将“计数”列表从第一个函数拉入第二个函数以更新选项卡

Question

尝试将“counts”列表从第一个函数拉入第二个函数以更新表。你知道吗

我在使用Python，SQLAlchemy，sqlite。我觉得问题出在我的Python身上。是名单吗，塔普？我该怎么做？你知道吗

def countCurrentOccupancy():
    p = session.query(Puppy.shelter_id, func.count(Puppy.shelter_id))
    grouped_count = p.group_by (Puppy.shelter_id).all()
    return grouped_count


def updateCurrentOccupancy():
    l = countCurrentOccupancy()
    list_c = list(l)
    s = session.query(Shelter.shelter_id, Shelter.current_occupancy)
    print 'updateCurrentOccupancy:'
    print '----------------------'
    for row_s in s:
        row_s = list_c #Trying to pull matching data from countCurrentOccupancy(), will update to an integer that I choose but that is not helpful.
        print row_s
    session.commit(row_s)
    showShelter()

结果：

TypeError: commit() takes exactly 1 argument (2 given)

Answer 1

您可以将列表转换为dict{shelter\u id:count}，并通过shelter_id获得count

def updateCurrentOccupancy():
    counts_list = countCurrentOccupancy()
    counts_dict = dict(counts_list)

    shelters = session.query(Shelter)
    for shelter in shelters:
        shelter.current_occupancy = counts_dict.get(shelter.id)
        session.add(shelter)

    session.commit()