Argument is URL or path

Question

What is the standard practice in Python when I have a command-line application taking one argument which is

URL to a web page

or

path to a HTML file somewhere on disk

(only one)

is sufficient the code?

if "http://" in sys.argv[1]:
  print "URL"
else:
  print "path to file"

jassinm · Answer 1 · 2013-03-30T01:55:30.490

22

import urlparse

def is_url(url):
    return urlparse.urlparse(url).scheme != ""
is_url(sys.argv[1])

edited Mar 30 '13 at 01:55

answered Mar 30 '13 at 01:43

jassinm

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1

Python 3 version: `import urllib urllib.parse.urlparse(url).scheme != ""` – JustAC0der Feb 01 '17 at 10:54
7

This returns true for Windows file paths like `c:\users\user\foo.txt`. – Paulo Raposo May 31 '17 at 01:49
7

Better to check if `urlparse(uri).scheme in ('http', 'https',)` because of Windows uri or uri starts with `file://`. – Mikhail Gerasimov Jun 21 '17 at 13:52

score 3 · Accepted Answer · answered Oct 21 '11 at 13:14

3

Depends on what the program must do. If it just prints whether it got a URL, sys.argv[1].startswith('http://') might do. If you must actually use the URL for something useful, do

from urllib2 import urlopen

try:
    f = urlopen(sys.argv[1])
except ValueError:  # invalid URL
    f = open(sys.argv[1])

answered Oct 21 '11 at 13:14

Fred Foo

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The `open() ` throws exception as well. – rplnt Oct 21 '11 at 13:33
Don't forget `except IndexError:` as the user might not specify an argument, which will throw an index error. Or am I wrong? – Griffin Oct 21 '11 at 13:55
@Griffin: I've considered that a separate problem for the purpose of this answer. – Fred Foo Oct 21 '11 at 15:04
@rplnt: yes, and the OP might or might not want to check for `IOError`. I'm just showing how `urlopen` and `open` may be combined, not how to tackle the larger problem. This snippet is enough for writing a generic `open_url_or_file` function that simply re-raises what it gets from `open`. – Fred Foo Oct 21 '11 at 15:05
@larsmans That may be, but from the looks of it the OP doesn't know how to use exception handlers. I don't see any reason not to include it since it won't work if an argument isn't specified. – Griffin Oct 21 '11 at 15:06
@FredFoo's implementation is the most correct exception handling. Only handle the exceptions you know how to handle, otherwise let the caller handle exceptions. In this case, if there's a file open or read or permissions error, etc. Let the caller know rather than catching and hiding the exception – xaviersjs Aug 27 '18 at 19:09
Note that if argument is url with 404 error, then the code slows down. – Chris P May 02 '20 at 09:05

score 1 · Answer 3 · answered Oct 21 '11 at 14:00

Larsmans might work, but it doesn't check whether the user actually specified an argument or not.

import urllib
import sys

try:
    arg = sys.argv[1]
except IndexError:
    print "Usage: "+sys.argv[0]+" file/URL"
    sys.exit(1)

try:
    site = urllib.urlopen(arg)
except ValueError:
    file = open(arg)

Argument is URL or path

3 Answers3

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