Eliminar los parámetros utm_ * de la URL en Python

Question

He intentado eliminar todos los parámetros utm_ * de una lista de URL. Lo más parecido que he encontrado es esto: https://gist.github.com/626834.

¿Alguna idea?

Answer 1

import re 
from urlparse import urlparse, urlunparse 

url = 'http://www.someurl.com/page.html?foo=bar&utm_medium=qux&baz=qoo' 
parsed_url = list(urlparse(url)) 
parsed_url[4] = '&'.join(
    [x for x in parsed_url[4].split('&') if not re.match(r'utm_', x)]) 
utmless_url = urlunparse(parsed_url) 

print utmless_url # 'http://www.someurl.com/page.html?foo=bar&baz=qoo' 

Answer 2

Es un poco largo pero usa los módulos url * y evita las re.

from urllib import urlencode 
from urlparse import urlparse, parse_qs, urlunparse 

url = 'http://whatever.com/somepage?utm_one=3&something=4&utm_two=5&utm_blank&something_else' 

parsed = urlparse(url) 
qd = parse_qs(parsed.query, keep_blank_values=True) 
filtered = dict((k, v) for k, v in qd.iteritems() if not k.startswith('utm_')) 
newurl = urlunparse([ 
    parsed.scheme, 
    parsed.netloc, 
    parsed.path, 
    parsed.params, 
    urlencode(filtered, doseq=True), # query string 
    parsed.fragment 
]) 

print newurl 
# 'http://whatever.com/somepage?something=4&something_else' 

Answer 3

simple, y las obras, y se basan en el enlace que has publicado, pero es re ... así, no estoy seguro de que no se romperá por alguna razón que no puedo pensar en :)

import re 

def trim_utm(url): 
    if "utm_" not in url: 
     return url 
    matches = re.findall('(.+\?)([^#]*)(.*)', url) 
    if len(matches) == 0: 
     return url 
    match = matches[0] 
    query = match[1] 
    sanitized_query = '&'.join([p for p in query.split('&') if not p.startswith('utm_')]) 
    return match[0]+sanitized_query+match[2] 

if __name__ == "__main__": 
    tests = [ "http://localhost/index.php?a=1&utm_source=1&b=2", 
       "http://localhost/index.php?a=1&utm_source=1&b=2#hash", 
       "http://localhost/index.php?a=1&utm_source=1&b=2&utm_something=no#hash", 
       "http://localhost/index.php?a=1&utm_source=1&utm_a=yes&b=2#hash", 
       "http://localhost/index.php?utm_a=a", 
       "http://localhost/index.php?a=utm_a", 
       "http://localhost/index.php?a=1&b=2", 
       "http://localhost/index.php", 
       "http://localhost/index.php#hash2" 
      ] 

    for t in tests: 
     trimmed = trim_utm(t) 
     print t 
     print trimmed 
     print 

Answer 4

Qué tal esto. Agradable y simple:

url = 'https://searchengineland.com/amazon-q3-ad-revenues-surpass-1-billion-roughly-2x-early-2016-285763?utm_source=feedburner&utm_medium=feed&utm_campaign=feed-main' 

print url[:url.find('?utm')] 

https://searchengineland.com/amazon-q3-ad-revenues-surpass-1-billion-roughly-2x-early-2016-285763