当我得到其他返回行的null = 0值时,如何显示其他查询行

时间:2017-03-02 02:46:00

标签: mysql database-design left-join coalesce

我有三个表,我曾经离开外部联接查看我的所有诊所名称,包括我需要的数据的总和和计数,我在这里有下面的表截图:

my tables

现在,我的查询是这样的:

SELECT
tbl_clinics.clinic_name,
COALESCE(COUNT(tbl_check_up.check_up_id),0) AS totalpatient,
COALESCE(SUM(tbl_bill.bill_amt),0) AS totalearn
FROM
tbl_clinics
LEFT OUTER JOIN tbl_check_up ON tbl_clinics.clinic_id = tbl_check_up.clinic_id
LEFT OUTER JOIN tbl_bill ON tbl_bill.bill_id = tbl_check_up.bill_id
WHERE tbl_clinics.user_id = 102

我的查询输出是:

myresult

问题: 现在,我的问题是我想通过显示零到零值来显示我的其他诊所。在我的数据中,我有3个诊所,其user_id = 102,但为什么只有诊所1显示? 。我想要一些如下所示的输出,其中显示tbl_check_up中没有值,其诊所ID为2和3与账单相同。

+---------------+---------------+------------+
|  Clinic Name  | Total Patient | Total Earn |
+---------------+---------------+------------+
| Clinic 1      |       4       |   800.00   |
+---------------+---------------+------------+
| Clinic 2      |       0       |   0        |
+---------------+---------------+------------+
| Clinic 3      |       0       |   0        |
+---------------+---------------+------------+

1 个答案:

答案 0 :(得分:2)

这是您的查询(COALESCE()不需要COUNT()):

SELECT c.clinic_name,
       COUNT(cu.check_up_id) AS totalpatient,
       COALESCE(SUM(b.bill_amt), 0) AS totalearn
FROM tbl_clinics c LEFT OUTER JOIN
     tbl_check_up cu
     ON c.clinic_id = cu.clinic_id LEFT OUTER JOIN
     tbl_bill b
     ON b.bill_id = cu.bill_id
WHERE c.user_id = 102
GROUP BY c.clinic_name;

我认为你只需要GROUP BY