File Open Function with Try & Except Python 2.7.1

Question

def FileCheck(fn):       
       try:
           fn=open("TestFile.txt","U") 
       except IOError: 
           print "Error: File does not appear to exist."
       return 0 

I'm trying to make a function that checks to see if a file exists and if doesn't then it should print the error message and return 0 . Why isn't this working???

Answer 1

You'll need to indent the return 0 if you want to return from within the except block. Also, your argument isn't doing much of anything. Instead of assigning it the filehandle, I assume you want this function to be able to test any file? If not, you don't need any arguments.

def FileCheck(fn):
    try:
      open(fn, "r")
      return 1
    except IOError:
      print "Error: File does not appear to exist."
      return 0

result = FileCheck("testfile")
print result

Answer 2

This is likely because you want to open the file in read mode. Replace the "U" with "r".

Of course, you can use os.path.isfile('filepath') too.