跨乘法表

Question

所以我有这样的SQL代码，我想显示其收费单，并在他们收到的每个项目的总和：

select item_description, sum(receipt_qty) as Samp1, sum(chargeSlip_qty) as Samp2 
from Items inner join Receipt_Detail on (Receipt_Detail.item_number = 
    Items.item_number) 
inner join ChargeSlip_Detail on (ChargeSlip_Detail.item_number = 
    Items.item_number) 
group by item_description 

它产生这样的输出：

Acetazolamide 2681 1730 
Ascorbic Acid 1512 651 
Paracetamol 1370 742 
Silk   576 952 

但它应该是：

Acetazolamide 383 173 
Ascorbic Acid 216 93 
Paracetamol 274 106 
Silk   96 238 

我的代码出了什么问题？

Answer 1

由于您正在加入表格，因此当您获取sum()时，您可能会遇到一对一的关系，从而导致问题。所以你可以使用子查询来获得结果。这将得到sum()为receipt和chargeslip每个item_number，然后你加入该回你items表得到最终的结果是：

select i.item_description, 
    r.Samp1, 
    c.Samp2 
from Items i 
inner join 
(
    select sum(receipt_qty) Samp1, 
    item_number 
    from Receipt_Detail 
    group by item_number 
) r 
    on r.item_number = i.item_number 
inner join 
(
    select sum(chargeSlip_qty) Samp2, 
    item_number 
    from ChargeSlip_Detail 
    group by item_number 
) c 
    on c.item_number = i.item_number 

Answer 2

每 Item_Number执行分组依据，第一，

，让你不乘以Receipt_Detail和ChargeSlip_Detail。也就是说，你加入回Items

select 
    I.item_description, 
    R.Samp1, 
    C.Samp2 
from 
    Items I 
    inner join 
    (SELECT item_number, sum(receipt_qty) as Samp1 
     FROM Receipt_Detail 
     GROUP BY item_number 
    ) R 
     on (R.item_number = I.item_number) 
    inner join 
    (SELECT item_number, sum(chargeSlip_qty) as Samp2 
     FROM ChargeSlip_Detail 
     GROUP BY item_number 
    ) C 
     on (C.item_number = I.item_number) 

Answer 3

左前生成每个Item_Number值的总和左表联接返回行，并在左边的表的每一行，在右边的表中的所有匹配的行。

因此，例如：

create table Customers (name varchar(50)); 
insert Customers values 
    ('Tim'), 
    ('John'), 
    ('Spike'); 
create table Orders (customer_name varchar(50), product varchar(50)); 
insert Orders values (
    ('Tim', 'Guitar'), 
    ('John', 'Drums'), 
    ('John', 'Trumpet'); 
create table Addresses (customer_name varchar(50), address varchar(50)); 
insert Addresses values (
    ('Tim', 'Penny Lane 1'), 
    ('John', 'Abbey Road 1'), 
    ('John', 'Abbey Road 2'); 

然后，如果你运行：

select c.name 
,  count(o.product) as Products 
,  count(a.address) as Addresses 
from Customers c 
left join Orders o on o.customer_name = c.name 
left join Addresses a on a.customer_name = c.name 
group by name 

你得到：

name Products Addresses 
Tim  1   1 
John 4   4 
Spike 0   0 

但约翰没有4级的产品！
如果没有group by运行，你可以看到为什么计数是关闭：

select * 
from Customers c 
left join Orders o on o.customer_name = c.name 
left join Addresses a on a.customer_name = c.name 

你得到：

name customer_name product  customer_name address 
Tim  Tim    Guitar  Tim    Penny Lane 1 
John John   Drums  John   Abbey Road 1 
John John   Drums  John   Abbey Road 2 
John John   Trumpet  John   Abbey Road 1 
John John   Trumpet  John   Abbey Road 2 
Spike NULL   NULL  NULL   NULL 

正如你所看到的，加入最终重复对方。对于每个产品，重复地址列表。这给你错误的数字。为了解决这个问题，使用优秀的其他答案之一：

select c.name 
,  o.order_count 
,  a.address_count 
from Customers c 
left join 
     (
     select customer_name 
     ,  count(*) as order_count 
     from Orders 
     group by 
       customer_name   
     ) o 
on  o.customer_name = c.name 
left join 
     (
     select customer_name 
     ,  count(*) as address_count 
     from Addresses 
     group by 
       customer_name   
     ) a 
on  a.customer_name = c.name 

子查询确保只有一行每个客户加入。结果好多了：

name order_count address_count 
Tim  1   1 
John 2   2 
Spike NULL  NULL