從數據庫查詢創建類實例 - Python

Question

我無法弄清楚這一點，並花了相當多的時間在這裏查看問題和答案，但似乎沒有什麼比較合適的。

我有一個類定義爲傳感器創建一個實例，持有各種屬性和功能。

在數據庫中，我已定義連接的傳感器。現在，如果有三個傳感器，我不得不這麼像這樣...

sensor1 = Sensor(1) 
sensor2 = Sensor(2) 
sensor3 = Sensor(3) 

我想這樣做的是通過迭代數據庫，返回所有定義的傳感器（很容易的），然後創建基於該列表的類的實例。

我無法弄清楚如何採取一個SQLite查詢的結果，並使其成爲類實例的名稱...

con = sqlite.connect('temp.db') 
with connection: 
    cur = connection.cursor() 
    cur.exectute('SELECT Sensor_Id FROM Sensors') 
    sensors = cur.fetchall() 

# I want to do something like this. The right side of the statement is correct, but how 
# can I make the left side of the assignment be dynamic based on SQL result? 
for n in sensors: 
    ***sensorN*** = ClassName(n[0]) 

基本上我需要創建一個類的實例的X號，其中X是數據庫表中定義每個傳感器的行數。

這一個讓我感到莫名其妙 - 先謝謝了！最後一個選項的

Answer 1

con = sqlite.connect('temp.db') 
with connection: 
    cur = connection.cursor() 
    cur.exectute('SELECT Sensor_Id FROM Sensors') 
    sensor_ids = cur.fetchall() 

# I would recommend this approach 
sensor = { sid : Sensor(sid) for sid in sensor_ids } 
# Which allows you to access them as 
sensor['Main-hall-3'], sensor['Flap-track-fairing-7'] # etc. 

# If the sensor ids are integers, you might prefer 
sensor = { int(sid) : Sensor(sid) for sid in sensor_ids } 
# Which allows you to access them as 
sensor[3], sensor[7] # etc 

# If you know that the ids are going to be CONSECUTIVE INTEGERS 
# STARTING AT 0, then you could do 
sensor = [ Sensor(sid) for sid in sensor_ids ] 
# Which allows you to access them as 
sensor[0], sensor[1] # etc 

# Now, if you *really* insist on having the sensors bound to names, 
# then you could try 
import sys 
current_module = sys.modules[__name__] 
for sid in sensor_ids: 
    setattr(current_module, 'sensor{}'.format(sid), Sensor(sid)) 
# Which would allow you to access them as 
sensor1, sensor2 # etc 

一個缺點是，它讓你與參考傳感器和那些沒有全局變量之間沒有明確的劃分。基於字典的方法（前兩個建議）和基於列表的方法（第三個建議）可以很容易地訪問所有傳感器，而只是傳感器。例如（前三種情況），很容易在所有傳感器上循環;在最後一種情況下，它更加棘手。

獎勵：請注意，我拒絕使用名稱id（而不是sid）的誘惑，因爲這會影響內建。