定义函数

Question

定义此函数时出现语法错误。

def questionfilehandler("filename.txt"):
    with open("filename.txt", "r") as file:
        print(file.read)
        return input()
        file.close()

我查了一下语法，这一切似乎都是正确的 This is the error message I got
And this is the code with the error highlighted by IDLE.

感谢所有阅读并试图回答此问题的人。非常感谢您的时间=）。

Answer 1

假设您的缩进得到修复，这很明显......您不能直接将字符串作为函数参数调用。你需要一个变量：

def questionfilehandler(filename):
    with open(filename, "r") as file:
        print(file.read())
        return input()
        # file.close() - not needed

然后......你可以用字符串作为参数来调用函数：

questionfilehandler("filename.txt")

Answer 2

def questionfilehandler(filename = "data.txt"): # filename has default value so it will take your input if you provide with function call.
    with open(filename) as filedata:
        # print(file.read()) # no need for this method as all work done by "open()".
        for data in filedata:
            print data

questionfilehandler() # You can pass file name if you want to else keep the default.

不需要filename.close（）作为“with”操作符自动处理它。