为什么 super() 只调用 class Parent1 的构造函数而不调用 class Parent2 的构造函数？

Question

class Parent1:
    def __init__(self):
        self.data1 = 10
        print("parent1")

class Parent2:
    def __init__(self):
        self.data2 = 20 
        print("parent2")

class child(Parent2,Parent1):
    def __init__(self):
        print("this is child class")
        super().__init__()
        print(self.data1)
        print(self.data2)
        
obj = child()

Output:

this is child class
parent2
Traceback (most recent call last):
  File "f:/Project2020/Rough Work/rough2.py", line 18, in <module>
    obj = child()
  File "f:/Project2020/Rough Work/rough2.py", line 15, in __init__
    print(self.data1)
AttributeError: 'child' object has no attribute 'data1'

Answer 1

Answer to your question: in python there is an inheritance chain. In your case this inheritance chain looks like [Child, Parent2, Parent1],
so kinda for the python compiler there's no such thing as multiple inheritance. super() is just a short-hand to walk one step up the inheritance chain. Therefore your statement is equal to Parent2.__init__() and indeed super().__init__() calls Parent2.__init__() as it should.
So you also need to call Parent1.__init__() from class Child .

I suggest writing:

class Parent1:
    def __init__(self):
        self.data1 = 10
        print("parent1")

class Parent2:
    def __init__(self):
        self.data2 = 20 
        print("parent2")

class Child(Parent2,Parent1):
    def __init__(self):
        print("this is child class")
        Parent2.__init__(self)
        Parent1.__init__(self)
        print(self.data1)
        print(self.data2)

In this situation the code is clearer if you don't use super() . This code may look odd but here using super() would be just confusing anyway.