Count records for every month in a year
SqlSql ServerSql Server-2008Sql Problem Overview
I have a table with total no of 1000 records in it.It has the following structure:
EMP_ID EMP_NAME PHONE_NO ARR_DATE
1 A 545454 2012/03/12
I want to calculate no of records for every month in year-2012
Is there any way that should solve my issue in a single shot?
I tried:
select count(*)
from table_emp
where year(ARR_DATE) = '2012' and month(ARR_DATE) = '01'
Sql Solutions
Solution 1 - Sql
SELECT COUNT(*)
FROM table_emp
WHERE YEAR(ARR_DATE) = '2012'
GROUP BY MONTH(ARR_DATE)
Solution 2 - Sql
This will give you the count per month for 2012;
SELECT MONTH(ARR_DATE) MONTH, COUNT(*) COUNT
FROM table_emp
WHERE YEAR(arr_date)=2012
GROUP BY MONTH(ARR_DATE);
Solution 3 - Sql
Try This query:
SELECT
SUM(CASE datepart(month,ARR_DATE) WHEN 1 THEN 1 ELSE 0 END) AS 'January',
SUM(CASE datepart(month,ARR_DATE) WHEN 2 THEN 1 ELSE 0 END) AS 'February',
SUM(CASE datepart(month,ARR_DATE) WHEN 3 THEN 1 ELSE 0 END) AS 'March',
SUM(CASE datepart(month,ARR_DATE) WHEN 4 THEN 1 ELSE 0 END) AS 'April',
SUM(CASE datepart(month,ARR_DATE) WHEN 5 THEN 1 ELSE 0 END) AS 'May',
SUM(CASE datepart(month,ARR_DATE) WHEN 6 THEN 1 ELSE 0 END) AS 'June',
SUM(CASE datepart(month,ARR_DATE) WHEN 7 THEN 1 ELSE 0 END) AS 'July',
SUM(CASE datepart(month,ARR_DATE) WHEN 8 THEN 1 ELSE 0 END) AS 'August',
SUM(CASE datepart(month,ARR_DATE) WHEN 9 THEN 1 ELSE 0 END) AS 'September',
SUM(CASE datepart(month,ARR_DATE) WHEN 10 THEN 1 ELSE 0 END) AS 'October',
SUM(CASE datepart(month,ARR_DATE) WHEN 11 THEN 1 ELSE 0 END) AS 'November',
SUM(CASE datepart(month,ARR_DATE) WHEN 12 THEN 1 ELSE 0 END) AS 'December',
SUM(CASE datepart(year,ARR_DATE) WHEN 2012 THEN 1 ELSE 0 END) AS 'TOTAL'
FROM
sometable
WHERE
ARR_DATE BETWEEN '2012/01/01' AND '2012/12/31'
Solution 4 - Sql
select count(*)
from table_emp
where DATEPART(YEAR, ARR_DATE) = '2012' AND DATEPART(MONTH, ARR_DATE) = '01'