跨乘法表

Question

所以我有這樣的SQL代碼，我想顯示其收費單，並在他們收到的每個項目的總和：

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 

它產生這樣的輸出：

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

但它應該是：

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

我的代碼出了什麼問題？

Answer 1

由於您正在加入表格，因此當您獲取sum()時，您可能會遇到一對一的關係，從而導致問題。所以你可以使用子查詢來獲得結果。這將得到sum()爲receipt和chargeslip每個item_number，然後你加入該回你items表得到最終的結果是：

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執行分組依據，第一，

，讓你不乘以Receipt_Detail和ChargeSlip_Detail。也就是說，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

左前生成每個Item_Number值的總和左表聯接返回行，並在左邊的表的每一行，在右邊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'); 

然後，如果你運行：

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 

但約翰沒有4級的產品！
如果沒有group by運行，你可以看到爲什麼計數是關閉：

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，加入最終重複對方。對於每個產品，重複地址列表。這給你錯誤的數字。爲了解決這個問題，使用優秀的其他答案之一：

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 

子查詢確保只有一行每個客戶加入。結果好多了：

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