SQLAlchemy 查詢所有孩子匹配過濾器的父母

Question

我正在嘗試創建一個 sqlalchemy 查詢（flask-sqlalchemy），如果所有子項都匹配查詢而不是 ANY，則該查詢僅返回父對象。 IE

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

將返回所有具有column == True的子對象的所有父對象，我該如何編寫它以使所有子對象都必須具有column == True ？

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

上面的部分是從這個實際代碼中抽象出來的。 在哪里

家長 == 學生和孩子 == StudentApplication

model 層次結構如下：

用戶 ---> 學生 ---> StudentApplication

我正在嘗試檢索有學生但尚未提交申請的用戶，同時忽略至少有 1 名學生提交申請的用戶。

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

這是它生產的 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

如果您正在尋找所有孩子都有column = TRUE的父母，那么這相當於沒有孩子有column = FALSE的所有父母。 如果您使用 JOIN，您會遇到每個孩子而不是每個父母擁有一行的問題。 因此，我建議改用WHERE EXISTS() 。 像這樣的查詢可以解決問題：

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

在 Flask-SQLAlchemy 中，這變成：

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

既然你在談論User s，所有Student s 都完成了所有StudentApplication s，我認為它應該變成

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)
)

這將返回所有用戶。 如果您隨后想要該用戶的學生，我會將其放入單獨的查詢中 - 並在那里應用營銷標志。

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)
                 )