的raw_input的Python的文件寫入

Question

我有以下代碼：

print "We're going to write to a file you'll be prompted for" 
targetfile = raw_input('Enter a filename: ') 
targetfilefound = open('targetfile' , 'w') 
print "What do we write in this file?" 
targetfilefound.write("hello this is working!") 
targetfilefound.close() 

我創建的腳本應該能夠寫入到用戶的raw_input通過定義文件。以上可能是核心問題，可以提出建議。

Answer 1

通過腳本打印你的東西來看可能要用戶輸入什麼應該被打印到文件，以便：

print "We're going to write to a file you'll be prompted for" 
targetfile = raw_input('Enter a filename: ') 
targetfilefound = open(targetfile , 'w') 
print "What do we write in this file?" 
targetfilefound.write(raw_input()) 
targetfilefound.close() 

注意：此方法將創建如果新文件它不存在。如果您想檢查文件是否存在，你可以使用os模塊，像這樣：

import os 

print "We're going to write to a file you'll be prompted for" 
targetfile = raw_input('Enter a filename: ') 
if os.path.isfile(targetfile) == True: 
    targetfilefound = open(targetfile , 'w') 
    print "What do we write in this file?" 
    targetfilefound.write(raw_input()) 
    targetfilefound.close() 
else: 
    print "File does not exist, do you want to create it? (Y/n)" 
    action = raw_input('> ') 
    if action == 'Y' or action == 'y': 
     targetfilefound = open(targetfile , 'w') 
     print "What do we write in this file?" 
     targetfilefound.write(raw_input()) 
     targetfilefound.close() 
    else: 
     print "No action taken" 

Answer 2

正如其他人指出的那樣，從目標文件中刪除引號，因爲您已經將其分配給變量。

但實際上，而不是寫的代碼，你可以用開放使用下面

with open('somefile.txt', 'a') as the_file: 
    the_file.write('hello this is working!\n') 

給出在上述情況下，你不需要做任何異常處理，而處理文件。當發生錯誤時，文件遊標對象會自動關閉，我們不需要明確地關閉它。即使寫入文件成功，它也會自動關閉文件指針引用。

Explanation of efficient use of with from Pershing Programming blog