How to extract a filename from a URL & append a word to it?

I have the following url:

url = http://photographs.500px.com/kyle/09-09-201315-47-571378756077.jpg

I would like to extract the file name in this url: 09-09-201315-47-571378756077.jpg

Once I get this file name, I'm going to save it with this name to the Desktop.

filename = **extracted file name from the url**     
download_photo = urllib.urlretrieve(url, "/home/ubuntu/Desktop/%s.jpg" % (filename))

After this, I'm going to resize the photo, once that is done, I've going to save the resized version and append the word "_small" to the end of the filename.

downloadedphoto = Image.open("/home/ubuntu/Desktop/%s.jpg" % (filename))               
resize_downloadedphoto = downloadedphoto.resize.((300, 300), Image.ANTIALIAS)
resize_downloadedphoto.save("/home/ubuntu/Desktop/%s.jpg" % (filename + _small))

From this, what I am trying to achieve is to get two files, the original photo with the original name, then the resized photo with the modified name. Like so:

09-09-201315-47-571378756077.jpg

09-09-201315-47-571378756077_small.jpg

How can I go about doing this?

You can use urllib.parse.urlparse with os.path.basename:

import os
from urllib.parse import urlparse

url = "http://photographs.500px.com/kyle/09-09-201315-47-571378756077.jpg"
a = urlparse(url)
print(a.path)                    # Output: /kyle/09-09-201315-47-571378756077.jpg
print(os.path.basename(a.path))  # Output: 09-09-201315-47-571378756077.jpg

Your URL might contain percent-encoded characters like %20 for space or %E7%89%B9%E8%89%B2 for "特色". If that's the case, you'll need to unquote (or unquote_plus) them. You can also use pathlib.Path().name instead of os.path.basename, which could help to add a suffix in the name (like asked in the original question):

from pathlib import Path
from urllib.parse import urlparse, unquote

url = "http://photographs.500px.com/kyle/09-09-2013%20-%2015-47-571378756077.jpg"
urlparse(url).path

url_parsed = urlparse(url)
print(unquote(url_parsed.path))  # Output: /kyle/09-09-2013 - 15-47-571378756077.jpg
file_path = Path("/home/ubuntu/Desktop/") / unquote(Path(url_parsed.path).name)
print(file_path)        # Output: /home/ubuntu/Desktop/09-09-2013 - 15-47-571378756077.jpg

new_file = file_path.with_stem(file_path.stem + "_small")
print(new_file)         # Output: /home/ubuntu/Desktop/09-09-2013 - 15-47-571378756077_small.jpg

Also, an alternative is to use unquote(urlparse(url).path.split("/")[-1]).

os.path.basename(url)

Why try harder?

In [1]: os.path.basename("https://example.com/file.html")
Out[1]: 'file.html'

In [2]: os.path.basename("https://example.com/file")
Out[2]: 'file'

In [3]: os.path.basename("https://example.com/")
Out[3]: ''

In [4]: os.path.basename("https://example.com")
Out[4]: 'example.com'

Note 2020-12-20

Nobody has thus far provided a complete solution.

A URL can contain a ?[query-string] and/or a #[fragment Identifier] (but only in that order: ref)

In [1]: from os import path

In [2]: def get_filename(url):
   ...:     fragment_removed = url.split("#")[0]  # keep to left of first #
   ...:     query_string_removed = fragment_removed.split("?")[0]
   ...:     scheme_removed = query_string_removed.split("://")[-1].split(":")[-1]
   ...:     if scheme_removed.find("/") == -1:
   ...:         return ""
   ...:     return path.basename(scheme_removed)
   ...:

In [3]: get_filename("a.com/b")
Out[3]: 'b'

In [4]: get_filename("a.com/")
Out[4]: ''

In [5]: get_filename("https://a.com/")
Out[5]: ''

In [6]: get_filename("https://a.com/b")
Out[6]: 'b'

In [7]: get_filename("https://a.com/b?c=d#e")
Out[7]: 'b'

filename = url[url.rfind("/")+1:]
filename_small = filename.replace(".", "_small.")

maybe use ".jpg" in the last case since a . can also be in the filename.

You could just split the url by "/" and retrieve the last member of the list:

    url = "http://photographs.500px.com/kyle/09-09-201315-47-571378756077.jpg"
    filename = url.split("/")[-1] 
    #09-09-201315-47-571378756077.jpg

Then use replace to change the ending:

    small_jpg = filename.replace(".jpg", "_small.jpg")
    #09-09-201315-47-571378756077_small.jpg

Use urllib.parse.urlparse to get just the path part of the URL, and then use pathlib.Path on that path to get the filename:

from urllib.parse import urlparse
from pathlib import Path


url = "http://example.com/some/long/path/a_filename.jpg?some_query_params=true&some_more=true#and-an-anchor"
a = urlparse(url)
a.path             # '/some/long/path/a_filename.jpg'
Path(a.path).name  # 'a_filename.jpg'

With python3 (from 3.4 upwards) you can abuse the pathlib library in the following way:

from pathlib import Path

p = Path('http://example.com/somefile.html')
print(p.name)
# >>> 'somefile.html'

print(p.stem)
# >>> 'somefile'

print(p.suffix)
# >>> '.html'

print(f'{p.stem}-spamspam{p.suffix}')
# >>> 'somefile-spamspam.html'

❗️ WARNING

^{The pathlib module is NOT meant for parsing URLs — it is designed to work with POSIX paths and not with URLs. Don't use it in production code! It's a dirty quick hack for non-critical code. The code is only provided as an example of what you can do but probably should not do. If you need to parse URLs then go with urllib.parse or alternatives.}

Sometimes there is a query string:

filename = url.split("/")[-1].split("?")[0] 
new_filename = filename.replace(".jpg", "_small.jpg")

A simple version using the os package:

import os

def get_url_file_name(url):
    url = url.split("#")[0]
    url = url.split("?")[0]
    return os.path.basename(url)

Examples:

print(get_url_file_name("example.com/myfile.tar.gz"))  # 'myfile.tar.gz'
print(get_url_file_name("example.com/"))  # ''
print(get_url_file_name("https://example.com/"))  # ''
print(get_url_file_name("https://example.com/hello.zip"))  # 'hello.zip'
print(get_url_file_name("https://example.com/args.tar.gz?c=d#e"))  # 'args.tar.gz'

Sometimes the link you have can have redirects (that was the case for me). In that case you have to solve the redirects

import requests
url = "http://photographs.500px.com/kyle/09-09-201315-47-571378756077.jpg"
response = requests.head(url)
url = response.url

then you can continue with the best answer at the moment (Ofir's)

import os
from urllib.parse import urlparse


a = urlparse(url)
print(a.path)                    # Output: /kyle/09-09-201315-47-571378756077.jpg
print(os.path.basename(a.path))  # Output: 09-09-201315-47-571378756077.jpg

it doesn't work with this page however, as the page isn't available anymore

https://stackoverflow.com/questions/10552188/python-split-url-to-find-image-name-and-extension

helps you to extract the image name. to append name :

imageName =  '09-09-201315-47-571378756077'

new_name = '{0}_small.jpg'.format(imageName) 

We can extract filename from a url by using ntpath module.

import ntpath
url = 'http://photographs.500px.com/kyle/09-09-201315-47-571378756077.jpg'
name, ext = ntpath.splitext(ntpath.basename(url))
# 09-09-201315-47-571378756077	.jpg


print(name + '_small' + ext)
09-09-201315-47-571378756077_small.jpg