Question

想知道是否/如何使用SQLAlchemy返回两个博客帖子之间共享的标签列表，或者我是否需要单独拉出每个标签并在Python端进行比较。以此为例（借鉴SQLAlchemy how to filter by children in many to many）：

import sqlalchemy as alc
Session = alc.sessionmaker(bind=engine)
session = Session()
Base = alc.declarative_base(bind=engine)

class Tag(Base):
    __tablename__ = 'tags'
    id = alc.Column(alc.Integer, primary_key=True)
    name = alc.Column(alc.String)
    def __repr__(self): 
        return '<Tag "%s">' % self.name

class Post(Base):
    __tablename__ = 'posts'
    id = alc.Column(alc.Integer, primary_key=True)
    text = alc.Column(alc.String)
    def __repr__(self): 
        return '<Post id %d>' % self.id


class PostTag(Base):
    __tablename__ = 'post_tags'
    post_id = alc.Column(alc.Integer, alc.ForeignKey('posts.id'), primary_key=True)
    tag_id = alc.Column(alc.Integer, alc.ForeignKey('tags.id'), primary_key=True)
    post = alc.relationship("Post", backref="ptags")
    tag = alc.relationship("Tag", backref="tposts")
    tagging_user_name = alc.Column(alc.String)

我试图弄清楚如何构建一个查询，给定两个帖子，将返回两者共享的标签列表。例如，如果我们将post_1与标签＆＃34;食谱＆＃34;，＆＃34;来宾＆＃34;，＆＃34;视频＆＃34;和＆＃34;素食＆＃34;和post_2与标签＆＃34;食谱＆＃34;，＆＃34;音频＆＃34;，＆＃34;甜点＆＃34;，＆＃34;素食＆＃34;，我想要一个返回[<Tag "recipe">, <Tag "vegetarian">]的查询。< / p>

我已经想通了：

session.query(Post).join(PostTag).filter(PostTag.tag.has(name='joke')).all()

将返回标记为＆＃34;笑话＆＃34;在列表中，但我不知道如何找到我可能希望从tags表开始的预感之外的两个帖子之间共享的标签列表。任何帮助将非常感激。

Answer 1

这对我有用：

from sqlalchemy.sql import exists, and_

pt1 = exists().where(and_(PostTag.tag_id == Tag.id, PostTag.post_id == id1))
pt2 = exists().where(and_(PostTag.tag_id == Tag.id, PostTag.post_id == id2))
tags = session.query(Tag.name).filter(pt1, pt2).all()

应该生成类似于以下内容的SQL：

SELECT name FROM tags
WHERE EXISTS(SELECT 1 FROM post_tags WHERE tag_id=id AND post_id=<post id 1>)
AND EXISTS(SELECT 1 FROM post_tags WHERE tag_id=id AND post_id=<post id 2>)

SQLAlchemy：返回两个给定记录共享的关联记录列表

1 个答案: