Question

如果我有这样的桌子：

jobId, jobName, Priority

其中，优先级可以是1到5之间的整数。

由于我需要此查询来生成报表上的图表，我需要显示jobid，jobname和名为Priority1，Priority2，Priority3，Priority4的5个字段。 Priority5。

Priority1应计算优先级字段值为1的行数。

Priority2应计算优先级字段值为2的行数。

Priority3应计算优先级字段值为3的行数。

等

我如何以快速和高效的方式做到这一点？

非常感谢，卡瓦

Answer 1

我想你可能会追求

select 
    jobID, JobName,
    sum(case when Priority = 1 then 1 else 0 end) as priority1,
    sum(case when Priority = 2 then 1 else 0 end) as priority2,
    sum(case when Priority = 3 then 1 else 0 end) as priority3,
    sum(case when Priority = 4 then 1 else 0 end) as priority4,
    sum(case when Priority = 5 then 1 else 0 end) as priority5
from
    Jobs
group by 
    jobID, JobName

但是我不确定你是否需要在结果中使用jobID和JobName，如果删除它们并删除该组，

Answer 2

使用COUNT而不是SUM将删除对ELSE语句的要求：

SELECT jobId, jobName,
    COUNT(CASE WHEN Priority=1 THEN 1 END) AS Priority1,
    COUNT(CASE WHEN Priority=2 THEN 1 END) AS Priority2,
    COUNT(CASE WHEN Priority=3 THEN 1 END) AS Priority3,
    COUNT(CASE WHEN Priority=4 THEN 1 END) AS Priority4,
    COUNT(CASE WHEN Priority=5 THEN 1 END) AS Priority5
FROM TableName
GROUP BY jobId, jobName

Answer 3

IIF不是标准的SQL构造，但如果数据库支持它，您可以获得更优雅的语句，从而产生相同的结果：

SELECT JobId, JobName,

COUNT(IIF (Priority=1, 1, NULL)) AS Priority1,
COUNT(IIF (Priority=2, 1, NULL)) AS Priority2,
COUNT(IIF (Priority=3, 1, NULL)) AS Priority3,
COUNT(IIF (Priority=4, 1, NULL)) AS Priority4,
COUNT(IIF (Priority=5, 1, NULL)) AS Priority5

FROM TableName
GROUP BY JobId, JobName

Answer 4

使用ANSI SQL-92 CASE语句，您可以执行以下操作（派生表加案例）：

 SELECT jobId, jobName, SUM(Priority1)
 AS Priority1, SUM(Priority2) AS
 Priority2, SUM(Priority3) AS
 Priority3, SUM(Priority4) AS
 Priority4,  SUM(Priority5) AS
 Priority5 FROM (
     SELECT jobId, jobName,
     CASE WHEN Priority = 1 THEN 1 ELSE 0 END AS Priority1,
     CASE WHEN Priority = 2 THEN 1 ELSE 0 END AS Priority2,
     CASE WHEN Priority = 3 THEN 1 ELSE 0 END AS Priority3,
     CASE WHEN Priority = 4 THEN 1 ELSE 0 END AS Priority4,
     CASE WHEN Priority = 5 THEN 1 ELSE 0 END AS Priority5
     FROM TableName


)

Answer 5

SELECT  Priority, COALESCE(cnt, 0)
FROM    (
        SELECT  1 AS Priority
        UNION ALL
        SELECT  2 AS Priority
        UNION ALL
        SELECT  3 AS Priority
        UNION ALL
        SELECT  4 AS Priority
        UNION ALL
        SELECT  5 AS Priority
        ) p
LEFT JOIN
        (
        SELECT  Priority, COUNT(*) AS cnt
        FROM    jobs
        GROUP BY
                Priority
        ) j
ON      j.Priority = p.Priority

Answer 6

你可以自己加入桌子：

select
   t.jobId, t.jobName,
   count(p1.jobId) as Priority1,
   count(p2.jobId) as Priority2,
   count(p3.jobId) as Priority3,
   count(p4.jobId) as Priority4,
   count(p5.jobId) as Priority5
from
   theTable t
   left join theTable p1 on p1.jobId = t.jobId and p1.jobName = t.jobName and p1.Priority = 1
   left join theTable p2 on p2.jobId = t.jobId and p2.jobName = t.jobName and p2.Priority = 2
   left join theTable p3 on p3.jobId = t.jobId and p3.jobName = t.jobName and p3.Priority = 3
   left join theTable p4 on p4.jobId = t.jobId and p4.jobName = t.jobName and p4.Priority = 4
   left join theTable p5 on p5.jobId = t.jobId and p5.jobName = t.jobName and p5.Priority = 5
group by
   t.jobId, t.jobName

或者您可以在总和中使用案例：

select
   jobId, jobName,
   sum(case Priority when 1 then 1 else 0 end) as Priority1,
   sum(case Priority when 2 then 1 else 0 end) as Priority2,
   sum(case Priority when 3 then 1 else 0 end) as Priority3,
   sum(case Priority when 4 then 1 else 0 end) as Priority4,
   sum(case Priority when 5 then 1 else 0 end) as Priority5
from
   theTable
group by
   jobId, jobName

Answer 7

我需要显示jobid，jobname和名为Priority1，Priority2，Priority3，Priority4的5个字段。 Priority5。

您的查询设计出了问题。您在每一行中都显示了一个特定的作业，因此您可能会遇到行有四个优先级列为'0'和一个优先级列为'1'（该作业的优先级）的情况或者你最终会重复计算每一行的所有优先级。

你真的想在这里展示什么？

Answer 8

试试这个：

SELECT Count(Student_ID) as 'StudentCount' 
FROM CourseSemOne
where Student_ID=3 
Having Count(Student_ID) < 6 and Count(Student_ID) > 0;

字段上的条件计数

8 个答案: