Question

为什么这对我不起作用？我试图选择每个结算单位（OJ）中超过2名员工执行的工作。

select W.NAME as NAME,
    W.OJ as OJ,
    W.JOB as JOB,
    W.MONTHLY_PAY as MONTHLY_PAY ,
    SUM(W.OJ) AS "SUM"
 from WORKER W
 WHERE W.OJ > 2
 GROUP BY W.IME, W.POSAO, W.MJESECNA_PLACA

Answer 1

这就是你需要的：

select W.NAME as NAME,
    W.OJ as OJ,
    W.JOB as JOB,
    W.MONTHLY_PAY as MONTHLY_PAY ,
    SUM(W.OJ) AS "SUM"
 from WORKER W
having count( W.OJ)  > 2
 GROUP By NAME,
    W.OJ  ,
    W.JOB  ,
    W.MONTHLY_PAY

Answer 2

我想你想要：

select W.OJ, count(*) as num_workers
from WORKER W
group by W.O
having count(*) > 2;

Answer 3

SELECT job
FROM
(
SELECT 
    w.job, w.oj, count(*) workerCount
FROM worker w
GROUP BY w.job, w.oj) t
GROUP BY job
HAVING min(workerCount) > 2

Answer 4

SELECT子句中的所有字段（聚合字段除外）都应存在于GROUP BY子句中。

尝试以下：

select W.IME, W.POSAO, W.MJESECNA_PLACA
    SUM(W.OJ) AS "SUM"
 from WORKER W
 WHERE W.OJ > 2
 GROUP BY W.IME, W.POSAO, W.MJESECNA_PLACA

SQL SUM，GROUP BY问题

4 个答案: