MySQL]多表关联查询技巧 - Flume

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MySQL]多表关联查询技巧

2019-05-11 02:07:13 【大中小】浏览:57次

示例表A:

author_id	author_name
1	Kimmy
2	Abel
3	Bill
4	Berton

示例表B:

book_id	author_id	start_date	end_date
9	1	2017-09-25 21:16:04	2017-09-25 21:16:06
10	3
11	2	2017-09-25 21:21:46	2017-09-25 21:21:47
12	1
13	8

示例表C:

order_id	book_id	price	order_date
1	9	0.2	2017-09-24 21:21:46
2	9	0.6	2017-09-25 21:16:04
3	11	0.1	2017-09-25 21:21:46

在以上表中执行AB表关联

1 2	`SELECT``authors`.*, `books`.book_id`FROM``authors` `LEFTJOIN``books``ON``authors`.author_id = `books`.author_id

结果

author_id	author_name	book_id
1	Kimmy	9
3	Bill	10
2	Abel	11
1	Kimmy	12
4	Berton

结果出现了2条author_id为1的记录，因为右表中存在了两条关联author_id=1的行

右边出现N条关联左边的记录，结果就会相应出现N条关联了右表出现的记录

在以上表中执行ABC表关联

SELECT`authors`.*, `books`.book_id, `orders`.order_id, `orders`.priceFROM`authors`

LEFTJOIN`books`ON`authors`.author_id = `books`.author_id

LEFTJOIN`orders`ON`books`.book_id = `orders`.book_id

结果

author_id	author_name	book_id	order_id	order_price
1	Kimmy	9	1	0.2
1	Kimmy	9	2	0.6
2	Abel	11	3	0.1
3	Bill	10
1	Kimmy	12
4	Berton

结果出现了3条author_id=1的记录，因为authors第一次关联了books表book_id为9和12的book关联了author_id为1的作者，而book_id为9的书本则关联了两个orders记录，所以结果集包含3条author_id为1的记录

可以运用

1	`count(),sum()`

等函数通过

groupby

来统计结果

SELECT`authors`.*,sum(`orders`.price)FROM`authors`

LEFTJOIN`books`ON`authors`.author_id = `books`.author_id

LEFTJOIN`orders`ON`books`.book_id = `orders`.book_id

GROUPBY`books`.book_id

结果集会基于book_id来统计每一本书的订单总额

author_id	author_name	book_id	sum(order_price)
4	Berton
1	Kimmy	9	0.80
3	Bill	10
2	Abel	11	0.10
1	Kimmy	12

book_id为9的订单总额为0.80，并且9的记录从多条合并为1条。

多条件join

SELECT`authors`.*, `books`.book_id, `orders`.order_id,sum(`orders`.price)FROM`authors`

LEFTJOIN`books`ON`authors`.author_id = `books`.author_id

LEFTJOIN`orders`ON`books`.book_id = `orders`.book_idAND`orders`.order_date >= `books`.start_dateAND`orders`.order_date <= `books`.end_date

GROUPBY`books`.book_id

选取在一定时间区间范围内的order订单，可以看到订单order_id为1的订单不再纳入book_id为9的统计当中，因为它的时间区间不符合join条件

author_id	author_name	book_id	order_id	sum(`order`.price)
4	Berton
1	Kimmy	9	2	0.60
3	Bill	10
2	Abel	11	3	0.10
1	Kimmy	12

关于where的使用，看下面示范

SELECT`authors`.*, `books`.book_id, `orders`.order_id,sum(`orders`.price)ASpricesFROM`authors`

LEFTJOIN`books`ON`authors`.author_id = `books`.author_id

LEFTJOIN`orders`ON`books`.book_id = `orders`.book_idAND`orders`.order_date >= `books`.start_dateAND`orders`.order_date <= `books`.end_date

WHEREpricesisnotNULL

GROUPBY`books`.book_id

以上语句假设选取price不为空的记录，导致了一个错误的出现

1	`[Err] 1054 - Unknowncolumn'prices'in'where clause'`

因为where不能用于选取列的AS别名判断，MYSQL的处理机制是先进行选取，再进行筛选，在选取阶段就启用了where条件，因为这时并不存在prices的筛选结果后才产生的字段，所以这里会抛出错误

我们可以这样做

SELECT`authors`.*, `books`.book_id, `orders`.order_id,sum(`orders`.price)ASpricesFROM`authors`

LEFTJOIN`books`ON`authors`.author_id = `books`.author_id

LEFTJOIN`orders`ON`books`.book_id = `orders`.book_idAND`orders`.order_date >= `books`.start_dateAND`orders`.order_date <= `books`.end_date

WHERE`orders`.priceisnotNULL

GROUPBY`books`.book_id

选取阶段order表是存在price字段的，所以只有price不为空的记录才会被选取

author_id	author_name	book_id	order_id	prices
2	Abel	11	3	0.10
1	Kimmy	9	2	0.60

运用

having

对那些无法进行 WHERE 的AS别名的字段进行一些筛选查询

SELECT`authors`.*, `books`.book_id,sum(`orders`.price)ASpricesFROM`authors`

LEFTJOIN`books`ON`authors`.author_id = `books`.author_id

LEFTJOIN`orders`ON`books`.book_id = `orders`.book_id

GROUPBY`books`.book_id

HAVINGprices > 0.1

这时只有sum为0.8的结果被选中

author_id	author_name	book_id	sum(order_price)
1	Kimmy	9	0.80

对于组合其他语法查询，也是没问题的

SELECT`authors`.*, `books`.book_id,sum(`orders`.price)ASpricesFROM`authors`

LEFTJOIN`books`ON`authors`.author_id = `books`.author_id

LEFTJOIN`orders`ON`books`.book_id = `orders`.book_id

GROUPBY`books`.book_id

HAVINGprices >= 0.1

ORDERBYpricesasc

LIMIT 1,1


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