與案件

Question

我試圖讓選擇了在Postgres以下諮詢：

 Managers | Clerks | Presidents | Analysts | Salesmans 
    -------------+---------+--------------+ -----------+------------- 

     3   4    1    2    4 

到目前爲止，我能做到：

select CASE WHEN lower(job)='salesman' THEN count(job) as "SALESMAN" 
WHEN lower(job)='clerk' THEN count(job) as "CLERK" 
WHEN lower(job)='manager' THEN count(job) as "MANAGER" 
WHEN lower(job)='analyst' THEN count(job) as "ANALYST" 
WHEN lower(job)='president' THEN count(job) as "PRESIDENT" 
from emp 
group by job; 

但我似乎無法運行它，它出現錯誤：

ERROR: syntax error at or near "as"
LINE 1: ... CASE WHEN lower(job)='salesman' THEN count(job) as "SALESM...

如何在select中創建單獨的列？

Answer 1

你需要把計數周圍的case：

select count(CASE WHEN lower(job)='salesman' THEN 1 END) as "SALESMAN" 
     count(CASE WHEN lower(job)='clerk' THEN 1 END) as "CLERK" 
     count(case WHEN lower(job)='manager' THEN 1 END) as "MANAGER" 
     count(case WHEN lower(job)='analyst' THEN 1 END) as "ANALYST" 
     count(case WHEN lower(job)='president' THEN 1 END) as "PRESIDENT" 
from emp; 

像count()聚合函數忽略空值。對於與條件不匹配的值，CASE表達式會返回NULL，因此不計算這些值。

或者使用filter條款簡單：

select count(*) filter (where lower(job)='salesman') as "SALESMAN" 
     count(*) filter (where lower(job)='clerk') as "CLERK" 
     count(*) filter (where lower(job)='manager') as "MANAGER" 
     count(*) filter (where lower(job)='analyst') as "ANALYST" 
     count(*) filter (where lower(job)='president') as "PRESIDENT" 
from emp;