需要一些小帮助来解决这种情况。我看到codeigniter不支持UNION,我想找出这个方法来获得两个表的总和,具有相同的乘积id。
问题是,在sum 2表上,值是重复的或乘法的。
链接到SQL Fiddle
结构:
create table products
(id int, name varchar(9));
insert into products
(id, name)
values
(1,'Product 1'),
(2,'Product 2'),
(3,'Product 3');
create table p_items
(puid int, prid int, quantity int);
insert into p_items
(puid, prid, quantity)
values
(11,1,100),
(11,2,100),
(11,2,100),
(14,2,100),
(14,3,100),
(15,3,100);
create table s_items
(puid int, prid int, quantity int);
insert into s_items
(puid, prid, quantity)
values
(11,1,1),
(11,2,1),
(11,2,1),
(13,2,1),
(15,3,1),
(13,3,1);
在正常SQL:中执行
select a.puid, b.name, a.prid, sum(a.quantity) as P_ITEMS, sum(c.quantity) as S_ITEMS
from p_items a
join products b
on b.id = a.prid
join s_items c
on c.prid = a.prid
group by a.prid;
代码点火器功能:
$this->alerts
->select('products.id as productid, products.name, sumt(p_items.quantity), sum(s_items.quantity)', FALSE)
->from('products')
->join('p_items', 'products.id = p_items.prid', 'left')
->join('s_items', 'products.id = s_items.prid')
->group_by("products.id");
echo $this->alerts->generate();
非常感谢您的帮助。
您的问题是生成笛卡尔乘积,从而得到重复的和。以产品ID 3为例。您将其与p_items prid=3相关联。就其本身而言,这将给你200英镑的回报。然而,一旦您在s_items prid=3上加入,现在对于s_items中的每一行,它都必须与p_items中每一行相匹配。含义:
14 3 100 15 3 1
14 3 100 15 3 1
15 3 100 15 3 1
15 3 100 15 3 1
---------------
400 4
除了产品表,哪张表是您的主表?如果使用p_items,则需要从s_items中获取第13行。同样,如果使用s_items,则不会从p_items获得第14行。
您可以使用子查询完成此操作:
select b.id,
b.name,
P_ITEMS,
S_ITEMS
from products b
left join
(SELECT prid, SUM(quantity) as P_Items
FROM p_items
GROUP BY prid)
a on b.id = a.prid
left join
(SELECT prid, SUM(quantity) as S_Items
FROM s_items
GROUP BY prid)
c on c.prid = a.prid
group by b.id, b.name
以及更新后的Fiddel:http://sqlfiddle.com/#!2/62b45/12
祝你好运。
如果获得所需答案的最佳方法是使用联合查询,而codeigniter不允许您编写联合查询,则不要使用codeignite器编写查询。将其放入存储过程中,并使用codeigniter调用存储过程。