Question

swimset_styletag

id          swimset_id  styletag_id
----------  ----------  -----------
1           1           1          
2           2           2          
3           3           1          
4           3           2

swimset

id          set_description        
----------  -----------------------
1           4 x 100 Free Interval A
2           4 x 100 Back Interval A
3           6 x 100 free/back on in

styletag

id          tag_name  
----------  ----------
1           Free      
2           back

我正在尝试查询没有styletag 2的游泳集。所以他们只设置返回应该是swimset_id 1.但是，当我尝试使用时：

SELECT swimset_id, set_description, styletag_id
FROM swimset, swimset_styletag, styletag
WHERE
swimset.id = swimset_styletag.swimset_id AND swimset_styletag.styletag_id = styletag.id AND
swimset.styletag_id !=2;

我得到swimset_id 1和3而不仅仅是swimset_id 1.我想这是因为在swimset_styletag中，swimset_id 3有两个条目，一个命中styletag_id 1，另一个命中styletag_id 2.有没有办法同时处理两个swimset_id条目在swimset_styletag作为一个条目？

如果这是应该通过不同的数据库设计解决的问题，我也会接受任何建议，因为我仍在开发精确的设计。当我在没有数据库的情况下对一些python脚本进行原型设计时，我最初使用模拟数据作为True / False值与每个样式标记的键配对。如果我只是将这些样式标签作为True / False条目添加到swimset表中，这将允许一个平面数据库，但似乎更丑陋？是不是那种糟糕的做法而不是有多对多的样式标签？

Answer 1

我认为这就是你所追求的目标：

SELECT swimset.id, swimset.set_description, swimset_styletag.styletag_id
FROM swimset_styletag INNER JOIN
    swimset on swimset_styletag.swimset_id=swimset.id INNER JOIN 
    styletag on swimset_styletag.styletag_id=styletag.id
WHERE
swimset_id NOT IN (select swimset_id from swimset_styletag WHERE styletag_id=2)

请参阅SQL Fiddle

中的结果

删除多个styletag_id：

SELECT swimset.id, swimset.set_description, swimset_styletag.styletag_id
FROM swimset_styletag INNER JOIN
    swimset on swimset_styletag.swimset_id=swimset.id INNER JOIN 
    styletag on swimset_styletag.styletag_id=styletag.id
WHERE
swimset_id NOT IN (select swimset_id from swimset_styletag WHERE styletag_id IN (2,3,4,5,6))

如果manytomany_id = x，则过滤掉SQL

1 个答案: