SQL生成多个计数列

Question

我开发了一些SQL来生成我们的本地故障单数据库表（SQL Server 2008）的统计信息。从我的代码中可以看到，我想从TICKETS中选择，加入组以获取组名称，按组代码以及年/月分组。

我想创建总计（计数）多少张门票是打开，关闭，在sla（过期日期）之外关闭的，最终是SLA％。

此代码有效，但我不满意编码所有嵌套（选择计数）;这似乎不是一个很好的策略与多次重新扫描。

是否有更好的设计来从桌面上的单个选择生成多个“计数”列......或者这是标准方法吗？

select g.group_name as [Group], 
     year(tm.date_open) as Year, 
     month(tm.date_open) as Month, 
     COUNT(*) as [Tickets Opened], 

     (select COUNT(*) 
     from TICKETS tm2 
     where tm2.group_code = tm.group_code 
     and year(tm2.COMPLETION_DATE) = year(tm.date_open) 
     and month(tm2.COMPLETION_DATE) = month(tm.date_open) 
     ) as [Tickets Closed], 

     (select COUNT(*) 
     from TICKETS tm2 
     where tm2.group_code = tm.group_code 
     and year(tm2.COMPLETION_DATE) = year(tm.date_open) 
     and month(tm2.COMPLETION_DATE) = month(tm.date_open) 
     and tm2.[COMPLETION_DATE] <= tm2.[DUE_DATE:] 
     ) as [Closed Within SLA], 

     (select COUNT(*) 
     from TICKETS tm2 
     where tm2.group_code = tm.group_code 
     and year(tm2.COMPLETION_DATE) = year(tm.date_open) 
     and month(tm2.COMPLETION_DATE) = month(tm.date_open) 
     and tm2.[COMPLETION_DATE] > tm2.[DUE_DATE:] 
     ) as [Closed Outside SLA] --service level agreement 


from TICKETS tm 
left join GROUPS g on g.group_code = tm.group_code 
where g.group_code in ('techs', 'reps', 'phone') 

and year(tm.date_open) = 2015 
--and month(tm.date_open) = 3 -- specific month 

group by tm.group_code, g.group_name, year(tm.date_open), month(tm.date_open) 

order by g.group_name, year(tm.date_open), month(tm.date_open) 

，我也要去想添加SLA％列将被（[关闭在SLA]/[关闭门票]）* 100。但是，当我看到它从我目前的设计，我会为另一列添加冗余嵌套SELECTS，例如...

(
    cast((select COUNT(*) 
    from TICKETS tm2 
    where tm2.group_code = tm.group_code 
    and year(tm2.COMPLETION_DATE) = year(tm.date_open) 
    and month(tm2.COMPLETION_DATE) = month(tm.date_open) 
    and tm2.[COMPLETION_DATE] <= tm2.[DUE_DATE:] 
    ) as decimal)/
    (select COUNT(*) 
    from TICKETS tm2 
    where tm2.group_code = tm.group_code 
    and year(tm2.COMPLETION_DATE) = year(tm.date_open) 
    and month(tm2.COMPLETION_DATE) = month(tm.date_open) 
    ) 
) * 100 as [SLA%] 

Answer 1

是的，你可以比一堆子查询做得更好。在单个聚合查询中，通过计算满足条件时评估为1的各种表达式，可以获得满足不同条件的单独行数，否则为零。或者，您可以使用COUNT()表达式计算出您想计算的那些行的非NULL。例如，它看起来像这样可能接近你之后：

SELECT 
    g.group_name AS [Group], 
    year(tm.date_open) as Year, 
    month(tm.date_open) as Month, 
    COUNT(*) AS [Tickets Opened], 
    COUNT(tm.completion_date) AS [Tickets Closed], 
    SUM(CASE WHEN tm.completion_date <= [DUE_DATE:] THEN 1 ELSE 0 END) 
    AS [Closed Within SLA], 
    SUM(CASE WHEN tm.completion_date > [DUE_DATE:] THEN 1 ELSE 0 END) 
    AS [Closed Outside SLA], --service level agreement 
    CAST(SUM(CASE WHEN tm.completion_date <= [DUE_DATE:] THEN 1 ELSE 0 END) 
    AS decimal)/COUNT(tm.completion_date)) AS [SLA%], 
FROM 
    tickets tm 
    LEFT JOIN GROUPS g 
    ON g.group_code = tm.group_code 
WHERE 
    g.group_code in ('techs', 'reps', 'phone') 
    and year(tm.date_open) = 2015 
    -- and month(tm.date_open) = 3 -- specific month 
GROUP BY 
    tm.group_code, 
    g.group_name, 
    year(tm.date_open), 
    month(tm.date_open) 
ORDER BY 
    g.group_name, 
    year(tm.date_open), 
    month(tm.date_open)