MySQL如何显示订单已付未付金额
How to display the paid and unpaid amount of Order in MySQL
客户TABLE
+------------+----------+-------+-------+-------------+-----------------------------+----------+--------------+
| CustomerID | Fname | Mname | Lname | Contact_no | Address | Valid_id | Credit_Limit |
+------------+----------+-------+-------+-------------+-----------------------------+----------+--------------+
| 7 | John | Dale | Doe | 09123654789 | Asan Sur, Sison, Pangasinan | NULL | 5000.000 |
| 8 | Jane | Dale | Doe | 09987654123 | Asan Sur, Sison, Pangasinan | NULL | 1500.000 |
| 91 | Kurdapya | Buang | Selos | 09741258963 | Paldit, Sison, Pangasinan | NULL | 5000.000 |
+------------+----------+-------+-------+-------------+-----------------------------+----------+--------------+
订单TABLE
+---------+------------+----------------+----------+------------+
| 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 |
+---------+------------+----------------+----------+------------+
付款TABLE
+-----------+------------+---------+------------+----------+
| 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 |
+-----------+------------+---------+------------+----------+
订单详情TABLE
+-------+---------+-----------+------+
| OR_ID | OrderID | ProductID | QTY |
+-------+---------+-----------+------+
| 173 | 82 | 5 | 1 |
| 174 | 82 | 9 | 1 |
| 184 | 86 | 5 | 1 |
| 185 | 86 | 9 | 1 |
| 186 | 86 | 13 | 1 |
| 187 | 86 | 17 | 1 |
| 224 | 109 | 3 | 3 |
| 225 | 109 | 6 | 3 |
| 292 | 145 | 20 | 2 |
| 293 | 145 | 12 | 1 |
| 294 | 145 | 8 | 2 |
| 295 | 146 | 14 | 1 |
| 296 | 146 | 11 | 1 |
| 297 | 146 | 12 | 1 |
| 298 | 146 | 3 | 1 |
| 299 | 146 | 6 | 1 |
| 300 | 146 | 7 | 1 |
| 301 | 146 | 16 | 1 |
+-------+---------+-----------+------+
我不知道是否需要，但这是我的 产品 TABLE:
+-----------+---------------+-----------------------+------------+-----------+------+
| ProductID | Pname | Pdesc | PUnitPrice | weight | Unit |
+-----------+---------------+-----------------------+------------+-----------+------+
| 2 | Pepsi | 1 Case Glass Bottle | 313.000 | 1 Litre | 12 |
| 3 | Mountain Dew | 1 Case Glass Bottle | 231.000 | 750 ML | 12 |
| 4 | Pepsi | 1 Case Plastic Bottle | 620.000 | 1.5 Litre | 12 |
| 5 | Mirinda | 1 Case Plastic Bottle | 620.000 | 1.5 Litre | 12 |
| 6 | Mountain Dew | 1 Case Plastic Bottle | 620.000 | 1.5 Litre | 12 |
| 7 | Mountain Dew | 1 Case Glass Bottle | 145.000 | 8 oz | 24 |
| 8 | Pepsi | 1 Case Glass Bottle | 145.000 | 8 oz | 24 |
| 9 | Mirinda | 1 Case Glass Bottle | 145.000 | 8 oz | 24 |
| 10 | 7up | 1 Case Glass Bottle | 145.000 | 8 oz | 24 |
| 11 | Sting | 1 Case Glass Bottle | 266.000 | 240 ml | 24 |
| 12 | Tropicana | 1 Case Glass Bottle | 266.000 | 240 ml | 24 |
| 13 | Cobra | 1 Case Glass Bottle | 266.000 | 240 ml | 24 |
| 14 | Sting | 1 Case Plastic Bottle | 166.000 | 300 ml | 12 |
| 15 | Cobra | 1 Case Plastic Bottle | 166.000 | 300 ml | 12 |
| 16 | Mountain Dew | 1 Case Plastic Bottle | 135.000 | 295 ml | 12 |
| 17 | Mirinda | 1 Case Plastic Bottle | 135.000 | 295 ml | 12 |
| 18 | Pepsi | 1 Case Plastic Bottle | 135.000 | 295 ml | 12 |
| 19 | Ginebra | 1 Case Glass Bottle | 129.000 | 350 ml | 24 |
| 20 | San Mig Light | 1 Case Glass Bottle | 1070.000 | 330 ml | 24 |
| 21 | Red Horse | 1 Case Glass Bottle | 535.000 | 500 ml | 12 |
| 22 | Red Horse | 1 Case Glass Bottle | 545.000 | 1 Litre | 6 |
+-----------+---------------+-----------------------+------------+-----------+------+
我想发生什么
我只想在一行中显示customerID 7 的所有已付款订单和他的未付款订单。我不知道如何从查询开始。我应该使用 If() 条件吗？如何正确查询它以实现我想要的输出？
我想要的输出：
+--------+------+-------+-------+-------+-------------+-----------------------------+--------------+----------+
| unpaid | paid | Fname | Mname | Lname | Contact_no | Address | Credit_Limit | total |
+--------+------+-------+-------+-------+-------------+-----------------------------+--------------+----------+
| 2995 | 6014 | John | Dale | Doe | 09123654789 | Asan Sur, Sison, Pangasinan | 5000.000 | 9009.000 |
+--------+------+-------+-------+-------+-------------+-----------------------------+--------------+----------+
编辑
这就是我到目前为止尝试过的方法
第一次尝试：
我正在尝试计算未付款订单，这是 output:
select
if(py.Amount IS NULL, sum(od.QTY * p.PUnitPrice), CONCAT(py.Amount - sum(od.QTY * p.PUnitPrice))) as remarks,
c.Fname, c.Mname, c.Lname, c.Contact_no, c.Address, c.Credit_Limit,
sum(od.QTY * p.PUnitPrice) as total
from customer c INNER JOIN orders r on r.CustomerID=c.CustomerID
INNER join order_details od on r.OrderID=od.OrderID
INNER JOIN product p on od.ProductID = p.ProductID
join payment py
where c.CustomerID=7
尝试 1 的结果：
+-------------+-------+-------+-------+-------------+-----------------------------+--------------+------------+
| remarks | Fname | Mname | Lname | Contact_no | Address | Credit_Limit | total |
+-------------+-------+-------+-------+-------------+-----------------------------+--------------+------------+
| -134335.000 | John | Dale | Doe | 09123654789 | Asan Sur, Sison, Pangasinan | 5000.000 | 135135.000 |
+-------------+-------+-------+-------+-------------+-----------------------------+--------------+------------+
第二次尝试：
SELECT IF(py.OrderID IS NULL, sum(od.QTY * p.PUnitPrice), 0) AS unpaid,
if(py.OrderID IS NOT NULL, sum(od.QTY * p.PUnitPrice), 0) as paid,
sum(od.QTY * p.PUnitPrice) as total,
O.OrderID, O.CustomerID, date_format(O.DateOfPurchase, '%M %d, %Y') AS DateOfPurchase, date_format(O.DueDate, '%M %d, %Y') AS DueDate
from Orders O INNER JOIN order_details od on od.OrderID=O.OrderID
INNER JOIN product p ON od.ProductID=p.ProductID
LEFT JOIN Payment py ON py.OrderID = O.OrderID
where O.CustomerID = 7
GROUP by O.OrderID
ORDER by O.OrderID desc
尝试 2 的结果：
+----------+----------+----------+---------+------------+----------------+--------------+
| unpaid | paid | total | OrderID | CustomerID | DateOfPurchase | DueDate |
+----------+----------+----------+---------+------------+----------------+--------------+
| 1829.000 | 0.000 | 1829.000 | 146 | 7 | April 27, 2022 | May 27, 2022 |
| 0.000 | 2696.000 | 2696.000 | 145 | 7 | April 27, 2022 | May 27, 2022 |
| 0.000 | 2553.000 | 2553.000 | 109 | 7 | April 25, 2022 | May 25, 2022 |
| 1166.000 | 0.000 | 1166.000 | 86 | 7 | April 17, 2022 | May 17, 2022 |
| 0.000 | 765.000 | 765.000 | 82 | 7 | April 17, 2022 | May 17, 2022 |
+----------+----------+----------+---------+------------+----------------+--------------+
注意：如何对已付和未付列中的所有行求和？
您接近解决方案的方式显然不起作用，因此您正在寻求帮助。您的第一个查询，您正在加入付款 table，但没有条件仅限于客户 7 的付款。这可能会巧合，因为只有一个付款记录。但是对于不止一个的客户，
您的总数将因笛卡尔积结果而发生偏差。
您应该做的是 pre-aggregate 购买与付款完全分开，这样您将最多拥有您正在寻找的一个客户的单一记录。现在，如果您想将结果应用到所有客户，那只会增加一点点，稍后会介绍。
每个单独，防止混淆多个杂乱。
在最短的时间里，由于您只关心购买和支付的总计，我不会关心每个订单的总计，只关心每个客户的总计。
select
c.Fname,
c.Mname,
c.Lname,
c.Contact_no,
c.Address,
c.Credit_Limit,
coalesce( PQ_Orders.TotalOrders, 0 ) TotalOrders,
coalesce( PQ_Payments.TotalPaid, 0 ) TotalPayments,
coalesce( PQ_Orders.TotalOrders, 0 )
- coalesce( PQ_Payments.TotalPaid, 0 ) BalanceDue
from
customer c
LEFT JOIN
( select
o.customerID,
sum( od.qty * p.pUnitPrice ) TotalOrders
from
orders o
join order_details od
on o.orderID = od.orderID
join product p
on od.productid = p.productid
where
o.customerid = 7
group by
o.customerID ) PQ_Orders
on c.customerID = PQ_Orders.customerID
LEFT JOIN
( select
py.customerid,
sum( py.Amount ) TotalPaid
from
payments py
where
py.customerid = 7
group by
py.customerid ) PQ_Payments
on c.customerID = PQ_Payments.CustomerID
where
c.CustomerID = 7
现在，如果您想要所有客户，只需删除所有 WHERE 子句，以便它获得所有客户。每个内部查询都将客户 ID 作为数据分组，因此如果一个人有 10 个订单和 3 个付款，您仍然有每个 PRE-QUERY (PQ) 聚合，每个客户只有 1 个订单条目和 1 个条目每个客户的付款。重新加入主要客户 table 并将其应用 across-the-board 给所有客户。
对O/P
的澄清
为了帮助澄清，让我们看看您的要求。对于单个客户，购买了多少东西。将其视为一个查询。根据订单详细信息和相应的产品价格获取该客户的所有订单，并按客户分组汇总。现在客户的 WHERE 子句 = 7 只是因为您是关于一个客户的。但是，如果没有 where 将向所有客户显示他们各自购买的所有商品。无论他们有 1 个订单还是 274 个订单，您最多（因为按客户 ID 分组）每个客户有 1 条记录。它是该客户的总计。
现在，付款的上下文完全相同。我不在乎他们购买了什么或何时购买。我只关心给定客户的付款。与购买说明中一样，有或没有客户 ID = 7 的 WHERE。同样，无论 1 笔还是 75 笔付款，每个客户只有 1 条记录，每个客户 1 条记录。
所以现在，您有一个 1:1:1 可能的比率，即 1 位客户对 1 次购买与可能的 1 次付款总额。
LEFT JOIN 的意思是，我想要左边的东西（客户），但可以选择性地在右边找到东西（购买 AND/OR 付款）。
因此 COALESCE() 可防止空值在计算中出错。如果给定 pre-query 的购买或付款有一个值，则获取它，否则，假设为零。因此，您可以只获取一个有问题的客户，或者通过删除 WHERE 子句，您可以获得所有客户的所有购买总额和所有付款总额，以获得所有客户的全部未结余额。
客户TABLE
+------------+----------+-------+-------+-------------+-----------------------------+----------+--------------+
| CustomerID | Fname | Mname | Lname | Contact_no | Address | Valid_id | Credit_Limit |
+------------+----------+-------+-------+-------------+-----------------------------+----------+--------------+
| 7 | John | Dale | Doe | 09123654789 | Asan Sur, Sison, Pangasinan | NULL | 5000.000 |
| 8 | Jane | Dale | Doe | 09987654123 | Asan Sur, Sison, Pangasinan | NULL | 1500.000 |
| 91 | Kurdapya | Buang | Selos | 09741258963 | Paldit, Sison, Pangasinan | NULL | 5000.000 |
+------------+----------+-------+-------+-------------+-----------------------------+----------+--------------+
订单TABLE
+---------+------------+----------------+----------+------------+
| 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 |
+---------+------------+----------------+----------+------------+
付款TABLE
+-----------+------------+---------+------------+----------+
| 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 |
+-----------+------------+---------+------------+----------+
订单详情TABLE
+-------+---------+-----------+------+
| OR_ID | OrderID | ProductID | QTY |
+-------+---------+-----------+------+
| 173 | 82 | 5 | 1 |
| 174 | 82 | 9 | 1 |
| 184 | 86 | 5 | 1 |
| 185 | 86 | 9 | 1 |
| 186 | 86 | 13 | 1 |
| 187 | 86 | 17 | 1 |
| 224 | 109 | 3 | 3 |
| 225 | 109 | 6 | 3 |
| 292 | 145 | 20 | 2 |
| 293 | 145 | 12 | 1 |
| 294 | 145 | 8 | 2 |
| 295 | 146 | 14 | 1 |
| 296 | 146 | 11 | 1 |
| 297 | 146 | 12 | 1 |
| 298 | 146 | 3 | 1 |
| 299 | 146 | 6 | 1 |
| 300 | 146 | 7 | 1 |
| 301 | 146 | 16 | 1 |
+-------+---------+-----------+------+
我不知道是否需要，但这是我的 产品 TABLE:
+-----------+---------------+-----------------------+------------+-----------+------+
| ProductID | Pname | Pdesc | PUnitPrice | weight | Unit |
+-----------+---------------+-----------------------+------------+-----------+------+
| 2 | Pepsi | 1 Case Glass Bottle | 313.000 | 1 Litre | 12 |
| 3 | Mountain Dew | 1 Case Glass Bottle | 231.000 | 750 ML | 12 |
| 4 | Pepsi | 1 Case Plastic Bottle | 620.000 | 1.5 Litre | 12 |
| 5 | Mirinda | 1 Case Plastic Bottle | 620.000 | 1.5 Litre | 12 |
| 6 | Mountain Dew | 1 Case Plastic Bottle | 620.000 | 1.5 Litre | 12 |
| 7 | Mountain Dew | 1 Case Glass Bottle | 145.000 | 8 oz | 24 |
| 8 | Pepsi | 1 Case Glass Bottle | 145.000 | 8 oz | 24 |
| 9 | Mirinda | 1 Case Glass Bottle | 145.000 | 8 oz | 24 |
| 10 | 7up | 1 Case Glass Bottle | 145.000 | 8 oz | 24 |
| 11 | Sting | 1 Case Glass Bottle | 266.000 | 240 ml | 24 |
| 12 | Tropicana | 1 Case Glass Bottle | 266.000 | 240 ml | 24 |
| 13 | Cobra | 1 Case Glass Bottle | 266.000 | 240 ml | 24 |
| 14 | Sting | 1 Case Plastic Bottle | 166.000 | 300 ml | 12 |
| 15 | Cobra | 1 Case Plastic Bottle | 166.000 | 300 ml | 12 |
| 16 | Mountain Dew | 1 Case Plastic Bottle | 135.000 | 295 ml | 12 |
| 17 | Mirinda | 1 Case Plastic Bottle | 135.000 | 295 ml | 12 |
| 18 | Pepsi | 1 Case Plastic Bottle | 135.000 | 295 ml | 12 |
| 19 | Ginebra | 1 Case Glass Bottle | 129.000 | 350 ml | 24 |
| 20 | San Mig Light | 1 Case Glass Bottle | 1070.000 | 330 ml | 24 |
| 21 | Red Horse | 1 Case Glass Bottle | 535.000 | 500 ml | 12 |
| 22 | Red Horse | 1 Case Glass Bottle | 545.000 | 1 Litre | 6 |
+-----------+---------------+-----------------------+------------+-----------+------+
我想发生什么
我只想在一行中显示customerID 7 的所有已付款订单和他的未付款订单。我不知道如何从查询开始。我应该使用 If() 条件吗？如何正确查询它以实现我想要的输出？
我想要的输出：
+--------+------+-------+-------+-------+-------------+-----------------------------+--------------+----------+
| unpaid | paid | Fname | Mname | Lname | Contact_no | Address | Credit_Limit | total |
+--------+------+-------+-------+-------+-------------+-----------------------------+--------------+----------+
| 2995 | 6014 | John | Dale | Doe | 09123654789 | Asan Sur, Sison, Pangasinan | 5000.000 | 9009.000 |
+--------+------+-------+-------+-------+-------------+-----------------------------+--------------+----------+
编辑
这就是我到目前为止尝试过的方法
第一次尝试： 我正在尝试计算未付款订单，这是 output:
select
if(py.Amount IS NULL, sum(od.QTY * p.PUnitPrice), CONCAT(py.Amount - sum(od.QTY * p.PUnitPrice))) as remarks,
c.Fname, c.Mname, c.Lname, c.Contact_no, c.Address, c.Credit_Limit,
sum(od.QTY * p.PUnitPrice) as total
from customer c INNER JOIN orders r on r.CustomerID=c.CustomerID
INNER join order_details od on r.OrderID=od.OrderID
INNER JOIN product p on od.ProductID = p.ProductID
join payment py
where c.CustomerID=7
尝试 1 的结果：
+-------------+-------+-------+-------+-------------+-----------------------------+--------------+------------+
| remarks | Fname | Mname | Lname | Contact_no | Address | Credit_Limit | total |
+-------------+-------+-------+-------+-------------+-----------------------------+--------------+------------+
| -134335.000 | John | Dale | Doe | 09123654789 | Asan Sur, Sison, Pangasinan | 5000.000 | 135135.000 |
+-------------+-------+-------+-------+-------------+-----------------------------+--------------+------------+
第二次尝试：
SELECT IF(py.OrderID IS NULL, sum(od.QTY * p.PUnitPrice), 0) AS unpaid,
if(py.OrderID IS NOT NULL, sum(od.QTY * p.PUnitPrice), 0) as paid,
sum(od.QTY * p.PUnitPrice) as total,
O.OrderID, O.CustomerID, date_format(O.DateOfPurchase, '%M %d, %Y') AS DateOfPurchase, date_format(O.DueDate, '%M %d, %Y') AS DueDate
from Orders O INNER JOIN order_details od on od.OrderID=O.OrderID
INNER JOIN product p ON od.ProductID=p.ProductID
LEFT JOIN Payment py ON py.OrderID = O.OrderID
where O.CustomerID = 7
GROUP by O.OrderID
ORDER by O.OrderID desc
尝试 2 的结果：
+----------+----------+----------+---------+------------+----------------+--------------+
| unpaid | paid | total | OrderID | CustomerID | DateOfPurchase | DueDate |
+----------+----------+----------+---------+------------+----------------+--------------+
| 1829.000 | 0.000 | 1829.000 | 146 | 7 | April 27, 2022 | May 27, 2022 |
| 0.000 | 2696.000 | 2696.000 | 145 | 7 | April 27, 2022 | May 27, 2022 |
| 0.000 | 2553.000 | 2553.000 | 109 | 7 | April 25, 2022 | May 25, 2022 |
| 1166.000 | 0.000 | 1166.000 | 86 | 7 | April 17, 2022 | May 17, 2022 |
| 0.000 | 765.000 | 765.000 | 82 | 7 | April 17, 2022 | May 17, 2022 |
+----------+----------+----------+---------+------------+----------------+--------------+
注意：如何对已付和未付列中的所有行求和？
您接近解决方案的方式显然不起作用，因此您正在寻求帮助。您的第一个查询，您正在加入付款 table，但没有条件仅限于客户 7 的付款。这可能会巧合，因为只有一个付款记录。但是对于不止一个的客户， 您的总数将因笛卡尔积结果而发生偏差。
您应该做的是 pre-aggregate 购买与付款完全分开，这样您将最多拥有您正在寻找的一个客户的单一记录。现在，如果您想将结果应用到所有客户，那只会增加一点点，稍后会介绍。
每个单独，防止混淆多个杂乱。
在最短的时间里，由于您只关心购买和支付的总计，我不会关心每个订单的总计，只关心每个客户的总计。
select
c.Fname,
c.Mname,
c.Lname,
c.Contact_no,
c.Address,
c.Credit_Limit,
coalesce( PQ_Orders.TotalOrders, 0 ) TotalOrders,
coalesce( PQ_Payments.TotalPaid, 0 ) TotalPayments,
coalesce( PQ_Orders.TotalOrders, 0 )
- coalesce( PQ_Payments.TotalPaid, 0 ) BalanceDue
from
customer c
LEFT JOIN
( select
o.customerID,
sum( od.qty * p.pUnitPrice ) TotalOrders
from
orders o
join order_details od
on o.orderID = od.orderID
join product p
on od.productid = p.productid
where
o.customerid = 7
group by
o.customerID ) PQ_Orders
on c.customerID = PQ_Orders.customerID
LEFT JOIN
( select
py.customerid,
sum( py.Amount ) TotalPaid
from
payments py
where
py.customerid = 7
group by
py.customerid ) PQ_Payments
on c.customerID = PQ_Payments.CustomerID
where
c.CustomerID = 7
现在，如果您想要所有客户，只需删除所有 WHERE 子句，以便它获得所有客户。每个内部查询都将客户 ID 作为数据分组，因此如果一个人有 10 个订单和 3 个付款，您仍然有每个 PRE-QUERY (PQ) 聚合，每个客户只有 1 个订单条目和 1 个条目每个客户的付款。重新加入主要客户 table 并将其应用 across-the-board 给所有客户。
对O/P
的澄清为了帮助澄清，让我们看看您的要求。对于单个客户，购买了多少东西。将其视为一个查询。根据订单详细信息和相应的产品价格获取该客户的所有订单，并按客户分组汇总。现在客户的 WHERE 子句 = 7 只是因为您是关于一个客户的。但是，如果没有 where 将向所有客户显示他们各自购买的所有商品。无论他们有 1 个订单还是 274 个订单，您最多（因为按客户 ID 分组）每个客户有 1 条记录。它是该客户的总计。
现在，付款的上下文完全相同。我不在乎他们购买了什么或何时购买。我只关心给定客户的付款。与购买说明中一样，有或没有客户 ID = 7 的 WHERE。同样，无论 1 笔还是 75 笔付款，每个客户只有 1 条记录，每个客户 1 条记录。
所以现在，您有一个 1:1:1 可能的比率，即 1 位客户对 1 次购买与可能的 1 次付款总额。
LEFT JOIN 的意思是，我想要左边的东西（客户），但可以选择性地在右边找到东西（购买 AND/OR 付款）。
因此 COALESCE() 可防止空值在计算中出错。如果给定 pre-query 的购买或付款有一个值，则获取它，否则，假设为零。因此，您可以只获取一个有问题的客户，或者通过删除 WHERE 子句，您可以获得所有客户的所有购买总额和所有付款总额，以获得所有客户的全部未结余额。