MySQL 具有多个表的多个内部联接

声明:本页面是StackOverFlow热门问题的中英对照翻译,遵循CC BY-SA 4.0协议,如果您需要使用它,必须同样遵循CC BY-SA许可,注明原文地址和作者信息,同时你必须将它归于原作者(不是我):StackOverFlow 原文地址: http://stackoverflow.com/questions/7150088/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me): StackOverFlow

提示:将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间:2020-08-31 20:54:45  来源:igfitidea点击:

Multiple inner joins with multiple tables

mysqlsqljoininner-join

提问by ATMathew

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.

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。

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  

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.

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具有哪个 ID。基本上,我想知道表 1 中的哪些值与表 4 中的值相关联。

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?

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。那有意义吗?

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.

无论如何,我想知道使用 4 个内部连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。

回答by Narnian

Inner joins are probably the best method, and you only need 3.

内连接可能是最好的方法,您只需要 3 个。

This will give you a result set with two columns: company and associated values.

这将为您提供一个包含两列的结果集:公司和相关值。

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