我有一个Rails应用程序,它有 Stations (气象站)和 Observations 。该应用程序在地图上显示当前风速和风向的许多气象站。
我有一个方法用于stations#index
方法,该方法选择站点并加入每个站点的最新观测值。
class Station < ActiveRecord::Base
has_many :observations
def self.with_observations(limit = 1)
eager_load(:observations).where(
observations: { id: Observation.pluck_from_each_station(limit) }
)
end
end
Observation.pluck_from_each_station
返回一组id。 observations
表包含数千行,因此必须保持rails不要急于加载数千条记录。
此方法应返回所有站 - 是否有观察。但目前情况并非如此。
it "includes stations that have no observations" do
new_station = create(:station)
stations = Station.with_observations(2)
expect(stations).to include new_station # fails
end
根据我的理解,LEFT OUTER JOIN
应返回连接表中是否有任何结果的所有行。为什么这不按预期工作?
这是SQL生成的一个示例:
SELECT "stations"."id" AS t0_r0,
"stations"."name" AS t0_r1,
"stations"."hw_id" AS t0_r2,
"stations"."latitude" AS t0_r3,
"stations"."longitude" AS t0_r4,
"stations"."balance" AS t0_r5,
"stations"."timezone" AS t0_r6,
"stations"."user_id" AS t0_r7,
"stations"."created_at" AS t0_r8,
"stations"."updated_at" AS t0_r9,
"stations"."slug" AS t0_r10,
"stations"."speed_calibration" AS t0_r11,
"stations"."firmware_version" AS t0_r12,
"stations"."gsm_software" AS t0_r13,
"stations"."description" AS t0_r14,
"stations"."sampling_rate" AS t0_r15,
"stations"."status" AS t0_r16,
"observations"."id" AS t1_r0,
"observations"."station_id" AS t1_r1,
"observations"."speed" AS t1_r2,
"observations"."direction" AS t1_r3,
"observations"."max_wind_speed" AS t1_r4,
"observations"."min_wind_speed" AS t1_r5,
"observations"."temperature" AS t1_r6,
"observations"."created_at" AS t1_r7,
"observations"."updated_at" AS t1_r8,
"observations"."speed_calibration" AS t1_r9
FROM "stations"
LEFT OUTER JOIN
"observations"
ON "observations"."station_id" = "stations"."id"
WHERE "observations"."id" IN (450, 500, 550, 600, 650, 700, 750, 800);
答案 0 :(得分:1)
我认为会发生这种情况,因为你排除"observations"."id"
之后left join
为空的记录:
eager_load(:observations).where(
'"observations"."id" is null or "observations"."id" in (?)', Observation.pluck_from_each_station(limit)
)
在两个条件下逻辑与left join
相同,但由于rails没有此功能,您可以使用where
子句解决它。