python - 如何在带有条件if的sqlalchemy中使用条件运算符or_？

Question

现有代码片段:

if sup_usr_only:
    query_ob = query_ob.filter(
        or_(
            and_(
                Department.id.in_(login_user.department_ids),  # logic is OK - checked.
                model.visibility == VISIBLE_DEPT
            ),

            and_(
                model.visibility == VISIBLE_COMPANY,
                model.company_id == login_user.company_id
            ))
    )
else:
    query_ob = query_ob.filter(
        or_(
            and_(
                Department.id.in_(login_user.department_ids),  # logic is OK - checked.
                model.visibility == VISIBLE_DEPT
            ),
            model.visibility == VISIBLE_GLOBAL,
            and_(
                model.visibility == VISIBLE_COMPANY,
                model.company_id == login_user.company_id
            ))
    )

是否有一种方法可以通过 in line if check 或任何其他优化来最小化代码片段？想要像下面这样(这在语法上是错误的):

query_ob = query_ob.filter(
            or_(
                and_(
                    Department.id.in_(login_user.department_ids),
                    model.visibility == VISIBLE_DEPT
                ),
                model.visibility == VISIBLE_GLOBAL if not sup_usr_only,
                and_(
                    model.visibility == VISIBLE_COMPANY,
                    model.company_id == login_user.company_id
                ))
        )

Answer 1

您可以预先在列表中为 or_ 创建参数，然后将它们应用于 or_() 函数:

options = [
    and_(
        Department.id.in_(login_user.department_ids),
        model.visibility == VISIBLE_DEPT
    ),
    and_(
        model.visibility == VISIBLE_COMPANY,
        model.company_id == login_user.company_id
    )]

if not sup_usr_only:
    options.append(model.visibility == VISIBLE_GLOBAL)

query_ob = query_ob.filter(or_(*options))

对于数据库而言，OR 语句中选项的排列顺序无关紧要，但如果您觉得顺序对您的应用程序很重要，则可以使用 options.insert( 1、model.visibility == VISIBLE_GLOBAL)也是。