php - 使用 CASE WHEN 时，如果 LEFT JOIN 中找不到匹配行，则忽略列

Question

我想获取医院账单的收款报告。有一张表记录了患者支付的预付款金额。并非所有患者都预付费用。

这是我现在使用的查询

"SELECT SUM(
    CASE 
    WHEN bill_type = 3 THEN b.payable_amount + w.balance
    ELSE 0
 END) AS 'amount'

 FROM bills b 
    LEFT JOIN ip_patients ip on ip.id = b.ip_id
    LEFT JOIN advance w on w.id = (SELECT x.id FROM advance x WHERE x.ip_id = ip.id ORDER BY x.created_at DESC  LIMIT 1)
    WHERE b.doctor_id = '$d->id' AND CAST(b.created_at AS DATE) >= '$date1' AND CAST(b.created_at AS DATE) <= '$date2' AND b.is_cancelled = 0 AND b.is_deleted = 0"

如果患者未支付预付款，则 advance 表中不会插入任何行。上述查询发生的情况是，当预付表中没有患者的行时，他的付款将被完全忽略(即，bills 表中的值不会添加到总和中。

Answer 1

如果将 null 添加到非 null，则结果为 null。例如

MariaDB [sandbox]> set @a = null;
Query OK, 0 rows affected (0.00 sec)

MariaDB [sandbox]> set @b = 100;
Query OK, 0 rows affected (0.00 sec)

MariaDB [sandbox]>
MariaDB [sandbox]> select @a+@b amt;
+------+
| amt  |
+------+
| NULL |
+------+
1 row in set (0.00 sec)

你可以使用 ifnull 来处理这个问题

set @a = null;
set @b = 100;

select ifnull(@a,0) + ifnull(@b,0) amt;

+-----+
| amt |
+-----+
| 100 |
+-----+
1 row in set (0.00 sec)