Python passing user input of name of file to open

Question

I don't understand why something this simple isn't very intuitive.

This is what I have so far:

    print("Enter name of file on Desktop:")
    filename = sys.stdin.readline()


    directory = os.path.join(os.path.expanduser("~"),"Desktop")
        for subdir, dirs, files in os.walk(directory):
            for file in files:
                if file.startswith(filename):
                    f = open(os.path.join(subdir, file),'r+', encoding="Latin-1")
                    data = f.read()
                    print(data)
                    f.close()

Why can't I pass the sys.stdin that was assigned to the variable filename to the 'if' statement:

   `if file.startswith(filename):`?

I tried:

    if file.startswith(str(filename)):
    if file.startswith("'" + filename + "'")
    if file.startswith(filename):

no options seem to "go through" and no errors pop up.

It just pretends like I didn't pass anything. Any help? Thanks.

Format code by indenting by four spaces. Only use backticks for small pieces of code. — Carcigenicate
– Carcigenicate, Commented Jun 4, 2017 at 19:02

Daniel Roseman · Accepted Answer · 2017-06-04 19:06:25Z

2

A line, whether read from stdin or from a file, includes a newline character. It seems unlikely that any of your filenames end in a newline.

But don't use sys.stdin.readline to get input from the user. Use the appropriately-named input function:

filename = input("Enter name of file on Desktop:")

(Note, in Python 2.x you should use raw_input rather than input.)

answered Jun 4, 2017 at 19:06

Daniel Roseman

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1 Comment

Infinity Loop Over a year ago

I knew it was something stupid. Thanks. I got it to work.

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