Question

我有这个SELECT语句：

SELECT TASK_ID, INVOICED, DELETE_WEEK,
SUM(MON_BILL+TUE_BILL+WED_BILL+THU_BILL+FRI_BILL) AS hours
FROM b_report_week
WHERE INVOICE = 1 AND DELETE_WEEK = 0
GROUP BY TASK_ID

我需要在此查询中添加另一个带有SUM的SELECT语句以提供结果：

WHERE INVOICE = 0

而不是前一个声明中的1。它仍然需要按TASK_ID分组，DELETE_WEEK仍需要为0.我试过这个：

SELECT TASK_ID, INVOICED, DELETE_WEEK,
SUM(MON_BILL+TUE_BILL+WED_BILL+THU_BILL+FRI_BILL) AS hours,
(SELECT SUM(MON_BILL+TUE_BILL+WED_BILL+THU_BILL+FRI_BILL) AS hours2 
FROM b_report_week WHERE INVOICE = 0 AND DELETE_WEEK = 0 GROUP BY TASK_ID)
FROM b_report_week
WHERE INVOICE = 1 AND DELETE_WEEK = 0
GROUP BY TASK_ID

但这显然不正确。我需要使用小时和小时2在PHP while循环中输出正确的结果。帮助将不胜感激

Answer 1

您可以使用Case语句：

SELECT TASK_ID, INVOICED, DELETE_WEEK,
case when INVOICE = 1 then SUM(MON_BILL+TUE_BILL+WED_BILL+THU_BILL+FRI_BILL) else 0 end AS hours,
case when INVOICE = 0 then SUM(MON_BILL+TUE_BILL+WED_BILL+THU_BILL+FRI_BILL) else 0 end AS hours2,
FROM b_report_week
WHERE DELETE_WEEK = 0
GROUP BY TASK_ID

（注意 - 必须更改where子句以避免将结果限制为Invoice = 1.）

Answer 2

使用CASE

SELECT TASK_ID, INVOICED, DELETE_WEEK,
SUM(MON_BILL+TUE_BILL+WED_BILL+THU_BILL+FRI_BILL) AS hours , 
SUM(CASE WHEN INVOICE = 0 then MON_BILL+TUE_BILL+WED_BILL+THU_BILL+FRI_BILL else o end case) as custom_sum
FROM b_report_week
WHERE DELETE_WEEK = 0
GROUP BY TASK_ID

Answer 3

您也可以使用UNION子句

解决它

SELECT TASK_ID, INVOICED, DELETE_WEEK,
SUM(MON_BILL+TUE_BILL+WED_BILL+THU_BILL+FRI_BILL) AS hours
FROM b_report_week
WHERE INVOICE = 1 AND DELETE_WEEK = 0
GROUP BY TASK_ID
UNION
SELECT TASK_ID, INVOICED, DELETE_WEEK,
SUM(MON_BILL+TUE_BILL+WED_BILL+THU_BILL+FRI_BILL) AS hours
FROM b_report_week
WHERE INVOICE = 0 AND DELETE_WEEK = 0
GROUP BY TASK_ID

但我认为更容易阅读CASE WHEN条款

在SELECT不同的WHERE中选择

3 个答案: