Question

select *, COUNT(*), DATE_FORMAT(CREATE_DATE,'%m-%Y') AS form_date
from incident_view
where (create_month = month(NOW() - INTERVAL 1 MONTH)
       and (create_year =  year(NOW() - INTERVAL 1 MONTH)))  
   OR (create_month = month(NOW() - INTERVAL 2 MONTH)
       and (create_year = year(NOW() - INTERVAL 2 MONTH)))
AND CUSTOMER_COMPANY_NAME = "Company"
GROUP BY CREATE_MONTH

大家好，

我上面的查询工作正常。

我得到的结果是一些行，但重要的行是：

COUNT(*)  | form_date
667          01-16
1964         02-16

我想知道是否可以比较过去2个月的两个计数，无论是上个月（02-16）＆gt;上个月的第二个月（01-16）。如果02-16> 01-16我希望结果为真，如果不是则为假。

感谢任何帮助。

问候。

Answer 1

不是按月/年分组，而是在计算计数时在CASE或IF中使用。

SELECT SUM(IF(create_month = MONTH(NOW() - INTERVAL 1 MONTH)
              AND create_year = YEAR(NOW() - INTERVAL 1 MONTH), 1, 0)) > 
       SUM(IF(create_month = MONTH(NOW() - INTERVAL 2 MONTH)
              AND create_year = YEAR(NOW() - INTERVAL 2 MONTH), 1, 0)) AS count_higher
FROM incident_view
WHERE customer_company_name = "Company"

如果你想要3个不同的结果大于，等于或小于，最好的方法是计算子查询中的总和，这样你就可以命名它们并使用CASE为每个案例返回不同的值

SELECT CASE WHEN last_month > prev_month THEN 1
            WHEN last_month = prev_month THEN 2
            ELSE 0
       END AS diff
FROM (
    SELECT SUM(IF(create_month = MONTH(NOW() - INTERVAL 1 MONTH)
                  AND create_year = YEAR(NOW() - INTERVAL 1 MONTH), 1, 0)) AS last_month, 
           SUM(IF(create_month = MONTH(NOW() - INTERVAL 2 MONTH)
                  AND create_year = YEAR(NOW() - INTERVAL 2 MONTH), 1, 0)) AS prev_month
    FROM incident_view
    WHERE customer_company_name = "Company"
) AS subquery

Answer 2

您可以执行类似

的操作

Select max(viewcount)
FROM (select * , COUNT(*) AS viewcount,
DATE_FORMAT(CREATE_DATE,'%m-%Y') AS form_date
from incident_view
Where (create_month = month(NOW() - INTERVAL 1 MONTH) and (create_year =  year(NOW() - INTERVAL 1 MONTH)))  
OR (create_month = month(NOW() - INTERVAL 2 MONTH) and (create_year = year(NOW() - INTERVAL 2 MONTH)))
AND CUSTOMER_COMPANY_NAME = "Company"
GROUP BY CREATE_MONTH);

比较两个数字并接收真或假？

2 个答案: