用变量打开文件

Question

我尝试根据用户输入的输入打开文件。

这是我现在的代码，但它似乎总是直接进入except块，即使我输入了正确的文件名。

filename = input("Enter a filename: ")

try:
    open(filename.txt, "w")
    print("Succesfully opened", filename,".txt")

except:
    print("File cannot be found.")

任何帮助将不胜感激！

Answer 1

这样可行。

filename = input("Enter a filename: ")

try:
    # Access filename as a variable
    open(filename + ".txt", "w")
    print("Succesfully opened", filename,".txt")

# Catch the specific exception
except IOError:
    print("File cannot be found.")

Answer 2

将open(filename.txt, "w")更改为open(filename + '.txt', "w")

Answer 3

由@Bharel标记，这将起作用：

filename = input("Enter a filename: ")

try:
    open(filename + ".txt", "w")
    print("Succesfully opened", filename,".txt")

except:
    print("File cannot be found.")

问题出在open(filename.txt, "w")，因为.txt不是字符串，所以最简单的解决方案是将文件名与扩展名连接起来。