做php开发,遇到一个数据库处理问题:
现在要处理表inventory,这个表的一个字段goods_id.关联到表goods,但goods表里的多个字段又关连到其对应表;
这个表的另一个字段goods_level关联到表level。
不知道能否通过sql语句直接搞定,得到全部对应值?
如果不能通过sql语句搞定,要用什么方法来实现这个目的?请各位指点一二,谢!
还是上图好理解:
回复讨论(解决方案)
select * from inventory, goods, size, brand where inventory.goods_id=goods.goods_id and goods.goods_size=size.id and goods.goods_brand=brand.id
只要找出表与表之间的关系,对应表关系,不是难事,关键是你sql基础牢不牢。
xuzuning兄,你没考虑到level表。
zy205817,我就是基础不牢,要不然哪会再这提问。如果你知道怎么解决,麻烦指点下,感激不尽
select * from inventory, goods, size, brand, level where inventory.goods_id=goods.goods_id and goods.goods_size=size.id and goods.goods_brand=brand.id and inventory.goods_level=level.id
zy205817兄,按你写的执行报错了,说:
#1054-
unknow column inventory.goods_levle in where clause
确定表inventory有这个字段goods_level
给你一个复杂点的...
给你解释一下...分成两部分,第一个是先组成一个临时表为c,这个表是查询你的goods表然后左连接到你的size表跟brand表.
然后查询你的inventory表再左连接上面的c表跟level表...
select a.id,a.supplier,a.warehouse_id,b.level,c.goods_name,c.size,c.brand from inventory as a left join level as b on a.goods_level = b.id left join (select a.goods_id as goods_id, a.goods_name as goods_name,b.brand as brand,c.size as size from goods as a left join brand as b on a.goods_brand=b.id left join size as c on a.goods_size=c.id) as c on a.goods_id=c.goods_id
谢谢leungigi的解答!解释很到位,一看就明白了,你一定是个良师益友!
顺路学习,感觉也很好。

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