Question

我目前正在编写一个应用程序，允许用户存储图像，然后标记这些图像。我正在使用Python和Peewee ORM（http://charlesleifer.com/docs/peewee/），这与Django的ORM非常相似。

我的数据模型看起来像这样（简化）：

class Image(BaseModel):
    key = CharField()

class Tag(BaseModel):
    tag = CharField()

class TagRelationship(BaseModel):
    relImage = ForeignKeyField(Image)
    relTag   = ForeignKeyField(Tag)

现在，我从概念上理解如何查询具有给定标签集的所有图像：

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

但是，我也希望能够进行更复杂的搜索。具体来说，我希望能够指定一个“所有标签”列表 - 即图像必须包含要返回的所有指定标签，以及“any”列表和“none”列表。< / p>

编辑：我想澄清一点。具体来说，上述查询是“所有标签”式查询。它返回包含所有给定标签的图像。我希望能够指定类似的内容：“给我所有带有标签的图像（绿色，山峰），任何一个标签（背景，风景），但不是标签（数字，绘图）”。

现在，理想情况下，我希望这是一个SQL查询，因为使用LIMIT和OFFSET分页变得非常容易。我实际上有一个实现工作，我只是将所有内容加载到Python集中，然后使用各种交集运算符。我想知道的是，是否有一种方法可以同时执行此操作？

另外，对于那些感兴趣的人，我已经通过电子邮件向Peewee的作者发送了关于如何使用Peewee来表示上述查询的信息，他回答了以下解决方案：

Image.select(['key']).group_by('key').join(TagRelationship).join(Tag).where(tag__in=['tag1', 'tag2']).having('count(*) = 2')

或者，或者更短的版本：

Image.filter(tagrelationship_set__relTag__tag__in=['tag1', 'tag2']).group_by(Image).having('count(*) = 2')

提前感谢您的时间。

Answer 1

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

<强>更新

上述查询的一个更普遍接受的版本（使用CASE表达式将IN谓词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

或使用COUNT而不是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

Answer 2

上半部分获得与强制性标签匹配的单词。下半部分是必须存在至少1的标签。底部查询没有GROUP BY，因为我想知道图像是否出现两次。如果是这样，它既有背景又有风景。 ORDER BY计数（*）将使带有BOTH背景和横向标签的图片显示在顶部。所以绿色，山脉，背景景观将是最相关的。然后是绿色，山脉，背景或风景照片。

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

Answer 3

以下查询必须返回所有标有（'A'和'B'）和（'C'或'D'）但不是'E'和'F'

的图像

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')

搜索多对多关系中的项目

3 个答案: