用内部函数的返回值调用外部函数的返回值，而不检查内部函数的返回值

Question

考虑以下装饰器：

def wraper(function):
    @wraps(function)
    def wrap(*args,**kwargs):
        if function(*args, **kwargs):
             return function(*args, **kwargs)
        else:
            response = JsonResponse()
            response.status_code = 500
            return response 
        return wrap

装饰功能：

@wraper
def function(1):
    var = None
    try:
       var = 1            #Consider somehow an exception is raised
    except Exception as e:
       print(e)
    return var  

考虑其他功能：

def function2():
    print("In function")
    variable = function()  # Assign the variable and proceed normally or return from function2,
                           # with the JsonResponse in case of an exception
    print("Returning from function2")
    return variable

如果返回JsonResponse，我希望函数2的输出为：

In function

以及以下内容，如果一切正常：

In function

Returnning from function2

注意：我希望从修饰函数返回的值是function2的返回值，以防引发异常。在另一种情况下，（当引发异常时）我只想返回JsonResponse，而不执行任何function2的代码

Answer 1

您定义的装饰器从不返回任何内容。您错过了最后一行的回报。

def wraper(function):
    @wraps(function)
    def wrap(*args,**kwargs):
        result = function(*args, **kwargs)
        if result:
            return result
        else:
            response = JsonResponse()
            response.status_code = 500
            return response
    return wrap

如果装饰的函数返回False或等效值，则上面的装饰器将返回响应。否则，它将返回装饰函数返回的内容。

如果您只是想知道function是否返回没有错误，请执行以下操作：

def function():
    # Code that can crash goes here
    return 0 # return value if no errors were found

def function2():
    print("In function")
    try:
        variable = function()
    except Exception: # If an error occurs return the response
        response = JsonResponse()
        response.status_code = 500
        return response
    print("Returning from function2")
    return variable # If everything went well return the value from function

并且请删除所有装饰器。对于这些用例，您不需要它们。