Question

我在Django中有一个调用外部库/类的视图。问题是由于某些原因，Django保持缓存来自该类之前调用的结果。

请考虑以下简单示例：

Django观点：

from some_path import Demo
def test_view(request):
    demo = Demo()
    result = demo.do_something()
    return render(request, 'test.html',
                            { 'result':result }
                )

演示课程：

class Demo():
    result = []

    def do_something(self):
        self.result.append(1)
        self.result.append(2)
        self.result.append(3)
        return self.result

你期望结果是[1,2,3]，对吗？错误！

第一次加载页面时，您将获得正确的结果。但是在以下所有请求中它将继续递增：[1,2,3,1,2,3] ...... [1,2,3,1,2,3,1,2,3] ......

所以我的问题很明显 - 这里发生了什么？每当我在Django视图中调用一个类时，我如何收到[1,2,3]？

Django 1.7 / MacOS X。

Answer 1

将result中的__init__定义为实例属性。

class Demo():

    def __init__(self):
        self.result = []

    def do_something(self):
        self.result.append(1)
        self.result.append(2)
        self.result.append(3)
        return self.result

如果您在代码中打印result，那么result只会分配一次，

class Demo():
    result = []
    print result

    def ...
        .
        .

d = Demo()
print d.do_something()
print d.do_something()
e = Demo()
print e.do_something()
>>> 
[]
[1, 2, 3]
[1, 2, 3, 1, 2, 3]
[1, 2, 3, 1, 2, 3, 1, 2, 3]

result是一个可变对象，每当你创建一个类的实例时它将result引用到同一个引用。如果是Immutable对象： -

class Demo():
    result = 1

    def do_something(self):
        self.result += 1

d = Demo()
d.do_something()


d.do_something()
e = Demo()
e.do_something()
print d.result, e.result

输出： -

>>> 
3 2

Answer 2

为什么不尝试在result=[]方法中初始化do_something这样的内容？

def do_something(self):
        result = []
        result.append(1)
        result.append(2)
        result.append(3)
        return result

Answer 3

"""
Case 1:
Here we have declared the result and some_string as static Class variable 
"""
class Demo_1():
    result = []
    some_string = ""

    def do_something(self):
        Demo_1.some_string = Demo_1.some_string + "Python "
        self.result.append(1)
        self.result.append(2)
        self.result.append(3)
        return self.result

demo = Demo_1()
print Demo_1.result
result = demo.do_something()
print Demo_1.result

demo = Demo_1()
result = demo.do_something()
print Demo_1.result

demo = Demo_1()
result = demo.do_something()
print Demo_1.result

print result
print demo.some_string

print "Demo_1 class attributes :-", dir(Demo_1)

"""
Case1 Output :
    []
    [1, 2, 3]
    [1, 2, 3, 1, 2, 3]
    [1, 2, 3, 1, 2, 3, 1, 2, 3]
    [1, 2, 3, 1, 2, 3, 1, 2, 3]
    Python Python Python 
    Demo_1 class attributes :- ['__doc__', '__module__', 'do_something', 'result', 'some_string']

As you can see both result and some_string exists in Demo_1's attributes 
which you can access by both by Class name as well as it's instance 
"""

print "-----------------------------------------------------------------"

"""
Case 2:
Here we have declared the result variable as Class instance variable.
So each time we create an instance of class it creates new memory for that instance.
"""
class Demo_2():

    def __init__(self):
        self.result = []
        self.some_string = " "

    def do_something(self):
        self.result.append(1)
        self.result.append(2)
        self.result.append(3)
        return self.result

demo = Demo_2()
result = demo.do_something()
print demo.result

demo = Demo_2()
result = demo.do_something()
print demo.result

demo = Demo_2()
result = demo.do_something()
print demo.result

print result
print "Demo_1 class attributes :-", dir(Demo_2)
print Demo_2.some_string

"""
Case2 Output :
    [1, 2, 3]
    [1, 2, 3]
    [1, 2, 3]
    [1, 2, 3]
    Demo_1 class attributes :- ['__doc__', '__init__', '__module__', 'do_something']
    Traceback (most recent call last):
      File "bl.py", line 83, in <module>
        print Demo_2.some_string
    AttributeError: class Demo_2 has no attribute 'some_string'


As you can see Class Demo_2 has no attributes result or some_string as 
they are instance variable.
"""

详细了解静态类变量 Here

Django奇怪的行为

3 个答案: