SQLAlchemy 查询所有子级匹配过滤器的父级

Question

我正在尝试创建一个 sqlalchemy 查询 (flask-sqlalchemy)，如果所有子项都匹配查询而不是 ANY，则该查询仅返回父对象。浏览器

(Parent
 .query
 .join(Child)
 .filter(Child.column == True)
 ).all()

将返回所有具有column == True 的子对象的父对象，我如何编写它以使所有子对象都必须具有column == True？

------------------- 更新 --------------- ---------------

以上部分是从这段实际代码中抽象出来的。在哪里

家长 == 学生和孩子 == StudentApplication

模型层次结构如下：

用户 ---> 学生 ---> StudentApplication

我正在尝试检索有学生但申请尚未提交的用户，同时忽略至少有 1 名学生已提交申请的用户。

def _active_student_query(statement=None):
    import sqlalchemy as sa
    statement = statement or _active_user_query()
    statement = statement.join(Student).filter(Student.suspended == sa.false())
    return statement

def users_without_submitted_applications(exclude_contacted=False):
    import sqlalchemy as sa
    from sqlalchemy.orm.util import AliasedClass

# AppAlias = AliasedClass(StudentApplication)

has_any_approved_app = sa.exists(
    sa.select([])
        .select_from(Student)
        .where((StudentApplication.student_id == Student.id) &
               (StudentApplication.flags.op('&')(AppFlags.SUBMITTED) > 0),
               )
)

statement = _active_student_query()
statement = (statement
             # .join(StudentApplication)
             # .filter(User.students.any())
             # has a student and no submitted applications
             .filter(~has_any_approved_app)
             # .filter(StudentApplication.flags.op('&')(AppFlags.SUBMITTED) == 0)
             )
if exclude_contacted:
    statement = (statement
                 .join(AlertPreferences)
                 .filter(AlertPreferences.marketing_flags.op('&')(MarketingFlags.NO_APP_PING) == 0)
                 )

from lib.logger import logger
logger.info(statement.sql)
return statement

这是它生成的 SQL

SELECT users.is_active, users.flags, users.created_on, users.updated_on, users.id
FROM users JOIN student ON users.id = student.user_id 
WHERE users.is_active = true AND student.suspended = false AND NOT (EXISTS (SELECT  
FROM student_application 
WHERE student_application.student_id = student.id AND (student_application.flags & %(flags_1)s) > %(param_1)s))

Answer 1

如果您正在寻找所有孩子都有column = TRUE 的父母，那么这相当于所有孩子都没有column = FALSE 的父母。如果您使用 JOIN，您会遇到每个孩子而不是每个父母拥有一行的问题。因此，我建议改用WHERE EXISTS()。像这样的查询可以解决问题：

SELECT *
FROM parent
WHERE NOT EXISTS(
    SELECT
    FROM child
    WHERE child.parent_id = parent.id
      AND child.column = FALSE
)

在 Flask-SQLAlchemy 中变成：

import sqlalchemy as sa
has_any_false_children = sa.exists(
  sa.select([])
  .select_from(Child)
  .where((Child.parent_id == Parent.id) &
         (Child.column == sa.false()))
)

Parent.query.filter(~has_any_false_children)

---

更新

既然你说的是Users，所有Students 都完成了所有StudentApplications，我认为它应该变成

student_has_any_non_approved_app = sa.exists(
    sa.select([])
      .select_from(StudentApplication)
      .where((StudentApplication.student_id == Student.id) &
             (StudentApplication.flags.op('&')(AppFlags.SUBMITTED) > 0) &
             ((StudentApplication.flags.op('&')(AppFlags.APPROVED) == 0) |
              (StudentApplication.flags.op('&')(AppFlags.PARTIALLY_APPROVED) == 0)))
)

user_has_any_non_approved_students = sa.exists(
    sa.select([])
      .select_from(Student)
      .where((Student.user_id == User.id) &
                         (Student.suspended == sa.false()) &
             student_has_any_non_approved_app)
)

statement = (
        _active_user_query()
        .filter(User.students.any())
      # has a student and no submitted applications
      .filter(~user_has_any_non_approved_students)
)

这将返回所有用户。如果您随后想要该用户的学生，我会将其放入单独的查询中 - 并在那里应用营销标志。

statement = Student.query.filter(
    Student.user_id.in_(user_ids),
    Student.suspended == sa.false()
)

if exclude_contacted:
    statement = (statement
                 .join(AlertPreferences)
                 .filter(AlertPreferences.marketing_flags.op('&')(MarketingFlags.NO_APP_PING) == 0)
                 )