用變量打開一個文件

Question

我試圖根據用戶輸入的輸入打開一個文件。

這是我現在的代碼，但它似乎總是直接轉到except塊，即使在輸入正確的文件名時也是如此。

filename = input("Enter a filename: ") 

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

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

任何幫助，將不勝感激！

Answer 1

這將工作。

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

變化open(filename.txt, "w")到open(filename + '.txt', "w")

Answer 3

正如標誌着@Bharel，這將工作：

filename = input("Enter a filename: ") 

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

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

的問題是在open(filename.txt, "w")，因爲.TXT是不是字符串，那麼最簡單的方法是連接與文件名延期。