Suppose f is a string function such as concatenation by 0: f(0011)=00110.
Suppose mytable has columns string which is an index, and price.
I would like to do something like the following pseudo-code:
UPDATE mytable SET price(X)=0.5 FOR ALL THOSE ROWS X SUCH THAT
THE ROW Y SUCH THAT THE `string` OF Y equals f(the `string` FOR X)
HAS price(Y)=0
More generally, how can I reference
"the row such that [condition-holds-that-depends-on-the-current-row]"
in (My)SQL?
Setup demo
CREATE TABLE mytable
( id INT UNSIGNED NOT NULL PRIMARY KEY
, thestring VARCHAR(32)
, price DECIMAL(11,2)
) ENGINE=INNODB
;
INSERT INTO mytable (id, thestring, price) VALUES
( 1,'0011' , 123.45)
,( 2,'00110' , 45.67)
,( 3,'001100' , 0.00)
,( 4,'10' , 4.44)
,( 5,'100' , 5.55)
,( 6,'1000' , 0.00)
;
Write a query that identifies the Y rows in mytable
SELECT y.*
FROM `mytable` `y`
WHERE y.price = 0.0
ORDER BY y.id
Add a join to mytable to find the matching X rows
SELECT x.id AS x_id
, x.thestring AS x_thestring
, x.price AS x_price
, y.id AS y_id
, y.thestring AS y_thestring
, y.price AS y_price
FROM mytable `y`
JOIN mytable `x`
ON CONCAT(x.thestring,'0') = y.thestring
WHERE y.price = 0.0
ORDER BY y.id
Convert the SELECT into an UPDATE. (Replace SELECT ... FROM with UPDATE, and add a SET clause before the WHERE clause.
UPDATE mytable `y`
JOIN mytable `x`
ON CONCAT(x.thestring,'0') = y.thestring
SET x.price = 0.5
WHERE y.price = 0.0
Just replace the CONCAT(x.thestring,'0') with f(x.thestring).
Another option is to use a correlated subquery.
First, write a SELECT
SELECT x.*
FROM mytable `x`
WHERE EXISTS ( SELECT 1
FROM mytable `y`
WHERE y.thestring = f(x.thestring)
AND y.price = 0.00
)
And then convert that to an UPDATE. Replace SELECT ... FROM with the UPDATE keyword, and add a SET clause before the WHERE clause.
UPDATE mytable `x`
SET x.price = 0.5
WHERE EXISTS ( SELECT 1
FROM mytable `y`
WHERE y.thestring = f(x.thestring)
AND y.price = 0.00
)
Use an UPDATE with a self-join.
UPDATE mytable AS x
JOIN mytable AS y ON y.string = f(x.string)
SET x.price = 0.5
WHERE y.price = 0
I am new to StackOverflow and have got stuck with a query to print prime numbers from 2 to 1000.
I have used the below query need input if this is the most efficient way to code it.
WITH NUM AS (
SELECT LEVEL N
FROM DUAL CONNECT BY LEVEL <= 1000
)
SELECT LISTAGG(B.N,'-') WITHIN GROUP(ORDER BY B.N) AS PRIMES
FROM (
SELECT N,
CASE WHEN EXISTS (
SELECT NULL
FROM NUM N_INNER
WHERE N_INNER .N > 1
AND N_INNER.N < NUM.N
AND MOD(NUM.N, N_INNER.N)=0
) THEN
'NO PRIME'
ELSE
'PRIME'
END IS_PRIME
FROM NUM
) B
WHERE B.IS_PRIME='PRIME'
AND B.N!=1;
I know this question has been asked multiple times and I am requesting better solution if any. More over need input on how this works with MySQL/MS SQL/PostgreSQL.
Any help will make my understanding better.
In PostgreSQL probably the most fastest query that prints prime numbers up to 1000 is:
SELECT regexp_split_to_table('2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997',E',')::int
AS x
;
It took only 16 ms on my computer.
Note: a list of prime numbers was copied from https://en.wikipedia.org/wiki/Prime_number
and pasted into this long string
If you prefer SQL, then this works
WITH x AS (
SELECT * FROM generate_series( 2, 1000 ) x
)
SELECT x.x
FROM x
WHERE NOT EXISTS (
SELECT 1 FROM x y
WHERE x.x > y.x AND x.x % y.x = 0
)
;
It's two times slower - 31 ms.
Ans an equivalent version for Oracle:
WITH x AS(
SELECT level+1 x
FROM dual
CONNECT BY LEVEL <= 999
)
SELECT x.x
FROM x
WHERE NOT EXISTS (
SELECT 1 FROM x y
WHERE x.x > y.x AND remainder( x.x, y.x) = 0
)
;
The most obvious improvement is that instead of checking from 1 to n you can check from 1 to the square root of n.
A second major optimization would be to use a temporary table to store the results and check them first. This way you can iterate incrementally from 1 to n, and only check the known primes from 1 to square root of n (recursively doing that until you have a list). If you go about things this way you would probably want to set up the prime detection in a function and then do the same with your number series generator.
That second one though means extending SQL and so I don't know if that fits your requirements.
For postgresql I would use generate_series go generate the list of numbers. I would then create functions which would then either store the list of primes in a temporary table or pass them back in and out in an ordered array and then couple them like that
MariaDB (with sequence plugin)
Similar to kordirkos algorithm:
select 2 as p union all
select n.seq
from seq_3_to_1000_step_2 n
where not exists (
select 1
from seq_3_to_32_step_2 q
where q.seq < n.seq
and n.seq mod q.seq = 0
);
Using LEFT JOIN:
select 2 as p union all
select n.seq
from seq_3_to_1000_step_2 n
left join seq_3_to_32_step_2 q
on q.seq < n.seq
and n.seq mod q.seq = 0
where q.seq is null;
MySQL
There are no sequence generating helpers in MySQL. So the sequence tables have to be created first:
drop temporary table if exists n;
create temporary table if not exists n engine=memory
select t2.c*100 + t1.c*10 + t0.c + 1 as seq from
(select 0 c union all select 1 c union all select 2 c union all select 3 c union all select 4 c union all select 5 c union all select 6 c union all select 7 c union all select 8 c union all select 9 c) t0,
(select 0 c union all select 1 c union all select 2 c union all select 3 c union all select 4 c union all select 5 c union all select 6 c union all select 7 c union all select 8 c union all select 9 c) t1,
(select 0 c union all select 1 c union all select 2 c union all select 3 c union all select 4 c union all select 5 c union all select 6 c union all select 7 c union all select 8 c union all select 9 c) t2
having seq > 2 and seq % 2 != 0;
drop temporary table if exists q;
create temporary table if not exists q engine=memory
select *
from n
where seq <= 32;
alter table q add primary key seq (seq);
Now similar queries can be used:
select 2 as p union all
select n.seq
from n
where not exists (
select 1
from q
where q.seq < n.seq
and n.seq mod q.seq = 0
);
select 2 as p union all
select n.seq
from n
left join q
on q.seq < n.seq
and n.seq mod q.seq = 0
where q.seq is null;
sqlfiddle
Oracle and without inner select in getting part:
with tmp(id)
as (
select level id from dual connect by level <= 100
) select t1.id from tmp t1
JOIN tmp t2
on MOD(t1.id, t2.id) = 0
group by t1.ID
having count(t1.id) = 2
order by t1.ID
;
/* Below is my solution */
/* Step 1: Get all the numbers till 1000 */
with tempa as
(
select level as Num
from dual
connect by level<=1000
),
/* Step 2: Get the Numbers for finding out the factors */
tempb as
(
select a.NUm,b.Num as Num_1
from tempa a , tempa b
where b.Num<=a.Num
),
/*Step 3:If a number has exactly 2 factors, then it is a prime number */
tempc as
(
select Num, sum(case when mod(num,num_1)=0 then 1 end) as Factor_COunt
from tempb
group by Num
)
select listagg(Num,'&') within group (order by Num)
from tempc
where Factor_COunt=2
;
Tested on sqlite3
WITH nums(n) AS
(
SELECT 1
UNION ALL
SELECT n + 1 FROM nums WHERE n < 100
)
SELECT n
FROM (
SELECT n FROM nums
)
WHERE n NOT IN (
SELECT n
FROM nums
JOIN ( SELECT n AS n2 FROM nums )
WHERE n <> 1
AND n2 <> 1
AND n <> n2
AND n2 < n
AND n % n2 = 0
ORDER BY n
)
AND n <> 1
Tested on Vertica 8
WITH seq AS (
SELECT ROW_NUMBER() OVER() AS n
FROM (
SELECT 1
FROM (
SELECT date(0) + INTERVAL '1 second' AS i
UNION ALL
SELECT date(0) + INTERVAL '100 seconds' AS i
) _
TIMESERIES tm AS '1 second' OVER(ORDER BY i)
) _
)
SELECT n
FROM (SELECT n FROM seq) _
WHERE n NOT IN (
SELECT n FROM (
SELECT s1.n AS n, s2.n AS n2
FROM seq AS s1
CROSS JOIN seq AS s2
ORDER BY n, n2
) _
WHERE n <> 1
AND n2 <> 1
AND n <> n2
AND n2 < n
AND n % n2 = 0
)
AND n <> 1
ORDER BY n
This is what worked for me in the SQL server. I tried to reduce the order of my nested loops.
declare #var int
declare #i int
declare #result varchar (max)
set #var = 1
select #result = '2&3&5' --first few obvious prime numbers
while #var < 1000 --the first loop
begin
set #i = 3;
while #i <= #var/2 --the second loop which I attempted to reduce the order
begin
if #var%#i = 0
break;
if #i=#var/2
begin
set #result = #result + '&' + CAST(#var AS VARCHAR)
break;
end
else
set #i = #i + 1
end
set #var = #var + 1;
end
print #result
SELECT LISTAGG(PRIME_NUMBER,'&') WITHIN GROUP (ORDER BY PRIME_NUMBER)
FROM
(
SELECT L PRIME_NUMBER FROM
(
SELECT LEVEL L FROM DUAL CONNECT BY LEVEL <= 1000 ),
(
SELECT LEVEL M FROM DUAL CONNECT BY LEVEL <= 1000
) WHERE M <= L
GROUP BY L
HAVING COUNT(CASE WHEN L/M = TRUNC(L/M) THEN 'Y' END
) = 2
ORDER BY L
);
SELECT GROUP_CONCAT(NUMB SEPARATOR '&')
FROM (
SELECT #num:=#num+1 as NUMB FROM
information_schema.tables t1,
information_schema.tables t2,
(SELECT #num:=1) tmp
) tempNum
WHERE NUMB<=1000 AND NOT EXISTS(
SELECT * FROM (
SELECT #nu:=#nu+1 as NUMA FROM
information_schema.tables t1,
information_schema.tables t2,
(SELECT #nu:=1) tmp1
LIMIT 1000
) tatata
WHERE FLOOR(NUMB/NUMA)=(NUMB/NUMA) AND NUMA<NUMB AND NUMA>1
)
MySQL Code :
DECLARE
#i INT,
#a INT,
#count INT,
#p nvarchar(max)
SET #i = 1
WHILE (#i <= 1000)
BEGIN SET #count = 0
SET #a = 1
WHILE (#a <= #i)
BEGIN IF (#i % #a = 0) SET #count = #count + 1 SET #a = #a + 1
END IF (#count = 2) SET #P = CONCAT(#P,CONCAT(#i,'&')) SET #i = #i + 1
END
PRINT LEFT(#P, LEN(#P) - 1)
The below code works to find prime numbers in SQL
Tested on SampleDB of local server
CREATE procedure sp_PrimeNumber(#number int)
as
begin
declare #i int
declare #j int
declare #isPrime int
set #isPrime=1
set #i=2
set #j=2
while(#i<=#number)
begin
while(#j<=#number)
begin
if((#i<>#j) and (#i%#j=0))
begin
set #isPrime=0
break
end
else
begin
set #j=#j+1
end
end
if(#isPrime=1)
begin
SELECT #i
end
set #isPrime=1
set #i=#i+1
set #j=2
end
end
I have created the stored procedure which has a parameter #number to find the prime numbers up to that given number
In order to get the prime numbers we can execute the below stored procedure
EXECUTE sp_PrimeNumber 100 -- gives prime numbers up to 100
If you are new to stored procedures and want to find the prime numbers in SQL we can use the below code
Tested on master DB
declare #i int
declare #j int
declare #isPrime int
set #isPrime=1
set #i=2
set #j=2
while(#i<=100)
begin
while(#j<=100)
begin
if((#i<>#j) and (#i%#j=0))
begin
set #isPrime=0
break
end
else
begin
set #j=#j+1
end
end
if(#isPrime=1)
begin
SELECT #i
end
set #isPrime=1
set #i=#i+1
set #j=2
end
This code can give the prime numbers between 1 to 100. If we want to find more prime numbers edit the #i and #j arguments in the while loop and execute
Simple query in PostgreSQL:
SELECT serA.el AS prime
FROM generate_series(2, 100) serA(el)
LEFT JOIN generate_series(2, 100) serB(el) ON serA.el >= POWER(serB.el, 2)
AND serA.el % serB.el = 0
WHERE serB.el IS NULL
Enjoy! :)
For SQL Server We can use below CTE
SET NOCOUNT ON
;WITH Prim AS
(
SELECT 2 AS Value
UNION ALL
SELECT t.Value+1 AS VAlue
FROM Prim t
WHERE t.Value < 1000
)SELECT *
FROM Prim t
WHERE NOT EXISTS( SELECT 1 FROM prim t2
WHERE t.Value % t2.Value = 0
AND t.Value != t2. Value)
OPTION (MAXRECURSION 0)
One simple one can be like this
select level id1 from dual connect by level < 2001
minus
select distinct id1 from (select level id1 from dual connect by level < 46) t1 inner join (select level id2 from dual connect by level < 11) t2
on 1=1 where t1.id1> t2.id2 and mod(id1,id2)=0 and id2<>1
Simplest method For SQL Server
DECLARE #range int = 1000, #x INT = 2, #y INT = 2
While (#y <= #range)
BEGIN
while (#x <= #y)
begin
IF ((#y%#x) =0)
BEGIN
IF (#x = #y)
PRINT #y
break
END
IF ((#y%#x)<>0)
set #x = #x+1
end
set #x = 2
set #y = #y+1
end
MySQL QUERY SOLUTION
I have solved this problem in mysql which is following:
SET #range = 1000;
SELECT GROUP_CONCAT(R2.n SEPARATOR '&')
FROM (
SELECT #ctr2:=#ctr2+1 "n"
FROM information_schema.tables R2IS1,
information_schema.tables R2IS2,
(SELECT #ctr2:=1) TI
WHERE #ctr2<#range
) R2
WHERE NOT EXISTS (
SELECT R1.n
FROM (
SELECT #ctr1:=#ctr1+1 "n"
FROM information_schema.tables R1IS1,
information_schema.tables R1IS2,
(SELECT #ctr1:=1) I1
WHERE #ctr1<#range
) R1
WHERE R2.n%R1.n=0 AND R2.n>R1.n
)
Note: No. of information_schema.tables should be increased for more range e.g. if range is 100000 so set the info tables by checking yourself.
--Create Table prime_number_t
create table prime_number_t (
integervalue_c integer not null primary key
);
--Insert Data into table prime_number_t
INSERT ALL
into prime_number_t(integervalue_c) values (1)
into prime_number_t(integervalue_c) values (2)
into prime_number_t(integervalue_c) values (3)
into prime_number_t(integervalue_c) values (4)
into prime_number_t(integervalue_c) values (5)
into prime_number_t(integervalue_c) values (6)
into prime_number_t(integervalue_c) values (7)
into prime_number_t(integervalue_c) values (8)
into prime_number_t(integervalue_c) values (9)
into prime_number_t(integervalue_c) values (10)
SELECT 1 FROM DUAL;
COMMIT;
--Write an SQL statement to determine which of the below numbers are prime numbers
--same query works for REMAINDER function also instead of MOD function
WITH cte_prime_number_t AS
(
select integervalue_c
from prime_number_t
order by integervalue_c
),
cte_maxval AS
(
select max(integervalue_c) AS maxval FROM cte_prime_number_t
),
cte_level AS
(
select LEVEL+1 as lvl
from dual,
cte_maxval
CONNECT BY LEVEL <= cte_maxval.maxval
)
SELECT DISTINCT cpnt.integervalue_c as PrimeNumbers
FROM cte_prime_number_t cpnt
inner join cte_level cl on lvl <= (SELECT maxval FROM cte_maxval)
WHERE NOT EXISTS (
SELECT 1 FROM cte_level cpn
WHERE cpnt.integervalue_c > cpn.lvl AND mod(cpnt.integervalue_c,cpn.lvl) = 0
)
order by PrimeNumbers;
For MySQL 8 or above
/* create a table with one row and that starts with 2 ends at 1000*/
SET cte_max_recursion_depth = 1001; /* works for MySQL 8.0*/
;WITH RECURSIVE sequence AS (
SELECT 1 AS l
UNION ALL
SELECT l + 1 AS value
FROM sequence
WHERE sequence.l < 1000
),
/* create a caretesian product of a number to other numbers uptil this very number
so for example if there is a value 5 in a row then it creates these rows using the table below
(5,2), (5,3), (5,4), (5,5) */
J as (
SELECT (a.l) as m , (b.l) as n
FROM sequence a, sequence b
WHERE b.l <= a.l)
,
/*take a row from column 1 then divide it with other column values but group by column 1 first,
note the completely divisible count*/
f as
( SELECT m , SUM(CASE WHEN mod(m,n) = 0 THEN 1 END) as fact
FROM J
GROUP BY m
HAVING fact = 2
ORDER BY m ASC /*this view return numbers in descending order so had to use order by*/
)
/* this is for string formatting, converting a column to a string with separator &*/
SELECT group_concat(m SEPARATOR '&') FROM f;
This worked for me in MySql:
select '2&3&5&7&11&13&17&19&23&29&31&37&41&43&47&53&59&61&67&71&73&79&83&89&97&101&103&107&109&113&127&131&137&139&149&151&157&163&167&173&179&181&191&193&197&199&211&223&227&229&233&239&241&251&257&263&269&271&277&281&283&293&307&311&313&317&331&337&347&349&353&359&367&373&379&383&389&397&401&409&419&421&431&433&439&443&449&457&461&463&467&479&487&491&499&503&509&521&523&541&547&557&563&569&571&577&587&593&599&601&607&613&617&619&631&641&643&647&653&659&661&673&677&683&691&701&709&719&727&733&739&743&751&757&761&769&773&787&797&809&811&821&823&827&829&839&853&857&859&863&877&881&883&887&907&911&919&929&937&941&947&953&967&971&977&983&991&997';
Well, I know the above one is just hardcoded and you will be able to run the problem but it's not what we should go for as a programmer so here is my solution for oracle:
SELECT LISTAGG(L1,'&') WITHIN GROUP (ORDER BY L1) FROM (Select L1 FROM (SELECT LEVEL L1 FROM DUAL CONNECT BY LEVEL<=1000) Where L1 <> 1 MINUS select L1 from (SELECT LEVEL L1 FROM DUAL CONNECT BY LEVEL<=1000) A , (SELECT LEVEL L2 FROM DUAL CONNECT BY LEVEL<=1000) B Where L2<=L1 and MOD(L1,L2)=0 AND L1<>L2 AND L2<>1);
Worked in Oracle:
SELECT LISTAGG(a,'&')
WITHIN GROUP (ORDER BY a)
FROM(WITH x AS(
SELECT level+1 x
FROM dual
CONNECT BY LEVEL <= 999
)
SELECT x.x as a
FROM x
WHERE NOT EXISTS (
SELECT 1 FROM x y
WHERE x.x > y.x AND remainder( x.x, y.x) = 0
));
SELECT GROUP_CONCAT(distinct PRIME_NUMBER SEPARATOR '&')
FROM (SELECT #prime:=#prime + 1 AS PRIME_NUMBER
FROM information_schema.tables
CROSS JOIN (SELECT #prime:=1) r
WHERE #num <1000
) p
WHERE NOT EXISTS (
SELECT * FROM
(SELECT #divisor := #divisor + 1 AS DIVISOR FROM
information_schema.tables
CROSS JOIN (SELECT #divisor:=1) r1
WHERE #divisor <=1000
) d
WHERE MOD(PRIME_NUMBER, DIVISOR) = 0 AND PRIME_NUMBER != DIVISOR) ;
enter code here
Explanation:
The two inner SELECTs (SELECT #prime and SELECT #divisor) create two lists. Both of them contain numbers from 1 to 1000. The first list is the "list of potential primes" and the second is the "list of divisors"
Then, we iterate over the list of the potential primes (the outer SELECT), and for each number from this list we look for divisors (SELECT * FROM clause) that can divide the number without a reminder and are not equal to the number (WHERE MOD... clause). If at least one such divisor exists, the number is not prime and is not selected (WHERE NOT EXISTS... clause).
I have a MySQL table with the following structure and data:
Increments
id emp_id starting_salary increment_rate increment_frequency
2 340 5000 250 1
3 340 5000 250 4
I need to have aliases, a and b which will hold some value based on the following formula:
starting_salary + (increment_rate * increment_frequency)
To be precise, I want a = 5250 (based on a = (5000 + (250 * 1))) and b = 6000 (based on b = (5000 + (250 * 4)))
Now I have another table with the following data:
PaySlips
id employee_id salary_month arrear
173824 340 '2015-06-01' 2350
I want to join a and b that I got from the table Increments with table PaySlips. And I want to use a and b in the following way:
((a * 8) / 30 + (b * 22) / 30)
My alias will be basic_salary. So basic_salary will hold this value from the above calculation:
basic_salary = ((a * 8) / 30 + (b * 22) / 30)
= ((5250 * 8) / 30 + (6000 *22) / 30)
= (1400 + 4400)
= 5800
I've got no idea how to do this. Can anyone please help me?
All I got are the common columns in both tables - emp_id and employee_id and I can join both tables using these columns. I just can't figure out how I can store the above values and organize the calculation inside my query.
Sample query:
SELECT x.id, x.employee_id,
(*my calculation using a and b from table Increments*) AS basic_salary,
x.salary_month, x.arrear
FROM PaySlips x
JOIN Increments y
ON x.employee_id = y.emp_id
For determining a:
SELECT
(
starting_salary +
(increment_rate * increment_frequency)
) AS a
FROM Increments
WHERE id = 2
And for determining b:
SELECT
(
starting_salary +
(increment_rate * increment_frequency)
) AS b
FROM Increments
WHERE id = 3
MySQL version: 5.2
I'm not clear on all the details, for example what should happen if there are three rows for one employee in increments? Anyhow, here's a sketch to start with:
select employee_id
, ((a * 8) / 30 + (b * 22) / 30) as basic_salary
from (
select x.employee_id
, min(starting_salary + (increment_rate * increment_frequency)) as a
, max(starting_salary + (increment_rate * increment_frequency)) as b
, x.salary_month, x.arrear
from payslips x
join increments y
on x.employee_id = y.emp_id
group by x.employee_id, x.salary_month, x.arrear
) as t
If id 2 and 3 are the criteria (I assumed they are examples) you can use a case statement like:
select employee_id
, ((a * 8) / 30 + (b * 22) / 30) as basic_salary
from (
select x.employee_id
, max(starting_salary + (increment_rate * case when y.id = 2 then increment_frequency end )) as a
, max(starting_salary + (increment_rate * case when y.id = 3 then increment_frequency end)) as b
, x.salary_month
, x.arrear
from payslips x
join increments y
on x.employee_id = y.emp_id
group by x.employee_id, x.salary_month, x.arrear
) as t;
In this case it does not matter what aggregate you use, it is to get rid of the row that contains null.
based on the requirements you added i think something like this will solve your problems:
SELECT PS.id, PS.employee_id, ((A.value * 8) / 30 + (B.value * 22) / 30) AS basic_salary
FROM PaySlips AS PS
JOIN (
SELECT I.emp_id, I.starting_salary + (increment_rate * increment_frequency) AS VALUE
FROM Increments AS I
WHERE I.id = 2
) AS A
ON A.emp_id = PS.employee_id
JOIN (
SELECT I.emp_id, I.starting_salary + (increment_rate * increment_frequency) AS value
FROM Increments AS I
WHERE I.id = 3
) AS B
ON B.emp_id = PS.employee_id
This version might need some alteration if it's not working on your real scenario, but please feel free to tell if anything else needs amending.
Hope it helps.
For determining and setting #a variable:
SET #a := (SELECT
(
starting_salary +
(increment_rate * increment_frequency)
) AS a
FROM Increments
WHERE id = 2);
And for determining and setting #b variable:
SET #b := (SELECT
(
starting_salary +
(increment_rate * increment_frequency)
) AS b
FROM Increments
WHERE id = 3);
Then you can use #a and #b in your main query;
you can test simply by
SELECT #a as a;
SELECT #b as b;
SELECT
x.id,
x.employee_id,
(y.a * 8) / 30 + (y.b * 22) / 30 as basic_salary,
x.salary_month,
x.arrear
FROM PaySlips x
JOIN (
select t1.emp_id, t1.a, t2.b
from (
select
emp_id,
starting_salary + increment_rate * increment_frequency as a
from Increments
where id = 2
) as t1
join (
select
emp_id,
starting_salary + increment_rate * increment_frequency as b
from Increments
where id = 3
) as t2
on t1.emp_id = t2.emp_id
) as y
ON x.employee_id = y.emp_id