热门标签 | HotTags
当前位置:  开发笔记 > 编程语言 > 正文

Mysql语句-执行顺序_MySQL

1这样一个问题,作为一个开发人员需要掌握数据库的哪些东西?在开发中涉及到数据库,基本上只用到了sql语句,如何写sql以及对其进行优化就比较重要,那些mysql的厚本书籍针对的是DBA,我们只需要学习其中
1.这样一个问题,作为一个开发人员需要掌握数据库的哪些东西? 在开发中涉及到数据库,基本上只用到了sql语句,如何写sql以及对其进行优化就比较重要,那些mysql的厚本书籍针对的是DBA,我们只需要学习其中的sql就可以了。

2.既然会写sql是目标,那么怎么才能写好sql.学习下面几点:

1)Mysql的执行顺序,这个是写sql的核心,之前遇到的一些错误就是因为对其不了解;

2)如何进行多表查询,优化,这个是很重要的部分;

3)sql语句的函数,sql提供的函数方便了很多操作;

3.这篇对Mysql语句执行顺序的学习做了总结:

1)Mysql语法顺序,即当sql中存在下面的关键字时,它们要保持这样的顺序:

select[distinct]
from
join(如left join)
on
where
group by
having
union
order by
limit


2)Mysql执行顺序,即在执行时sql按照下面的顺序进行执行:

from
on
join
where
group by
having
select
distinct
union
order by

3)针对上面的Mysql语法顺序和执行顺序,循序渐进进行学习:

建立如下表格orders:

\

注:下面所有语句符合语法顺序(也不可能不符合,因为会报错^_^),只分析其执行顺序:(join和on属于多表查询,放在最后展示)

语句一:

select a.Customer
from orders a
where a.Customer='Bush' or a.Customer = 'Adams'

分析一:首先是from语句找到表格,然后根据where得到符合条件的记录,最后select出需要的字段,结果如下:

\
语句二groupby:groupby要和聚合函数一起使用

select a.Customer,sum(a.OrderPrice)
from orders a
where a.Customer='Bush' or a.Customer = 'Adams'
group by a.Customer

分析二:在from,where执行后,执行group by,同时也根据group by的字段,执行sum这个聚合函数。这样的话得到的记录对group by的字段来说是不重复的,结果如下:
\

语句三having:

select a.Customer,sum(a.OrderPrice)
from orders a
where a.Customer='Bush' or a.Customer = 'Adams'
group by a.Customer
having sum(a.OrderPrice) > 2000

分析三:由于where是在group之前执行,那么如何对group by的结果进行筛选,就用到了having,结果如下:

\

语句四distinct: (为测试,先把数据库中Adams那条记录的OrderPrice改为3000)

select distinct sum(a.OrderPrice)
from orders a
where a.Customer='Bush' or a.Customer = 'Adams' or a.Customer = 'Carter'
group by a.Customer
having sum(a.OrderPrice) > 1700

分析四:将得到一条记录(没有distinct,将会是两条同样的记录):

\

语句五union:完全是对select的结果进行合并(默认去掉重复的记录):

select distinct sum(a.OrderPrice) As Order1
from orders a
where a.Customer='Bush' or a.Customer = 'Adams' or a.Customer = 'Carter'
group by a.Customer
having sum(a.OrderPrice) > 1500
union
select distinct sum(a.OrderPrice) As Order1
from orders a
where a.Customer='Bush' or a.Customer = 'Adams' or a.Customer = 'Carter'
group by a.Customer
having sum(a.OrderPrice) > 2000

分析五:默认去掉重复记录(想保留重复记录使用union all),结果如下:

\

语句六order by:

select distinct sum(a.OrderPrice) As order1
from orders a
where a.Customer='Bush' or a.Customer = 'Adams' or a.Customer = 'Carter'
group by a.Customer
having sum(a.OrderPrice) > 1500
union
select distinct sum(a.OrderPrice) As order1
from orders a
where a.Customer='Bush' or a.Customer = 'Adams' or a.Customer = 'Carter'
group by a.Customer
having sum(a.OrderPrice) > 2000
order by order1

分析:升序排序,结果如下:

\

语句七limit:

select distinct sum(a.OrderPrice) As order1
from orders a
where a.Customer='Bush' or a.Customer = 'Adams' or a.Customer = 'Carter'
group by a.Customer
having sum(a.OrderPrice) > 1500
union
select distinct sum(a.OrderPrice) As order1
from orders a
where a.Customer='Bush' or a.Customer = 'Adams' or a.Customer = 'Carter'
group by a.Customer
having sum(a.OrderPrice) > 2000
order by order1
limit 1

分析七:取出结果中的前1条记录,结果如下:

\
语句八(上面基本讲完,下面是join 和 on):

select distinct sum(a.OrderPrice) As order1,sum(d.OrderPrice) As order2
from orders a
left join (select c.* from Orders c) d 
on a.O_Id = d.O_Id
where a.Customer='Bush' or a.Customer = 'Adams' or a.Customer = 'Carter'
group by a.Customer
having sum(a.OrderPrice) > 1500
union
select distinct sum(a.OrderPrice) As order1,sum(e.OrderPrice) As order2
from orders a
left join (select c.* from Orders c) e 
on a.O_Id = e.O_Id
where a.Customer='Bush' or a.Customer = 'Adams' or a.Customer = 'Carter'
group by a.Customer
having sum(a.OrderPrice) > 2000
order by order1
limit 1

分析八:上述语句其实join on就是多连接了一张表,而且是两张一样的表,都是Orders。 执行过程是,在执行from关键字之后根据on指定的条件,把left join指定的表格数据附在from指定的表格后面,然后再执行where字句。

注:

1)使用distinct要写在所有要查询字段的前面,后面有几个字段,就代表修饰几个字段,而不是紧随distinct的字段;

2)group by执行后(有聚合函数),group by后面的字段在结果中一定是唯一的,也就不需要针对这个字段用distinct;

以上就是Mysql语句-执行顺序_MySQL的内容,更多相关内容请关注PHP中文网(www.php1.cn)!

推荐阅读
author-avatar
平凡文学女
这个家伙很懒,什么也没留下!
PHP1.CN | 中国最专业的PHP中文社区 | DevBox开发工具箱 | json解析格式化 |PHP资讯 | PHP教程 | 数据库技术 | 服务器技术 | 前端开发技术 | PHP框架 | 开发工具 | 在线工具
Copyright © 1998 - 2020 PHP1.CN. All Rights Reserved | 京公网安备 11010802041100号 | 京ICP备19059560号-4 | PHP1.CN 第一PHP社区 版权所有