Count records for every month in a year

Question

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'

Answer 1

SELECT    COUNT(*) 
FROM      table_emp 
WHERE     YEAR(ARR_DATE) = '2012' 
GROUP BY  MONTH(ARR_DATE)

Answer 2

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);

Demo here.

Answer 3

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' 

Answer 4

select count(*) 
from table_emp 
 where DATEPART(YEAR, ARR_DATE) = '2012' AND DATEPART(MONTH, ARR_DATE) = '01'