Question

我有这个SQL代码，我想在收费单和收据上显示每个项目的总和：

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

它产生这个输出：

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，因此当您获得sum()时，您可能会遇到导致问题的一对多关系。因此，您可以使用子查询来获取结果。对于每个sum()，这将获得receipt和chargeslip的{{1}}，然后将其加入到您的item_number表中以获得最终结果：< / p>

items

Answer 2

首先按Item_Number执行GROUP BY，因此您不会将Receipt_Detail和ChargeSlip_Detail中的行相乘。也就是说，在加入Item_Number

之前，每Items生成一个SUM值

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

左连接从左表中返回行，对于左表中的每一行，返回右表中的所有匹配行。

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');

然后如果你跑：

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种产品！如果您在没有group by的情况下运行，您可以看到计数关闭的原因：

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

正如您所看到的，连接最终会相互重复。对于每个产品，重复地址列表。这给了你错误的数量。要解决这个问题，请使用其中一个优秀的答案：

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

子查询确保每个客户只加入一行。结果好多了：

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

交叉乘法表

3 个答案: