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Sqlalchemy 烧瓶-炼金术结合2个要求在一个_Sqlalchemy_Flask Sqlalchemy - Fatal编程技术网

Sqlalchemy 烧瓶-炼金术结合2个要求在一个

Sqlalchemy 烧瓶-炼金术结合2个要求在一个,sqlalchemy,flask-sqlalchemy,Sqlalchemy,Flask Sqlalchemy,我希望将这些请求合并为一个: items = Workspace.query.filter(Workspace.user_id==g.user.id).all() items += Workspace.query.join(Member).filter(Member.user_id==g.user.id).all() 这是我的当前代码,但工作不正常,返回错误结果: items = Workspace.query.join(Member)\ .filter(or_(Wor

我希望将这些请求合并为一个:

items = Workspace.query.filter(Workspace.user_id==g.user.id).all()
items += Workspace.query.join(Member).filter(Member.user_id==g.user.id).all()
这是我的当前代码,但工作不正常,返回错误结果:

    items = Workspace.query.join(Member)\
        .filter(or_(Workspace.user_id==g.user.id, Member.user_id==g.user.id))\
        .order_by(Workspace.created_at.desc()).all()
第一次查询中缺少项, 我错在哪里?

你的
。加入(会员)
可能是原因。尝试改用
outerjoin