Python 3.5中的压缩，需要一个整数（got type str）

Question

我正在尝试制作一个程序，允许用户输入数据以指定文件名和扩展名，然后压缩该文件并允许他们为其命名。但是我不断收到错误消息“需要一个整数（得到类型 str），它说它与 'rb' 点有关。关于如何解决这个问题的任何信息？

import zlib
first_answer = input("Please input file name")
print(first_answer)
second_answer = input("Please input file extension")
with open(first_answer, ".", second_answer, 'rb') as in_file:
compressed = zlib.compress(in_file.read(), 9)
third_answer = input("What would you like to call this new file?")

with open(third_answer, "wb") as out_file:
out_file.write(compressed)

print("File has been compressed!")

Answer 1

您正在将几个单独的参数传递给open()函数：

with open(first_answer, ".", second_answer, 'rb') as in_file:

open() 函数的第三个位置参数是 buffer 参数，如果指定，它必须始终是整数。

您需要将这些字符串与 + 连接起来，或者使用字符串格式将其变成一个参数：

with open(first_answer + "." + second_answer, 'rb') as in_file:

Python 3.5中的压缩，需要一个整数（got type str）

Compression in Python 3.5, an integer is required (got type str)

python

compression

python-3.x