在MySQL中有效地连接3个表

Question

I have 3 tables like this:

images (~10,000 rows)

id | place_id | filename 
---|----------|----------
 1 |     4    |  abc.jpg
 2 |     3    |  def.jpg
 3 |     4    |  ghi.jpg
 4 |     7    |  jkl.jpg

tags (~100 rows)

id |   name   |
---|----------|
 1 |   tagA   |
 2 |   tagB   |
 3 |   tagC   |
 4 |   tagD   |

tagsToImages (~30,000 rows)

id |  tag_id  | image_id
---|----------|----------
 1 |     1    |    4
 2 |     3    |    2
 3 |     2    |    4
 4 |     1    |    1

As an example, the last table shows that the tag with id = 1 is linked with the image with id = 4.

I've got two questions, one pretty simple (I think!), and one more difficult.

Simple question

Given a place_id, how can I list all the tags that have image in that place_id? An example would be:

Given place_id = 7, return tagA, tagB.

Hard question

What I'd like to do is to select all the images with a certain place_id, and associate each image with a list of tags like so:

Select all with place_id = 4, whilst joining to tags info.

 filename |    tags    
-----------------------
  abc.jpg | tagA       
  ghi.jpg | tagA, tagB 

Would this be better to do in PHP by performing multiple queries, or can I do this using MYSQL?

Answer 1

1. SELECT i.place_id, t.name as tag_name
FROM images i
INNER JOIN tagsToImages tti ON (tti.image_id = i.id)
INNER JOIN tags t ON (t.id = tti.tag_id)
WHERE i.place_id = 7

2. SELECT i.filename, GROUP_CONCAT(t.name SEPARATOR ',') AS tags
FROM images i
INNER JOIN tagsToImages tti ON (tti.image_id = i.id)
INNER JOIN tags t ON (t.id = tti.tag_id)
WHERE i.place_id = 4 GROUP BY i.filename

It's always better to avoid multiple queries if possible.
*Be aware that the maximum length of GROUP_CONCAT is controlled by group_concat_max_len variable. If you are expecting to get a long list of concatenated strings, you might need to change the default value.

UPDATE
If you want the queries to display images with no tags associated, change INNER JOIN s to LEFT JOIN s

Answer 2

Simple Question

SELECT name 
FROM images 
RIGHT JOIN tagsToImages on images.id = tagsToImages.image_id 
RIGHT JOIN tags ON tags.tag_id = tagsToImages.image_id 
WHERE place_id = '7'

Answer 3

For simple question, use join which will be more efficient:

select
    t.name as `Tag`
from
    images i
left join
    tagsToImages ti on ti.image_id = i.id
left join
    tags t on t.id = ti.tag_id
where
    place_id = $place_id

For hard question, see GROUP_CONCAT : http://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html#function_group-concat

Answer 4

使用join发出单个查询比添加循环，多个数据结构以及手动比较PHP中的值要好。