多个表的多个内连接
所以我有四张桌子.每个表都有一个对应于前一个表 id 的 id.所以我的点击表有一个 id 和它来自的广告的 id.在广告表中,它有一个广告 id 和它来自的广告系列的 id.所以这是一个例子.
So I have four tables. Each table has a single id for the previous table id. So my in click table has an id and an id for the ad from which it came. In the ad table, it has an id for the ad and one for the campaign it's from. So here's an example.
Table4 -
id company table_id
11 hp 20
12 apple 23
13 kohls 26
14 target 21
15 borders 28
Table3 -
id value table2_id
21 ks 53
22 al 54
23 tx 53
24 fl 55
25 co 51
Table2 -
id value table1_id
51 ks 34
52 al 34
53 tx 33
54 fl 35
55 co 31
Table1 -
id value
31 ks
32 al
33 tx
34 fl
35 co
因此,要找出表 4 中的值的来源,我需要返回每个表并检查它们的 ID.基本上,我想知道表 1 中的哪些值与表 4 中的值相关联.
So to find out where the values in Table 4 came from, I need to work back through each table and check which id they have. Basically, I want to know which values in table 1 are associated with the values in table 4.
表 4 为网站访问者,表 1 为互联网广告.我想知道哪些访问者来自哪些广告.不幸的是,数据的设置使我只能从访问者到来源再到广告组再到广告仅几步之遥.有意义吗?
This of table 4 as visitors to a website and Table 1 as internet ads. I want to know which visitors came from which ads. Unfortunately, the data is set up so that I can only take single steps back from visitor to source to ad group to ad. Does that make sense?
无论如何,我想知道使用 4 个内部联接是否是解决这个问题的最佳策略,或者是否有一些我不知道的更简单的 mysql 解决方案.
Anyways, I'm wondering if using 4 innner joins was the optimal strategy for this problem or is there some simpler mysql solution that i'm not aware of.
推荐答案
内连接可能是最好的方法,你只需要 3 个.
Inner joins are probably the best method, and you only need 3.
这将为您提供一个包含两列的结果集:公司和相关值.
This will give you a result set with two columns: company and associated values.
SELECT Table4.company, table1.id, table1.value
FROM Table1
INNER JOIN Table2
ON Table2.table1_id = Table1.id
INNER JOIN Table3
ON Table3.table2_id = Table2.id
INNER JOIN Table4
ON Table4.table3_id = Table3.id
相关文章