Get sum of column base on equality of values of one row From previous generate request( using SELECT and EXTRACT)

Mostafazs

Hello I can't find solution for this simple query.Want to get sum of column of `VISIT_FEE` that have colum `MONTH_` 2 and `YEAR_` 1396 My main table is like this:

ID VISIT_DATE VISIT_TIME VISIT_FEE IS_PAY
1 13960124 10:00 300000 1
6 13960208 10:50 2500000 1
7 13960208 11:00 210000 1
8 13960209 10:20 300000 1

and now i use below query:

SELECT
VISIT_FEE,
EXTRACT(YEAR FROM VISIT_DATE) AS YEAR_,
EXTRACT(MONTH FROM VISIT_DATE) AS MONTH_,
EXTRACT(DAY FROM VISIT_DATE) AS DAY_

FROM
MZS_VISIT_REQUEST
WHERE
IS_PAY = 1

and result is :

VISIT_FEE YEAR_ MONTH_ DAY_
300000 1396 1 1
2500000 1396 2 2
210000 1396 2 2
300000 1396 2 2

And now i want some of colum `VISIT_FEE` that have YEAR_ 1396 and MONTH_ 2 and 1 somethings like below table:

FEE YEAR_ MONTH_ DAY_
300000 1396 1 1
3010000 1396 2 2

Thanks.

Maciej Los

Please replace:

EXTRACT(MONTH FROM VISIT_DATE) AS DAY_

with:

EXTRACT(DAY FROM VISIT_DATE) AS DAY_

and everything should be OK. For further details, please see: [EXTRACT (datetime)](https://docs.oracle.com/cd/B28359\_01/server.111/b28286/functions052.htm#SQLRF00639)

Mostafazs

Thanks for replay. but this is just my misspelling and nothings changed on result. my show result table table is correct. I want to get sum of column `VISIT_FEE`

anarayane

SELECT
sum(VISIT_FEE) AS VISIT_FEE,
EXTRACT(YEAR FROM VISIT_DATE) AS YEAR_,
EXTRACT(MONTH FROM VISIT_DATE) AS MONTH_,

FROM
MZS_VISIT_REQUEST
WHERE
IS_PAY = 1
GROUP by
EXTRACT(YEAR FROM VISIT_DATE) AS YEAR_,
EXTRACT(MONTH FROM VISIT_DATE) AS MONTH_

Please try this one

Maciej Los

All you need to do is to use [SUM()](https://www.techonthenet.com/oracle/functions/sum.php) - one of [aggragate functions](https://docs.oracle.com/database/121/SQLRF/functions003.htm#SQLRF20035):

SELECT SUM(VISIT_FEE) AS FEE,
EXTRACT(YEAR FROM VISIT_DATE) AS YEAR_,
EXTRACT(MONTH FROM VISIT_DATE) AS MONTH_,
EXTRACT(DAY FROM VISIT_DATE) AS DAY_
FROM MZS_VISIT_REQUEST
WHERE IS_PAY = 1
GROUP BY EXTRACT(YEAR FROM VISIT_DATE),
EXTRACT(MONTH FROM VISIT_DATE) ,
EXTRACT(DAY FROM VISIT_DATE)

Good luck

Mostafazs

Thanks. :^) But it's not what i want. I want sum of each month like: FEE YEAR_ MONTH_ 300000 1396 1 3010000 1396 2 Sum of `Fee` column if `MONTH_` is equal. like above table.

Mostafazs

Thanks. But when running i get below error: `Err] 1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'AS AYEAR_, EXTRACT(MONTH FROM VISIT_DATE) AS BMONTH_' at line 10` When i delete last part(ORDER BY), query without any error with below result FEE YEAR_ MONTH_ 3310000 1396 1 is this any things we can do like --> ` SELECT SUM(VISIT_FEE) AS FEE, EXTRACT(YEAR FROM VISIT_DATE) AS YEAR_, EXTRACT(MONTH FROM VISIT_DATE) AS MONTH_ FROM MZS_VISIT_REQUEST WHERE IS_PAY = 1 AND DISTINCT MONTH_ `???