Opening a .txt file in Python

Question

I'm trying to open a .txt file in Python with the following function.

def get_my_string():
   """Returns a string of the text"""
   f = open("/home/Documents/text.txt", 'r')
   string = str(f.read())
   f.close()
   return string

I want "string" to be a string of the text from the opened file. However, after calling the function above, "string" is an empty list.

Solution

def get_my_string():
    """Returns the file inputFn"""

    inputFn = "/home/Documents/text.txt"

    try:
        with open(inputFn) as inputFileHandle:
            return inputFileHandle.read()

    except IOError:
        sys.stderr.write( "[myScript] - Error: Could not open %s\n" % (inputFn) )
        sys.exit(-1)

Comments

[Niklas B.]

The file will not be closed if its completely empty. It would be more compact and clear to use with open(inputFn, 'r') as f: return f.next()

[John Machin]

-1 (1) this reads only the first line, which is not what the OP said he wanted (2) if there is an IOError, it supresses the detailed error info (file doesn't exist? caller doesn't have permissions? etc?) and gratuitously calls sys.exit

[Alex Reynolds]

Thanks, I edited my answer to reflect the change for item 1. I only wanted to show the try...except block as a way to show how to think about error checking. Sorry this upset you.