Count records for every month in a year

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囚心锁ツ
囚心锁ツ 2020-12-02 20:16

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


        
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  • 2020-12-02 20:40

    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.

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  • 2020-12-02 20:47
    select count(*) 
    from table_emp 
     where DATEPART(YEAR, ARR_DATE) = '2012' AND DATEPART(MONTH, ARR_DATE) = '01'
    
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  • 2020-12-02 20:49
    SELECT    COUNT(*) 
    FROM      table_emp 
    WHERE     YEAR(ARR_DATE) = '2012' 
    GROUP BY  MONTH(ARR_DATE)
    
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  • 2020-12-02 20:54

    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' 
    
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