在SQLAlchemy/Flask中连接多张表格

Question

在SQLAlchemy/Flask中连接多张表格

36

我正在尝试在SQLAlchemy中找到正确的连接查询设置，但似乎无法理解。

我有以下表设置（简化了，省略了非关键字段）：

class Group(db.Model):
    id            = db.Column(db.Integer, primary_key = True)
    number        = db.Column(db.SmallInteger, index = True, unique = True)
    member        = db.relationship('Member', backref = 'groups', lazy = 'dynamic')

class Member(db.Model):
    id            = db.Column(db.Integer, primary_key = True)
    number        = db.Column(db.SmallInteger, index = True)
    groupid       = db.Column(db.Integer, db.ForeignKey('group.id'))
    item          = db.relationship('Item', backref = 'members', lazy = 'dynamic')

class Version(db.Model):
    id           = db.Column(db.Integer, primary_key = True)
    name         = db.Column(db.String(80), index = True)
    items        = db.relationship('Item', backref='versions', lazy='dynamic')  

class Item(db.Model):
    id           = db.Column(db.Integer, primary_key = True)
    member       = db.Column(db.Integer, db.ForeignKey('member.id'))
    version      = db.Column(db.Integer, db.ForeignKey('version.id'))

所以关系如下:

1:n 组成员
1:n 成员项
1:n 版本项

我想通过选择数据库中具有特定版本的所有Item行来构建查询。然后，我想按组和成员顺序对它们进行排序。使用Flask / WTForm的输出应该类似于这样:

* GroupA
  * MemberA
     * ItemA (version = selected by user)
     * ItemB ( dito )
  * Member B
     * ItemC ( dito )
  ....

我想出了下面这样的查询语句，但我相当确定它是不正确的（而且效率低下）

   session.query(Item,Member,Group,Version)
    .join(Member).filter(version.id==1)
    .order_by(Group).order_by(Member).all()

我的第一直觉是创建类似于

Item.query.join(Member, Item.member==Member.id)
    .filter(Member.versions.name=='MySelection')
    .order_by(Member.number).order_by(Group.number)

但显然这根本行不通。在版本表上的联接操作似乎没有产生我预期的两个表之间的联接类型。也许我完全误解了这个概念，但在阅读教程后，这对我来说应该是有意义的。

- user3347953

1个回答

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- van · Accepted Answer

以下代码可以一次性地获取您需要的多个对象：

q = (session.query(Group, Member, Item, Version)
        .join(Member)
        .join(Item)
        .join(Version)
        .filter(Version.name == my_version)
        .order_by(Group.number)
        .order_by(Member.number)
        ).all()
print_tree(q)

然而，您得到的结果将是一个元组列表(Group, Member, Item, Version)。现在轮到您以树形式显示它了。下面的代码可能会有所帮助：

def print_tree(rows):
    def get_level_diff(row1, row2):
        """ Returns tuple: (from, to) of different item positions.  """
        if row1 is None: # first row handling
            return (0, len(row2))
        assert len(row1) == len(row2)
        for col in range(len(row1)):
            if row1[col] != row2[col]:
                return (col, len(row2))
        assert False, "should not have duplicates"

    prev_row = None
    for row in rows:
        level = get_level_diff(prev_row, row)
        for l in range(*level):
            print 2 * l * " ", row[l]
            prev_row = row

更新-1：如果您愿意放弃在前两个关系上使用 lazy = 'dynamic'，您可以使用以下代码加载整个对象网络（而不是上面的元组）：

q = (session.query(Group)
        .join(Member)
        .join(Item)
        .join(Version)
        # @note: here we are tricking sqlalchemy to think that we loaded all these relationships,
        # even though we filter them out by version. Please use this only to get data and display,
        # but not to continue working with it as if it were a regular UnitOfWork
        .options(
            contains_eager(Group.member).
            contains_eager(Member.items).
            contains_eager(Item.version)
            )
        .filter(Version.name == my_version)
        .order_by(Group.number)
        .order_by(Member.number)
        ).all()

# print tree: easy navigation of relationships
for g in q:
    print "", g
    for m in g.member:
        print 2 * " ", m
        for i in m.items:
            print 4 * " ", i