从数据库查询创建类实例 - Python

Question

我无法弄清楚这一点，并花了相当多的时间在这里查看问题和答案，但似乎没有什么比较合适的。

我有一个类定义为传感器创建一个实例，持有各种属性和功能。

在数据库中，我已定义连接的传感器。现在，如果有三个传感器，我不得不这么像这样...

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

我想这样做的是通过迭代数据库，返回所有定义的传感器（很容易的），然后创建基于该列表的类的实例。

我无法弄清楚如何采取一个SQLite查询的结果，并使其成为类实例的名称...

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]) 

基本上我需要创建一个类的实例的X号，其中X是数据库表中定义每个传感器的行数。

这一个让我感到莫名其妙 - 先谢谢了！最后一个选项的

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 

一个缺点是，它让你与参考传感器和那些没有全局变量之间没有明确的划分。基于字典的方法（前两个建议）和基于列表的方法（第三个建议）可以很容易地访问所有传感器，而只是传感器。例如（前三种情况），很容易在所有传感器上循环;在最后一种情况下，它更加棘手。

奖励：请注意，我拒绝使用名称id（而不是sid）的诱惑，因为这会影响内建。