在SQLAlchemy / Flask中联接多个表

一尘不染

在SQLAlchemy / Flask中联接多个表

flask

我正在尝试找出SQLAlchemy中正确的联接查询设置，但似乎无法解决。

我有以下表格设置（简化后，我省略了非必要字段）：

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'))

因此，关系如下：

1:n Group Member
1:n Member Item
1:n Version Item

我想通过从数据库中选择具有特定版本的所有项目行来构建查询。然后，我想按组然后按成员订购。使用Flask / WTForm的输出应如下所示：

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

我想出了类似以下查询的内容，但是我很确定它是不正确的（效率低下）

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

我的第一个直观方法是创建类似

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

但显然，这根本不起作用。版本表上的联接操作似乎并未产生我期望的两个表之间的联接类型。也许我完全误解了这个概念，但是在阅读了教程之后，这对我来说很有意义。

阅读 568

2020-04-07

共1个答案

一尘不染

以下将在一个查询中为你提供所需的对象：

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)

但是，你得到的结果将是元组列表(Group, Member, Item, Version)。现在由你决定以树形形式显示它。下面的代码可能被证明是有用的：

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

Update-1：如果你愿意放弃lazy = 'dynamic'前两个关系，则可以执行查询以object network使用以下代码加载整个对象（与上述元组相对）：

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

2020-04-07