Question

我有一张看起来像这样的照片表：

id  title      rating  photopath
1   myself     7.0     /photopath1.jpg
2   cat        8.0     /photopath2.jpg
3   dog        6.0     /photopath3.jpg
4   girlfriend 5.0     /photopath4.jpg

和标签表：

id  tag_name   photo_id
1   selfie     1
2   sun        1
3   nature     2
4   relax      2
5   loyal      3
6   journal    3
7   selfie     4
8   sun        4
9   problems   4

我想打电话给所有带有标签的照片＆＃34; selfie＆＃34;和＆＃34; sun＆＃34;。我该怎么做？

Answer 1

这是＆＃34; set-within-sets＆＃34;查询。我喜欢使用聚合和having：

来解决这些问题

select p.id
from photos p join
     tags t
     on p.id = t.photo_id
group by p.id
having sum(case when tag_name = 'selfie' then 1 else 0 end) > 0 and
       sum(case when tag_name = 'sun' then 1 else 0 end) > 0;

这种方法的便利之处在于很容易添加更多条件（例如另一个标签）或者反转条件（例如＆＃34;自拍＆＃34; s没有＆＃34; sun＆＃34;）

Answer 2

select p.*
  from photo p
  join (select p.id
          from photo p
          join tag t
            on p.id = t.photo_id
         where t.tag_name in ('selfie', 'sun')
         group by p.id
        having count(*) = 2) x
    on p.id = x.id

小提琴： http://sqlfiddle.com/#!2/5c95c2/2/0

输出：

| ID |      TITLE | RATING |       PHOTOGRAPH |
|----|------------|--------|------------------|
|  1 |     myself |      7 | /photograph1.jpg |
|  4 | girlfriend |      5 | /photograph4.jpg |

Answer 3

SELECT * 
FROM Photos
WHERE EXISTS (SELECT 1
              FROM Tag_Table
              WHERE photo_id = Photos.id
               AND tag_name = 'selfie'
             )
  AND EXISTS (SELECT 1
              FROM Tag_Table
              WHERE photo_id = Photos.id
               AND tag_name = 'Sun'
             )

SQL数据库标记请求

3 个答案: