sql - 在 Postgres 中选择大小写

Question

我正在尝试进行以下咨询:

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

      3            4              1               2               4

到目前为止我设法做到了:

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;

但我似乎无法运行它，它会出错:

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

如何在选择中使用 case 来创建单独的列？

Answer 1

您需要将计数放在周围 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;

count() 之类的聚合函数会忽略空值。 CASE 表达式为不符合条件的值返回一个 NULL，因此不计算在内。

或者更简单地使用 filter 子句:

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;