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我现在到 Pony ORM，我已经创建了这样的数据库模型：

class RSSSource(db.Entity):
    title = Required(str, max_len=100)
    link = Required(str, max_len=800)
    key = Required(str, max_len=200)
    description = Required(str, max_len=800)
    is_deleted = Optional(bool, sql_default=False)
    rss_content = Set('RSS', reverse='source')
    subscriptions = Set("Subscription", reverse='source')

    def __str__(self):
        return self.title

class RSS(db.Entity):
    title = Optional(str, max_len=300)
    link = Optional(str, max_len=500)
    description = Required(str, max_len=500)
    source = Optional("RSSSource", reverse="rss_content")
    pub_date = Optional(datetime)
    is_deleted = Optional(datetime, sql_default=False)
    likes = Set("Like", reverse="rss")

    def __str__(self):
        return self.title

class User(db.Entity):
    first_name = Optional(str, max_len=100)
    last_name = Optional(str, max_len=100)
    username = Required(str, max_len=200, unique=True)
    password = Required(str, max_len=500)
    is_deleted = Optional(bool, sql_default=False)
    subscriptions = Set("Subscription", reverse="user")
    likes = Set("Like", reverse="user")

    def __str__(self):
        return self.username

class Subscription(db.Entity):
    source = Optional("RSSSource", reverse="subscriptions")
    user = Optional("User", reverse="subscriptions")
    is_deleted = Optional(bool, sql_default=False)

现在我想执行一个类似于 Django ORM 的选择查询。像这样的东西：

select user.id
from Subscription where source.key = $key and (is_deleted is null or is_deleted = FALSE)

我找不到办法做到这一点。有人可以帮忙吗？

4

1 回答 1

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我终于想出了这样的东西，它奏效了。

select user
from Subscription as s
left join RSSSource as rs on rs.id = s.source
where rs.key = $key and (s.is_deleted is null or s.is_deleted = FALSE)
于 2021-09-29T16:34:15.040 回答