订单表

+---------+------------+----------------+----------+------------+
| OrderID | CustomerID | DateOfPurchase | Discount | DueDate    |
+---------+------------+----------------+----------+------------+
|      82 |          7 | 2022-04-17     | 0        | 2022-05-17 |
|      83 |         91 | 2022-04-17     | 0        | 2022-05-17 |
|      84 |          8 | 2022-04-17     | 0        | 2022-05-17 |
|      85 |         91 | 2022-04-17     | 0        | 2022-05-17 |
|      86 |          7 | 2022-04-17     | 0        | 2022-05-17 |
|      87 |         91 | 2022-04-18     | 0        | 2022-05-18 |
|     109 |          7 | 2022-04-25     | 0        | 2022-05-25 |
+---------+------------+----------------+----------+------------+

Customer表

+------------+----------+-------+-------+-------------+-----------------------------+----------+--------------+
| CustomerID | Fname    | Mname | Lname | Contact_no  | Address                     | Valid_id | Credit_Limit |
+------------+----------+-------+-------+-------------+-----------------------------+----------+--------------+
|          7 | John     | Dale  | Doe   | 09123654789 |  | NULL     |     5000.000 |
|          8 | Jane     | Dale  | Doe   | 09987654123 | | NULL     |     1500.000 |
|         91 | Kurdapya | Buang | Selos | 09741258963 |   | NULL     |     5000.000 |
+------------+----------+-------+-------+-------------+-----------------------------+----------+--------------+

支付表

+-----------+------------+---------+------------+----------+
| PaymentID | CustomerID | OrderID | PayDate    | Amount   |
+-----------+------------+---------+------------+----------+
|        20 |          7 |      82 | 2022-04-25 |  800.000 |
|        21 |         91 |      83 | 2022-04-17 | 2500.000 |
|        22 |         91 |      85 | 2022-04-17 |  200.000 |
|        23 |         95 |      88 | 2022-04-18 | 2122.000 |
|        24 |         96 |      90 | 2022-04-25 |  577.000 |
|        25 |         97 |     111 | 2022-04-25 |    0.000 |
|        26 |         98 |     114 | 2022-04-25 |  166.000 |
|        27 |         99 |     115 | 2022-04-25 | 1740.000 |
+-----------+------------+---------+------------+----------+

我想知道客户Kurdapya(OrderID=91)的OrderID是已支付和未支付的

这是我到目前为止尝试过的查询

尝试1：

select if(py.OrderID=r.OrderID, 'paid','unpaid') as remarks, r.OrderID, r.CustomerID
from orders r,
     payment py
where py.OrderID = r.OrderID and r.CustomerID = 91
GROUP by r.OrderID;

尝试1的结果：

+---------+---------+------------+
| remarks | OrderID | CustomerID |
+---------+---------+------------+
| paid    |      83 |         91 |
| paid    |      85 |         91 |
+---------+---------+------------+

尝试2：

select if(py.OrderID=r.OrderID and py.OrderID=py.Amount!='null', 'paid','unpaid') as remarks, r.OrderID, r.CustomerID
from orders r,
     payment py
where r.CustomerID = 91
GROUP by r.OrderID;

尝试2的结果：

+---------+---------+------------+
| remarks | OrderID | CustomerID |
+---------+---------+------------+
| unpaid  |      83 |         91 |
| unpaid  |      85 |         91 |
| unpaid  |      87 |         91 |
+---------+---------+------------+

我希望得到的结果是：

+---------+---------+------------+
    | remarks | OrderID | CustomerID |
    +---------+---------+------------+
    | unpaid  |      83 |         91 |
    | unpaid  |      85 |         91 |
    | paid    |      87 |         91 |
    +---------+---------+------------+

3条答案

按热度按时间

r6l8ljro1#

You can try

select  CASE
    WHEN P.Amount > 0 then 'paid' else 'unpaid' end as remarks,
        O.orderID, O.CustomerID
from Customer C inner join OrderTable O on C.CustomerID  = O.CustomerID 
inner join Payment P on P.OrderID = O.OrderID
where O.CustomerID = 91

赞(0）回复(0）举报 2022-09-18

62lalag42#

Your second query didn't even compile. Try this

select if(py.Amount is not null, 'paid','unpaid') as remarks, r.OrderID, r.CustomerID
from orders r
left outer join payment py on r.OrderID=py.OrderID
where r.CustomerID = 91;

Points to note:

I have used the modern, explicit JOIN syntax - see the comment from @jarlh
I have used LEFT OUTER join
I have removed the quotation marks from 'null' - NULL is a specific value whereas 'null' is a string
The GROUP BY is unecessary

3bygqnnd3#

SELECT
  IF(py.OrderID IS NULL, 'unpaid', 'paid') AS remarks,
  r.OrderID,
  r.CustomerID
FROM orders AS r
LEFT OUTER JOIN payment AS py USING (OrderID)
WHERE r.CustomerID = 91

However, since the order table doesn't have an amount, I don't know if the amount paid is sufficient to pay for the full order.

如何在带有if()条件的MySQL中同时显示真和假行

3条答案

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