contacts (contact_id),
CONSTRAINT fk_contact_interest_interests_interest_id FOREIGN KEY (interest_id) REFERENCES interests (interest_id)
) SELECT my_contacts.contact_id, interests.interest_id
FROM my_contacts, tmp_table2, interests
WHERE my_contacts.contact_id = tmp_table2.id AND interests.interest = tmp_table2.value;
Because you have to keep adding "select number union all" in the example above which can be a issue if you need a large number of splits.
select
tablename.id,
SUBSTRING_INDEX(SUBSTRING_INDEX(tablename.name, ',', numbers.n), ',', -1) name
from
(select 1 n union all
select 2 union all select 3 union all
select 4 union all select 5) numbers INNER JOIN tablename
on CHAR_LENGTH(tablename.name)
-CHAR_LENGTH(REPLACE(tablename.name, ',', ''))>=numbers.n-1
order by
id, n
I decided a better way was this which only adds a number row for each digit. Example below is good for 1-1000 add another row makes it good for 1-10000 and so on.
select
tablename.id,
SUBSTRING_INDEX(SUBSTRING_INDEX(tablename.name, ',', numbers.n), ',', -1) name
from(SELECT #row := #row + 1 AS n FROM
(select 0 union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as t,
(select 0 union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as t2,
(select 0 union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as t3,
(SELECT #row:=0) as numbers)as numbers INNER JOIN tablename
on CHAR_LENGTH(tablename.name)
-CHAR_LENGTH(REPLACE(tablename.name, ',', ''))>=numbers.n-1
order by
id, n
CREATE PROCEDURE `getVal`()
BEGIN
declare r_len integer;
declare r_id integer;
declare r_val varchar(20);
declare i integer;
DECLARE found_row int(10);
DECLARE row CURSOR FOR select length(replace(val,"|","")),id,val from split;
create table x(id int,name varchar(20));
open row;
select FOUND_ROWS() into found_row ;
read_loop: LOOP
IF found_row = 0 THEN
LEAVE read_loop;
END IF;
set i = 1;
FETCH row INTO r_len,r_id,r_val;
label1: LOOP
IF i <= r_len THEN
insert into x values( r_id,SUBSTRING(replace(r_val,"|",""),i,1));
SET i = i + 1;
ITERATE label1;
END IF;
LEAVE label1;
END LOOP label1;
set found_row = found_row - 1;
END LOOP;
close row;
select * from x;
drop table x;
END
The original question was for MySQL and SQL in general. The example below is for the new versions of MySQL. Unfortunately, a generic query that would work on any SQL server is not possible. Some servers do no support CTE, others do not have substring_index, yet others have built-in functions for splitting a string into multiple rows.
--- the answer follows ---
Recursive queries are convenient when the server does not provide built-in functionality. They can also be the bottleneck.
The following query was written and tested on MySQL version 8.0.16. It will not work on version 5.7-. The old versions do not support Common Table Expression (CTE) and thus recursive queries.
with recursive
input as (
select 1 as id, 'a,b,c' as names
union
select 2, 'b'
),
recurs as (
select id, 1 as pos, names as remain, substring_index( names, ',', 1 ) as name
from input
union all
select id, pos + 1, substring( remain, char_length( name ) + 2 ),
substring_index( substring( remain, char_length( name ) + 2 ), ',', 1 )
from recurs
where char_length( remain ) > char_length( name )
)
select id, name
from recurs
order by id, pos;
Here is another trick. The number 20 is the maximum number of values in comma separated list.
We use single query, no procedures.
If tbl has more rows than the maximum number of values in single comma separated list you may remove 'inner join tbl a inner join tbl c' part from query. I added this because there are only 2 rows.
CREATE TABLE tbl(id int NOT NULL,name varchar(50),PRIMARY KEY (`id`));
insert into tbl values(1, 'a,b,c'), (2, 'd');
select id ,SUBSTRING_INDEX(SUBSTRING_INDEX(name, ',', k.n), ',', -1) as name
from tbl
INNER JOIN (
SELECT *
FROM (
SELECT #n:=#n+1 AS n
FROM tbl inner join tbl a inner join tbl c
INNER JOIN (SELECT #n:=0) AS _a
) AS _a WHERE _a.n <= 20
)AS k ON k.n <= LENGTH(name) - LENGTH(replace(name, ',','')) + 1
order by id
This is a trick to extract nth value in comma separated list:
SUBSTRING_INDEX(SUBSTRING_INDEX(name, ',', k.n), ',', -1)
Here is my solution
-- Create the maximum number of words we want to pick (indexes in n)
with recursive n(i) as (
select
1 i
union all
select i+1 from n where i < 1000
)
select distinct
s.id,
s.oaddress,
-- n.i,
-- use the index to pick the nth word, the last words will always repeat. Remove the duplicates with distinct
if(instr(reverse(trim(substring_index(s.oaddress,' ',n.i))),' ') > 0,
reverse(substr(reverse(trim(substring_index(s.oaddress,' ',n.i))),1,
instr(reverse(trim(substring_index(s.oaddress,' ',n.i))),' '))),
trim(substring_index(s.oaddress,' ',n.i))) oth
from
app_schools s,
n
Best Practice.
Result:
SELECT
SUBSTRING_INDEX(SUBSTRING_INDEX('ab,bc,cd',',',help_id+1),',',-1) AS oid
FROM
(
SELECT #xi:=#xi+1 as help_id from
(SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) xc1,
(SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) xc2,
(SELECT #xi:=-1) xc0
) a
WHERE
help_id < LENGTH('ab,bc,cd')-LENGTH(REPLACE('ab,bc,cd',',',''))+1
First, create a numbers table:
SELECT #xi:=#xi+1 as help_id from
(SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) xc1,
(SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) xc2,
(SELECT #xi:=-1) xc0;
| help_id |
| --- |
| 0 |
| 1 |
| 2 |
| 3 |
| ... |
| 24 |
Second, just split the str:
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX('ab,bc,cd',',',help_id+1),',',-1) AS oid
FROM
numbers_table
WHERE
help_id < LENGTH('ab,bc,cd')-LENGTH(REPLACE('ab,bc,cd',',',''))+1
| oid |
| --- |
| ab |
| bc |
| cd |
SELECT id, unnest(string_to_array(name, ',')) AS names
FROM datatable
Hope this helps :D
CREATE DEFINER=`root`#`localhost` PROCEDURE `apply_Leave_SP`(
in leavetypeid int ,
in empid int,
in reason varchar(100),
in startdate date,
in enddate date,
in startsession int,
in endsession int,
in compoffid int,
in mangerid int
)
BEGIN
declare leavestatus int(10) default 0;
declare optionalyHolidays int(10) default 0;
declare listofholidays int(10) default 0;
declare totalhours int (10) default 0;
declare hours int (10) default 0;
declare satsun int (10) default 0;
declare manger_id int(10) default 0;
declare paidleave int(10) default 0;
declare days int(10) default 0;
declare leaveappliedid int(10) default 0;
declare lossofpay int(10) default 0;
declare casualleave int (10) default 0;
declare sickleave int (10) default 0;
select count(holiday_id) into optionalyHolidays from Parabola.holidays where holiday_type_id=2 and
DATEDIFF(startdate,curdate())>=20;
select count(*) into listofholidays from Parabola.holidays where date between startdate and enddate;
SELECT
COUNT(*) AS total into hours
FROM
( SELECT ADDDATE(startdate, INTERVAL #i:=#i+1 DAY) AS DAY
FROM (
SELECT a.a
FROM (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS a
CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS b
CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS c
) a
JOIN (SELECT #i := -1) r1
WHERE
#i < DATEDIFF(enddate, startdate)
) AS dateTable
WHERE WEEKDAY(dateTable.Day) IN (5,6);
SELECT datediff(enddate,startdate) into totalhours;
set hours=(totalhours-listofholidays-satsun)*8;
if(leavetypeid=1)then
set leavestatus=5;
else
set leavestatus=1;
end if;
if(optionalyHolidays>=0) then
set leavestatus=5;
end if;
insert into leave_applied(leave_type_id,hours,employee_id,
created_at,updated_at,start_date,end_date,start_date_session,
end_date_session,reason,
status,reminder_count,personal_calendar_event_id,system_calendar_event_id)
values(leavetypeid,hours,empid,curdate(),curdate(),startdate,enddate,
startsession,endsession,
reason,leavestatus,1,'sdasdas','sadeew');
Select LAST_INSERT_ID() into leaveappliedid ;
if((leaveappliedid=1 or 4) and days >paidleave) then
set lossofpay=days-paidleave;
else if((leaveappliedid=1 or 4) and days> #CL) then
set sickleave=days-#CL;
set casualleave=#CL;
else if((leaveappliedid=1 or 4) and days>#SL) then
set casualleave=days-#SL;
set sickleave=#SL;
else
set mangerid=10;
end if;
end;
insert into leave_actual (leave_applied_id,
leave_type_id,hours,start_date,
end_date,created_at,updated_at) values(leaveappliedid,leavetypeid,
hours,start_date,end_date,curdate(),curdate());
insert into leave_approval (leave_applied_id,
manager_id,status ,created_at,updated_at) values
(leaveappliedid,mangerid,leavestatus,curdate(),curdate());
END
this my code when i try to apply else if condition then there is Syntax Error is coming while only with if there is no Error is coming but i have to apply else if condition please suggest me where am doing wrong it show else if Condition
The syntax of if, else if is like this:
IF condition THEN
... ...
ELSEIF condition THEN
... ...
END IF;
See manual here.
IF search_condition THEN statement_list
[ELSEIF search_condition THEN statement_list] ...
[ELSE statement_list]
END IF