條件計數

Question

如果我有像這樣的表：

jobId, jobName, Priority 

。由此可以優先於5.

1之間的整數。由於我需要此查詢生成的報告圖表，我需要顯示jobid，jobname和稱爲Priority1，Priority2，Priority3，Priority4的5個字段。 Priority5。

應的優先級爲計數，其中優先級字段具有1

的值的行的量

優先級2應計數，其中優先級字段具有2

的值的行的量

Priority3應該數的量行，其中優先級字段的值爲3

等價值

我會怎麼做，在一個快速，高性能的方式？

非常感謝， 卡瓦

Answer 1

我想你可能是後

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 

但我不確定是否你需要在你的結果中找到jobID和JobName，如果這樣刪除它們並刪除組，

Answer 2

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 3

我需要顯示作業ID，作業名和第5場被稱爲優先級爲1，優先級2，Priority3，Priority4。 Priority5。

你的查詢設計出了問題。你也在每一行顯示一個特定的工作，所以你可能會遇到這樣一種情況，行中有四個優先級列'0'和一個優先級列'1'（該作業的優先級）否則你最終會重複每行所有優先級的計數。

你真的想在這裏展示什麼？

Answer 4

使用ANSI SQL-92 CASE語句，你可以做這樣的事情（派生表加時）：

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

你可以自己加入表格：

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 

或者你可以使用情況之內：

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 6

使用COUNT代替SUM消除了一個ELSE語句的要求：

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 7

IIF不是一個標準的SQL構造，但如果它受到數據庫的支持，則可以實現更加優雅的語句，從而產生相同的結果：

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 8

SELECT Count（Student_ID）as'StudentCount'FROM CourseSemOne where Student_ID = 3有計數（Student_ID）< 6和Count（Student_ID）> 0;