SQL 计算一年中每个月的记录
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时间：2020-09-01 15:02:00 来源:igfitidea点击:
Count records for every month in a year
提问by manoj kumar singh
I have a table with total no of 1000 records in it.It has the following structure:
我有一个表，其中总共有 1000 条记录。它具有以下结构：
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
我想计算 2012 年每个月的记录数
Is there any way that should solve my issue in a single shot?
有什么方法可以一次性解决我的问题？
I tried:
我试过：
select count(*)
from table_emp
where year(ARR_DATE) = '2012' and month(ARR_DATE) = '01'
回答by Dan
SELECT COUNT(*)
FROM table_emp
WHERE YEAR(ARR_DATE) = '2012'
GROUP BY MONTH(ARR_DATE)
回答by Joachim Isaksson
回答by cichy
Try This query:
试试这个查询：
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'
回答by Darren
select count(*)
from table_emp
where DATEPART(YEAR, ARR_DATE) = '2012' AND DATEPART(MONTH, ARR_DATE) = '01'
