I am new in sql. I want to count something like:
Select count(*) from table where col1= x and col2=x and Col3=x.
I need to count the same value in all different column.
Any help will be appreciated.
You can use conditional aggregation :
Select sum(case when col1='x' then 1 else 0 end) as count_col1,
sum(case when col2='x' then 1 else 0 end) as count_col2,
sum(case when col3='x' then 1 else 0 end) as count_col3
from tab;
If you want to have sum of these count values, consider the above query as an inner and use the following :
Select q.*,
q.count_col1 + q.count_col2 + q.count_col3 whole_sum
from
(
Select sum(case when col1='x' then 1 else 0 end) as count_col1,
sum(case when col2='x' then 1 else 0 end) as count_col2,
sum(case when col3='x' then 1 else 0 end) as count_col3
from tab
) q
Rextester Demo
Related
Im trying to write a query that returns a count depending on the value of a feedback field that ranges from 0-5 (0 meaning that it was not rated).
I want:
Count of all rows ( anything rated 1 or greater)
Count of all rows rated as 1 (anything = 1)
And all rows rated as 1 and also is the first iteration of a given task (anything rated =1 and iteration = 0)
I have written this query but I am getting the same value for all counts:
select
DATE_FORMAT(created_at,'%M') as Month,
COUNT(CASE WHEN rate > 0 THEN 1 ELSE 0 END) AS total,
COUNT(CASE WHEN rate = 1 THEN 1 ELSE 0 END) AS Rated_1,
COUNT(CASE WHEN client_feedback = 1 AND index = 0 THEN 1 ELSE 0 END) AS first_iteration_rated_1
from tablexxx
where created_at between date('2022-04-01') and date('2022-10-01')
GROUP BY Month
Try to use SUM() instead of COUNT().
Count() will count up regardless of the value being 0 or 1.
you can have two approaches:
method 1: use NULL in else part of the CASE
select
DATE_FORMAT(created_at,'%M') as Month,
COUNT(CASE WHEN rate > 0 THEN 1 ELSE null END) AS total,
COUNT(CASE WHEN rate = 1 THEN 1 ELSE null END) AS Rated_1,
COUNT(CASE WHEN client_feedback = 1 AND index = 0 THEN 1 ELSE null END) AS first_iteration_rated_1
from tablexxx
where created_at between date('2022-04-01') and date('2022-10-01')
GROUP BY Month
method 2: use sum instead of count
select
DATE_FORMAT(created_at,'%M') as Month,
SUM(CASE WHEN rate > 0 THEN 1 ELSE 0 END) AS total,
SUM(CASE WHEN rate = 1 THEN 1 ELSE 0 END) AS Rated_1,
SUM(CASE WHEN client_feedback = 1 AND index = 0 THEN 1 ELSE 0 END) AS first_iteration_rated_1
from tablexxx
where created_at between date('2022-04-01') and date('2022-10-01')
GROUP BY Month
I have a table in my database from which i want to fetch data on the basis of count
my table name is syncbill and columns are Cancelled and Noofprints i am trying to write a query to fetch the count of cancelled and count of noofbills using single query
For cancelled i want to fetch count of cancelled where cancelled<>Y
for Noofprints i want to fetch count of noofprints where noofprints>1
I am using
select count(CANCELLED) as canceledbill,count(NOOFPRINTS) as duplicatebill
from syncbill where CANCELLED<>'Y' and NOOFPRINTS>1`
this query but it is giving me the same no of counts for both
any one out-here please guide me
you can use sum() with case expreseeion
SELECT
sum(CASE WHEN CANCELLED <> 'Y' THEN 1 else 0 END) AS canceledbill,
sum(CASE WHEN NOOFPRINTS > 1 THEN 1 else 0 END) AS duplicatebill
FROM table_name;
Use conditional aggregation:
SELECT
COUNT(CASE WHEN CANCELLED <> 'Y' THEN 1 END) AS cancelledbill,
COUNT(CASE WHEN NOOFPRINTS > 1 THEN 1 END) AS duplicatebill
FROM yourTable;
If you are actually using MySQL, then the above can be simplified to this:
SELECT
SUM(CANCELLED <> 'Y') AS cancelledbill,
SUM(NOOFPRINTS > 1) AS duplicatebill
FROM yourTable;
USE CASE WHEN Expression
select
sum(case when CANCELLED <> 'Y' then 1 else 0 end) AS cancelledbill,
sum(case when NOOFPRINTS > 1 then 1 else 0 end) AS duplicatebill
FROM tablename;
Given SQL: Is there a way to bring in the total of the result set?
SELECT
SUM(CASE
WHEN status = 3 THEN 1
ELSE 0
END) AS Open,
SUM(CASE
WHEN status = 4 THEN 1
ELSE 0
END) AS Close
FROM
Table1
WHERE
id = 2;
Result:
Open,Close
5,5
Desired Result:
Open,Close,Total
5,5,10
just add another case statement
SELECT
SUM(CASE
WHEN status = 3 THEN 1
ELSE 0
END) AS Open,
SUM(CASE
WHEN status = 4 THEN 1
ELSE 0
END) AS Close
SUM(CASE
WHEN status IN (3, 4) THEN 1
ELSE 0
END) AS Total
FROM
Table1
WHERE
id = 2;
You can use a CTE:
WITH sumCase AS (
SELECT
SUM(CASE
WHEN status = 3 THEN 1
ELSE 0
END) AS Open,
SUM(CASE
WHEN status = 4 THEN 1
ELSE 0
END) AS Close
FROM
Table1
WHERE
id = 2;)
SELECT Open,Close, Open + Close AS Total FROM Table1;
http://www.mysqltutorial.org/mysql-cte/
using sub-query
select open,close,open+close as total from
(
SELECT
SUM(CASE WHEN status = 3 THEN 1
ELSE 0
END ) AS Open,
SUM(CASE WHEN status = 4 THEN 1
ELSE 0
END) AS Close
FROM
Table1
WHERE id = 2 ) as T
I want to count the occurrences of a string within the column "diagnosis".
What I am doing now is simply this - it gets my needed results
SELECT COUNT(*), diagnosis
FROM patients
WHERE diagnosis like "%OS1%"
UNION
SELECT COUNT(*), diagnosis
FROM patients
WHERE diagnosis like "%OS2%"
... and so on
Sometimes in my table the strings can occur twice (e.g. OS1, OS2), I want to count every single occurence of the strings.
I think it would be a pretty easy task in another language but I want to do it in pure SQL.
Put the OS1-OS6 and CH1-CH6 values in a diagn table. JOIN and GROUP BY:
SELECT COUNT(*), d.diagnosis
FROM patients p
RIGHT JOIN diagn d
on p.diagnosis like concat('%', d.diagnosis, '%')
group by d.diagnosis
SELECT sum(OS1) as OS1_total, sum(OS2) as OS2_total, sum(OS3) as OS3_total,
sum(OS4) as OS4_total, sum(OS5) as OS5_total, sum(OS6) as OS6_total,
sum(CH1) as CH1_total, sum(CH2) as CH2_total, sum(CH3) as CH3_total,
sum(CH4) as CH4_total, sum(CH5) as CH5_total, sum(CH6) as CH6_total
FROM
(
SELECT (case when diagnosis like '%OS1%' then 1 else 0 end) as OS1,
(case when diagnosis like '%OS2%' then 1 else 0 end) as OS2,
(case when diagnosis like '%OS3%' then 1 else 0 end) as OS3,
(case when diagnosis like '%OS4%' then 1 else 0 end) as OS4,
(case when diagnosis like '%OS5%' then 1 else 0 end) as OS5,
(case when diagnosis like '%OS6%' then 1 else 0 end) as OS6,
(case when diagnosis like '%CH1%' then 1 else 0 end) as CH1,
(case when diagnosis like '%CH2%' then 1 else 0 end) as CH2,
(case when diagnosis like '%CH3%' then 1 else 0 end) as CH3,
(case when diagnosis like '%CH4%' then 1 else 0 end) as CH4,
(case when diagnosis like '%CH5%' then 1 else 0 end) as CH5,
(case when diagnosis like '%CH6%' then 1 else 0 end) as CH6
FROM patients
) as mytable
You can try this also:
SELECT COUNT(*), diagnosis
FROM patients
GROUP BY diagnosis;
Not sure if this works.
SELECT COUNT(*)
FROM patients
WHERE REGEXP_LIKE(diagnosis, "(OS[1-6]|CH[1-6])")
You may also check out link below if you're interested. Not sure if it's what you're looking for though:
https://dev.mysql.com/doc/refman/8.0/en/regexp.html
SELECT sum(case when (gender)=1 THEN 1 ELSE 0 END),
GROUP_CONCAT(sum(case when (gender)=2 THEN 1 ELSE 0 END) SEPARATOR ' ') as combine
from family_member_tbl
GROUP BY gender
NO, you can't nest grouping function like that. Rather get the sum first and then group_concat() like
select sum_1, sum_2, group_concat(sum_2) as combine
from (
SELECT gender,
sum(case when gender = 1 THEN 1 ELSE 0 END) as sum_1,
sum(case when gender = 2 THEN 1 ELSE 0 END) as sum_2
from family_member_tbl
GROUP BY gender ) xxx
group by gender;