故事是一个自由职业者程序员消失了,而未完成的工作留给了前者-包括一些MySQL查询。
我可以管理简单的东西,但是这超出了我的能力。
该查询返回商店中所有产品的排序方式
products_date_added
而应该只提取144个最新结果。
我尝试在其末尾添加限制144,但显然这种方式不起作用。
查询如下:
select p.products_id,
p.products_status,
p.products_quantity,
if(length(pd1.products_name),pd1.products_name, pd.products_name) as products_name,
if(p.products_image_lrg is null, p.products_image, p.products_image_lrg) as products_image,
p.products_model,
p.products_price,
pd.products_description_short,
p.products_tax_class_id,
p.products_date_added,
p.products_master,
p.products_master_status,
p.products_listing_status,
m.manufacturers_name
from products_description pd,
products p
left join products_description pd1 on pd1.products_id = p.products_id
and pd1.language_id='1'
and pd1.affiliate_id = '0'
left join products_prices pp on p.products_id = pp.products_id
and pp.groups_id = '0'
and pp.currencies_id = '0'
left join manufacturers m on (p.manufacturers_id = m.manufacturers_id)
where p.products_status > 0
and (p.products_master='0' or p.products_master is null or p.products_master='')
and p.products_master_status!='1'
and pd.affiliate_id = 0
and p.products_id = pd.products_id
and if(pp.products_group_price is null,
pp.products_group_price != -1 )
and pd.affiliate_id = 0
and pd.language_id = '1'
order by p.products_date_added DESC, products_name
我将不胜感激。
最佳答案
根据您的评论,这可能是您想要的。我只是将products
和products_description
之间的联接移动到添加LIMIT
的派生表(子查询)中:
select p.products_id,
p.products_status,
p.products_quantity,
if(length(pd1.products_name),pd1.products_name, p.products_name) as products_name,
if(p.products_image_lrg is null, p.products_image, p.products_image_lrg) as products_image,
p.products_model,
p.products_price,
p.products_description_short,
p.products_tax_class_id,
p.products_date_added,
p.products_master,
p.products_master_status,
p.products_listing_status,
m.manufacturers_name
from
(
select * -- you might have to list the actual columns to avoid a "duplicate name" error
from products_description pd
join products p on p.products_id = pd.products_id
where p.products_status > 0
and (p.products_master='0' or p.products_master is null
or p.products_master='')
and p.products_master_status!='1'
and pd.affiliate_id = 0
and if(pp.products_group_price is null,
pp.products_group_price != -1 )
and pd.affiliate_id = 0
and pd.language_id = '1'
order by p.products_date_added DESC, products_name
LIMIT 144
) p
left join products_description pd1 on pd1.products_id = p.products_id
and pd1.language_id='1'
and pd1.affiliate_id = '0'
left join products_prices pp on p.products_id = pp.products_id
and pp.groups_id = '0'
and pp.currencies_id = '0'
left join manufacturers m on (p.manufacturers_id = m.manufacturers_id)
order by p.products_date_added DESC, products_name
关于mysql - 巨大的多重连接,需要将结果限制为144,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/23660370/