我已经查看了文档,我似乎不知道如何在SQLAlchemy中执行OR查询。我只想做这个问题。

SELECT address FROM addressbook WHERE city='boston' AND (lastname='bulger' OR firstname='whitey')

应该是这样的

addr = session.query(AddressBook).filter(City == "boston").filter(????)

当前回答

这真的很有帮助。 以下是我对任何给定表的实现:

def sql_replace(self, tableobject, dictargs):

    #missing check of table object is valid
    primarykeys = [key.name for key in inspect(tableobject).primary_key]

    filterargs = []
    for primkeys in primarykeys:
        if dictargs[primkeys] is not None:
            filterargs.append(getattr(db.RT_eqmtvsdata, primkeys) == dictargs[primkeys])
        else:
            return

    query = select([tableobject]).where(and_(*filterargs))

    if self.r_ExecuteAndErrorChk2(query)[primarykeys[0]] is not None:
        # update
        filter = and_(*filterargs)
        query = tableobject.__table__.update().values(dictargs).where(filter)
        return self.w_ExecuteAndErrorChk2(query)

    else:
        query = tableobject.__table__.insert().values(dictargs)
        return self.w_ExecuteAndErrorChk2(query)

# example usage
inrow = {'eqmtvs_id': eqmtvsid, 'datetime': dtime, 'param_id': paramid}

self.sql_replace(tableobject=db.RT_eqmtvsdata, dictargs=inrow)

其他回答

SQLAlchemy重载了按位操作符&、|和~,因此,您可以使用这些操作符来代替or_()和and_()(就像Bastien的回答一样)的丑陋且难以阅读的前缀语法:

.filter((AddressBook.lastname == 'bulger') | (AddressBook.firstname == 'whitey'))

注意,由于按位操作符的优先级,括号不是可选的。

所以你的整个查询可以是这样的:

addr = session.query(AddressBook) \
    .filter(AddressBook.city == "boston") \
    .filter((AddressBook.lastname == 'bulger') | (AddressBook.firstname == 'whitey'))

这真的很有帮助。 以下是我对任何给定表的实现:

def sql_replace(self, tableobject, dictargs):

    #missing check of table object is valid
    primarykeys = [key.name for key in inspect(tableobject).primary_key]

    filterargs = []
    for primkeys in primarykeys:
        if dictargs[primkeys] is not None:
            filterargs.append(getattr(db.RT_eqmtvsdata, primkeys) == dictargs[primkeys])
        else:
            return

    query = select([tableobject]).where(and_(*filterargs))

    if self.r_ExecuteAndErrorChk2(query)[primarykeys[0]] is not None:
        # update
        filter = and_(*filterargs)
        query = tableobject.__table__.update().values(dictargs).where(filter)
        return self.w_ExecuteAndErrorChk2(query)

    else:
        query = tableobject.__table__.insert().values(dictargs)
        return self.w_ExecuteAndErrorChk2(query)

# example usage
inrow = {'eqmtvs_id': eqmtvsid, 'datetime': dtime, 'param_id': paramid}

self.sql_replace(tableobject=db.RT_eqmtvsdata, dictargs=inrow)

如果你需要应用一个或条件,如果条件是满足过滤器可以存储在变量和链接它们与管道。这里有一个例子

 class Student(db.Model):
     __tablename__ = "student"
     id = Column(Integer, primary_key=True)
     name = Column(String, nullable=False)

   def get_by_name(self, name):
      condition = # Code Here for condition
      filters = (Student.name == "Stack") | (Student.name == "Overflow") if 
condition else (Student.name == "Stack")
      query = Student.query.filter(filters).order_by(Student.id.asc())

or_()函数在OR查询组件数量未知的情况下很有用。

例如,让我们假设我们正在创建一个带有几个可选过滤器的REST服务,如果任何过滤器返回true,则应该返回记录。另一方面,如果请求中没有定义参数,则查询不应更改。如果没有or_()函数,我们必须这样做:

query = Book.query
if filter.title and filter.author:
    query = query.filter((Book.title.ilike(filter.title))|(Book.author.ilike(filter.author)))
else if filter.title:
    query = query.filter(Book.title.ilike(filter.title))
else if filter.author:
    query = query.filter(Book.author.ilike(filter.author))

使用or_()函数,它可以被重写为:

query = Book.query
not_null_filters = []
if filter.title:
    not_null_filters.append(Book.title.ilike(filter.title))
if filter.author:
    not_null_filters.append(Book.author.ilike(filter.author))

if len(not_null_filters) > 0:
    query = query.filter(or_(*not_null_filters))

对于SQLAlchemy ORM 2.0, |和or_都被接受。

文档

from sqlalchemy.future import select
from sqlalchemy.sql import or_


query = select(User).where(or_(User.name == 'ed', User.name == 'wendy'))
print(query)

# also possible:

query = select(User).where((User.name == 'ed') | (User.name == 'wendy'))
print(query)