Question

我有一张数据表，其中记录了过去几周的通过和失败。

我想计算每周的尝试次数以及通过和失败的次数，然后取这些通过和失败的总和作为检查。

SELECT week, 
    COUNT (Week) as weekcount, 

    sum(case when result = 'pass' then 1 else 0 end) as 'passcount' ,
    sum(case when result = 'fail' then 1 else 0 end) as 'failcount',
        sum(case when result = 'pass' then 1 else 0 end) +
        sum(case when result = 'fail' then 1 else 0 end) as 'sum'

FROM table1
group by week
order by week

上面的代码段可以满足我的要求，但是我一直在阅读有关CTE的信息，以代替在查询中进行重复操作。

我希望能够在查询中使用别名“ passcount”和“ failcount”，而无需将它们作为列添加到表中。我尝试先在下面设置CTE。

with sums as
(
    select result,
    sum(case when result = 'pass' then 1 else 0 end) as 'passcount' ,
    sum(case when result = 'fail' then 1 else 0 end) as 'failcount',
    from table1 
)

以上报告了右括号附近的语法错误，但我不明白为什么。

任何人都可以阐明这种方法或更好的方法吗？

Answer 1

在SQL Server中，我认为横向联接（即^compose）比子查询和CTE更清洁。在您的情况下，您可以在聚合之前生成标志，因此如下所示：

cross apply

Answer 2

您仅需再进行一次选择就可以了

using namespace std::chrono;
system_clock::time_point tp{system_clock::duration{i}};

Answer 3

您实际上不需要同时编写两个CASE表达式，而只需使用IN：

SELECT Week,
       COUNT(Week) AS weekcount,
       COUNT(CASE Result WHEN 'pass' THEN 1 END) AS PassCount,
       COUNT(CASE Result WHEN 'fail' THEN 1 END) AS FailCount,
       COUNT(CASE WHEN Result IN ('pass','fail') THEN 1 END) AS PassFailCount
FROM Table1
GROUP BY Week
ORDER BY Week;

是的，这没有告诉您如何使用CTE，但是，与重复两个CASE表达式相比，逻辑是“更干净”（更简洁）。

Answer 4

我不确定您为什么要考虑CTE，您可以将它设置为一组：

SELECT 
        [Week], 
        COUNT([Week]) AS weekcount
    ,   SUM(CASE WHEN result = 'pass' THEN 1 ELSE 0 END) AS passcount   
    ,   SUM(CASE WHEN result = 'fail' THEN 1 ELSE 0 END) AS failcount
    ,   COUNT(*) AS TotalSum
FROM table1
GROUP BY [Week]

在SQL Server中，您不能引用同一SELECT中的别名，例如密码，周数等。这样，您就需要一种变通方法，例如使用子查询，CTE，JOIN，CROSS APPLY或任何其他可用于此目的的方法。

以下是一些常用方法：

使用子查询：

SELECT 
    [Week]
,   SUM(weekcount) weekcount
,   SUM(passcount) passcount
,   SUM(failcount) failcount
,   SUM(passcount + failcount) AS TotalSum
FROM (
    SELECT 
        [Week], 
        COUNT([Week]) AS weekcount
    ,   SUM(CASE WHEN result = 'pass' THEN 1 ELSE 0 END) AS passcount   
    ,   SUM(CASE WHEN result = 'fail' THEN 1 ELSE 0 END) AS failcount
    FROM table1
    GROUP BY [Week]
) D 
GROUP BY 
    [Week]

使用CTE 方法：

;WITH CTE AS (
    SELECT 
        [Week], 
        COUNT([Week]) AS weekcount
    ,   SUM(CASE WHEN result = 'pass' THEN 1 ELSE 0 END) AS passcount   
    ,   SUM(CASE WHEN result = 'fail' THEN 1 ELSE 0 END) AS failcount
    FROM table1
    GROUP BY [Week]
)
SELECT      
        [Week]
    ,   SUM(weekcount) weekcount
    ,   SUM(passcount) passcount
    ,   SUM(failcount) failcount
    ,   SUM(passcount + failcount) AS TotalSum
FROM CTE 
GROUP BY 
    [Week]

使用交叉应用方法：请参阅@Gordon Linoff答案（基本而直接）。

这些是您可以在案例中使用的最简单的方法。您也可以在案例中使用PIVOT，但是为了简单起见，我没有提到它。

尝试使用通用表表达式对列标题的别名求和

4 个答案: