在多对多关系中搜索项目
我现在正在写一个应用程序,可以存储图片,并给这些图片打标签。我使用的是Python和Peewee ORM(http://charlesleifer.com/docs/peewee/),它和Django的ORM非常相似。
我的数据模型看起来是这样的(简化版):
class Image(BaseModel):
key = CharField()
class Tag(BaseModel):
tag = CharField()
class TagRelationship(BaseModel):
relImage = ForeignKeyField(Image)
relTag = ForeignKeyField(Tag)
现在,我大致明白如何查询所有带有特定标签的图片:
SELECT Image.key
FROM Image
INNER JOIN TagRelationship
ON Image.ID = TagRelationship.ImageID
INNER JOIN Tag
ON TagRelationship.TagID = Tag.ID
WHERE Tag.tag
IN ( 'A' , 'B' ) -- list of multiple tags
GROUP BY Image.key
HAVING COUNT(*) = 2 -- where 2 == the number of tags specified, above
不过,我还想进行更复杂的搜索。具体来说,我希望能够指定一个“所有标签”的列表——也就是说,返回的图片必须包含所有指定的标签,同时还要有一个“任意标签”的列表和一个“无标签”的列表。
编辑:我想稍微澄清一下。具体来说,上面的查询是一个“所有标签”的查询。它返回的是所有包含给定标签的图片。我想能够指定类似于:“给我所有带有标签(绿色,山脉)的图片,任意一个标签(背景,风景),但不包含标签(数字,绘画)”。
理想情况下,我希望这能成为一个SQL查询,因为这样使用LIMIT和OFFSET进行分页就非常简单了。实际上,我已经有一个实现方法,就是把所有内容加载到Python的集合中,然后使用各种交集运算符。我想知道是否有办法一次性完成这一切?
另外,对于感兴趣的人,我已经给Peewee的作者发了邮件,询问如何用Peewee表示上述查询,他给出了以下解决方案:
Image.select(['key']).group_by('key').join(TagRelationship).join(Tag).where(tag__in=['tag1', 'tag2']).having('count(*) = 2')
或者,另外一个更简短的版本:
Image.filter(tagrelationship_set__relTag__tag__in=['tag1', 'tag2']).group_by(Image).having('count(*) = 2')
谢谢你们的时间。
3 个回答
这个查询的意思是:你需要找出所有的图片,这些图片同时被标记为'A'和'B',并且还要被标记为'C'或者'D',但是不能被标记为'E'和'F'。
SELECT Image.key
FROM Image
INNER JOIN TagRelationship
ON Image.ID = TagRelationship.ImageID
INNER JOIN Tag tag1
ON TagRelationship.TagID = tag1.ID
INNER JOIN Tag tag2
ON TagRelationship.TagID = tag2.ID
WHERE tag1.tag
IN ( 'A' , 'B' )
AND tag2.tag NOT IN ('E', 'F')
GROUP BY Image.key
HAVING COUNT(*) = 2
UNION
SELECT Image.key
FROM Image
INNER JOIN TagRelationship
ON Image.ID = TagRelationship.ImageID
INNER JOIN Tag tag1
ON TagRelationship.TagID = tag1.ID
INNER JOIN Tag tag2
ON TagRelationship.TagID = tag2.ID
WHERE tag1.tag
IN ( 'C' , 'D' )
AND tag2.tag NOT IN ('E', 'F')
上半部分是找出那些符合必须标签的词。下半部分则是找出至少要有一个标签的情况。下半部分的查询没有使用GROUP BY,因为我想知道一张图片是否出现了两次。如果出现了两次,那就说明它同时有背景和风景这两个标签。使用ORDER BY count(*)可以让同时带有背景和风景标签的图片排在最前面。所以像“绿色、山脉、背景风景”这样的图片会是最相关的。接下来是“绿色、山脉、背景或风景”的图片。
SELECT Image.key, count(*) AS 'relevance'
FROM
(SELECT Image.key
FROM
--good image candidates
(SELECT Image.key
FROM Image
WHERE Image.key NOT IN
--Bad Images
(SELECT DISTINCT(Image.key) --Will reduce size of set, remove duplicates
FROM Image
INNER JOIN TagRelationship
ON Image.ID = TagRelationship.ImageID
INNER JOIN Tag
ON TagRelationship.TagID = Tag.ID
WHERE Tag.tag
IN ('digital', 'drawing' )))
INNER JOIN TagRelationship
ON Image.ID = TagRelationship.ImageID
INNER JOIN Tag
ON TagRelationship.TagID = Tag.ID
WHERE Tag.tag
IN ('green', 'mountain')
GROUP BY Image.key
HAVING COUNT(*) = count('green', 'mountain')
--we need green AND mountain
UNION ALL
--Get all images with one of the following 2 tags
SELECT *
FROM
(SELECT Image.key
FROM Image
INNER JOIN TagRelationship
ON Image.ID = TagRelationship.ImageID
INNER JOIN Tag
ON TagRelationship.TagID = Tag.ID
WHERE Tag.tag
IN ( 'background' , 'landscape' ))
)
GROUP BY Image.key
ORDER BY relevance DESC
SELECT Image.key
FROM Image
JOIN TagRelationship
ON Image.ID = TagRelationship.ImageID
JOIN Tag
ON TagRelationship.TagID = Tag.ID
GROUP BY Image.key
HAVING SUM(Tag.tag IN (mandatory tags )) = N /*the number of mandatory tags*/
AND SUM(Tag.tag IN (optional tags )) > 0
AND SUM(Tag.tag IN (prohibited tags)) = 0
更新
上面这个查询的一个更普遍接受的版本(使用CASE表达式将IN条件的布尔结果转换为整数):
SELECT Image.key
FROM Image
JOIN TagRelationship
ON Image.ID = TagRelationship.ImageID
JOIN Tag
ON TagRelationship.TagID = Tag.ID
GROUP BY Image.key
HAVING SUM(CASE WHEN Tag.tag IN (mandatory tags ) THEN 1 ELSE 0 END) = N /*the number of mandatory tags*/
AND SUM(CASE WHEN Tag.tag IN (optional tags ) THEN 1 ELSE 0 END) > 0
AND SUM(CASE WHEN Tag.tag IN (prohibited tags) THEN 1 ELSE 0 END) = 0
或者用COUNT代替SUM:
SELECT Image.key
FROM Image
JOIN TagRelationship
ON Image.ID = TagRelationship.ImageID
JOIN Tag
ON TagRelationship.TagID = Tag.ID
GROUP BY Image.key
HAVING COUNT(CASE WHEN Tag.tag IN (mandatory tags ) THEN 1 END) = N /*the number of mandatory tags*/
AND COUNT(CASE WHEN Tag.tag IN (optional tags ) THEN 1 END) > 0
AND COUNT(CASE WHEN Tag.tag IN (prohibited tags) THEN 1 END) = 0