在 SQLAlchemy 中执行联接时出错"请显式指定此联接的'on子句'。

Question

我有以下使用烧瓶-sqlalchemy的sqlalchemy模型。我的日程安排模型中有 3 个人才偏好项目。总需要有 3 个，而且不少。

class TalentPref(db.Model):
    __tablename__ = 'talentpref'
    id = db.Column(db.Integer, primary_key=True)
    firstName = db.Column(db.String(80), unique=True)
    lastName = db.Column(db.String(80), unique=True)
    def __init__(self, firstName, lastName):
        self.firstName = firstName
        self.lastName = lastName
    def __repr__(self):
        return '<talentpref %r %r>' % (self.firstName, self.lastName)
class Schedule(db.Model):
    __tablename__ = 'schedule'
    id = db.Column(db.Integer, primary_key=True)
    talentpref1_id = db.Column(db.Integer, db.ForeignKey('talentpref.id'))
    talentpref2_id = db.Column(db.Integer, db.ForeignKey('talentpref.id'))
    talentpref3_id = db.Column(db.Integer, db.ForeignKey('talentpref.id'))
    talentpref1 = db.relationship("TalentPref", uselist=False, foreign_keys=talentpref1_id)
    talentpref2 = db.relationship("TalentPref", uselist=False, foreign_keys=talentpref2_id)
    talentpref3 = db.relationship("TalentPref", uselist=False, foreign_keys=talentpref3_id)

我正在使用烧瓶不安分将计划模型作为 api 资源提供。当我按计划执行查询并要求按talentpref1__lastName对查询进行排序时，我收到一个错误，该错误与我有多个引用"TalentPref"表的实例有关：

我可以在 id 列上成功使用查询字符串，如下所示：

/api/schedule?id=id&page=1&q={"order_by":[{"field":"id","direction":"desc"}]}

但是使用以下 http GET 查询字符串的查询失败：

/api/schedule?id=id&page=1&q={"order_by":[{"field":"talentpref1__lastName","direction":"desc"}]}

跟：

Traceback (most recent call last):
  File "/Users/myuser/Documents/pythonenvironments/venv/lib/python2.7/site-packages/flask_restless/views.py", line 1172, in _search
    result = search(self.session, self.model, search_params)
  File "/Users/myuser/Documents/pythonenvironments/venv/lib/python2.7/site-packages/flask_restless/search.py", line 587, in search
    query = create_query(session, model, search_params, _ignore_order_by)   
  File "/Users/myuser/Documents/pythonenvironments/venv/lib/python2.7/site-packages/flask_restless/search.py", line 549, in create_query
    _ignore_order_by)
  File "/Users/myuser/Documents/pythonenvironments/venv/lib/python2.7/site-packages/flask_restless/search.py", line 498, in create_query
    query = query.join(relation_model)
  File "/Users/myuser/Documents/pythonenvironments/venv/lib/python2.7/site-packages/sqlalchemy/orm/query.py", line 1971, in join
    from_joinpoint=from_joinpoint)
  File "<string>", line 2, in _join
  File "/Users/myuser/Documents/pythonenvironments/venv/lib/python2.7/site-packages/sqlalchemy/orm/base.py", line 201, in generate
    fn(self, *args[1:], **kw)
  File "/Users/myuser/Documents/pythonenvironments/venv/lib/python2.7/site-packages/sqlalchemy/orm/query.py", line 2115, in _join
    outerjoin, full, create_aliases, prop)
  File "/Users/myuser/Documents/pythonenvironments/venv/lib/python2.7/site-packages/sqlalchemy/orm/query.py", line 2188, in _join_left_to_right
    self._join_to_left(l_info, left, right, onclause, outerjoin, full)   
  File "/Users/myuser/Documents/pythonenvironments/venv/lib/python2.7/site-packages/sqlalchemy/orm/query.py", line 2317, in _join_to_left
    "Tried joining to %s, but got: %s" % (right, ae))
InvalidRequestError: Could not find a FROM clause to join from.  Tried
joining to <class 'app.model.TalentPref'>, but got: Can't determine
join between 'schedule' and 'talentpref'; tables have more than one
foreign key constraint relationship between them. Please specify the
'onclause' of this join explicitly.

有没有办法成功查询此关系？

Answer 1

这似乎是烧瓶不安分本身的限制。当传递<关系名称>__<字段名称>形式的<字段名称>时，它将拆分名称并使用第一部分作为关系。它使用关系属性来查找要联接到的相关模型类：

if '__' in field_name:
    field_name, field_name_in_relation = 
        field_name.split('__')
    relation = getattr(model, field_name)
    relation_model = relation.mapper.class_
    field = getattr(relation_model, field_name_in_relation)
    direction = getattr(field, val.direction)
    query = query.join(relation_model)
    #                  ^^^^^^^^^^^^^^ TalentPref model class
    query = query.order_by(direction())

https://github.com/jfinkels/flask-restless/blob/0.17.0/flask_restless/search.py#L498

在您的情况下，这实际上是有效的

query = query.join(TalentPref)

由于您有多个联接路径，SQLAlchemy 无法确定要执行的操作。如果 flask-restless 使用简单的关系连接而不是加入目标实体，或者使 relationship() 驱动的 ON 子句适应目标实体，这将不是问题。

您可以修补烧瓶不安分来修复此特定查询：

--- search.py   2017-03-29 09:56:00.439981932 +0300
+++ search_joinfix.py   2017-03-29 09:56:41.375851375 +0300
@@ -495,7 +495,7 @@
                         relation_model = relation.mapper.class_
                         field = getattr(relation_model, field_name_in_relation)
                         direction = getattr(field, val.direction)
-                        query = query.join(relation_model)
+                        query = query.join(relation)
                         query = query.order_by(direction())
                     else:
                         field = getattr(model, val.field)