mysql - Complex 加入 Peewee

Question

问这个问题很尴尬，因为它似乎微不足道，但我找不到有效的解决方案。

我有以下功能:

def planner(departure_id, arrival_id):
    departure = Stop.get(Stop.id == departure_id)
    arrival = Stop.get(Stop.id == arrival_id)
    buses = Bus.select().join(RideStopRelationship).join(Stop).where(Stop.id == departure)
    for bus in buses:
        print bus.line
        for stop in bus.stops:
            print stop.time, stop.stop.name

基于以下模型:

class Stop(BaseModel):
    name = CharField()
    #lat = FloatField()
    #lng = FloatField()

class Bus(BaseModel):
    line = IntegerField()
    number = IntegerField()
    direction = IntegerField()

class RideStopRelationship(BaseModel):
    bus = ForeignKeyField(Bus, related_name = "stops")
    stop = ForeignKeyField(Stop, related_name = "buses")
    time = TimeField()

关键行是 Bus.select().join(RideStopRelationship).join(Stop).where(Stop.id == department)。我正在尝试让所有将在 departure 和 arrival 停靠的公共(public)汽车。但是，上述查询返回所有停靠在 departure 的公交车。我如何获得在“出发”和“到达”都停靠的巴士？

如果我把它弄得太复杂(要么是我的模型太复杂，要么是我的查询)，请随时纠正我。

编辑: 有一种方法确实有效:

buses_departure = Bus.select().join(RideStopRelationship).join(Stop).where(Stop.id == departure)
buses_arrival = Bus.select().join(RideStopRelationship).join(Stop).where(Stop.id == arrival)
buses = Bus.select().where(Bus.id << buses_departure & Bus.id << buses_arrival)

但是对于应该是简单查询的内容来说它相当长...

Answer 1

你可以尝试这样的事情:

departure = Stop.get(...)
arrival = Stop.get(...)
query = (Bus
         .select(Bus)
         .join(RideStopRelationship)
         .where(RideStopRelationship.stop << [departure, arrival])
         .group_by(Bus)
         .having(fn.Count(Bus.id) == 2))

无关，但需要注意的一件事是，由于 python 评估运算符的方式，您需要在 in 查询中加上括号:

buses = Bus.select().where(
    (Bus.id << buses_departure) & 
    *Bus.id << buses_arrival))