I already tried to extract this html data with BeautifulSoup but it's only limited with tags. What I need to do is to get the trailing something.html or some/something.html after the prefix www.example.com/products/ while eliminating the parameters like ?search=1. I prefer to use regex with this but I don't know the exact pattern for this.
input:
System","urlKey":"ppath","value":[],"hidden":false,"locked":false}],"bizData":"Related+Categories=Mobiles","pos":0},"listItems":[{"name":"Sam-Sung B309i High Precision Smooth Keypad Mobile Phone ","nid":"250505808","icons":[],"productUrl":"//www.example.com/products/sam-sung-b309i-high-precision-smooth-keypad-mobile-phone-i250505808-s341878516.html?search=1", "image": ["//www.example.com/products/site/ammaxxllx.html], "https://www.example.com/site/kakzja.html
prefix = "www.example.com/products/"
# do something
# expected output: ['sam-sung-b309i-high-precision-smooth-keypad-mobile-phone-i250505808-s341878516.html', 'site/ammaxxllx.html']
I guess you want to use re here - with a trick since I "?" will follow the "html" in a URI:
import re
L = ["//www.example.com/products/ammaxxllx.html", "https://www.example.com/site/kakzja.html", "//www.example.com/products/sam-sung-b309i-high-precision-smooth-keypad-mobile-phone-i250505808-s341878516.html?search=1"]
prefix = "www.example.com/products/"
>>> [re.search(prefix+'(.*)html', el).group(1) + 'html' for el in L if prefix in el]
['ammaxxllx.html', 'sam-sung-b309i-high-precision-smooth-keypad-mobile-phone-i250505808-s341878516.html']
Though the above answer by using re module is just awesome. You could also work around without using the module. Like this:
prefix = 'www.example.com/products/'
L = ['//www.example.com/products/sam-sung-b309i-high-precision-smooth-keypad-mobile-phone-i250505808-s341878516.html?search=1', '//www.example.com/products/site/ammaxxllx.html', 'https://www.example.com/site/kakzja.html']
ans = []
for l in L:
input_ = l.rsplit(prefix, 1)
try:
input_ = input_[1]
ans.append(input_[:input_.index('.html')] + '.html')
except Exception as e:
pass
print ans
['sam-sung-b309i-high-precision-smooth-keypad-mobile-phone-i250505808-s341878516.html', 'site/ammaxxllx.html']
Another option is to use urlparse instead of/along with re
It will allow you to split a URL like this:
import urlparse
my_url = "http://www.example.com/products/ammaxxllx.html?spam=eggs#sometag"
url_obj = urlparse.urlsplit(my_url)
url_obj.scheme
>>> 'http'
url_obj.netloc
>>> 'www.example.com'
url_obj.path
>>> '/products/ammaxxllx.html'
url_obj.query
>>> 'spam=eggs'
url_obj.fragment
>>> 'sometag'
# Now you're able to work with every chunk as wanted!
prefix = '/products'
if url_obj.path.startswith(prefix):
# Do whatever you need, replacing the initial characters. You can use re here
print url_obj.path[len(prefix) + 1:]
>>>> ammaxxllx.html
Related
I've a URL like this:
url = 'https://hp.wd5.myworkdayjobs.com/en-US/ExternalCareerSite/job/Enterprise-Business-Planning-Analyst_3103928-1'
x= 'Enterprise-Business-Planning-Analyst_3103928-1'
I want to extract id at the last of url you can say the x part from the above string to get the unique id.
Any help regarding this will be highly appreciated.
_parsed_url.path.split("/")[-1].split('-')[-1]
I am using this but it is giving error.
Python's urllib.parse and pathlib builtin libraries can help here.
url = 'https://hp.wd5.myworkdayjobs.com/en-US/ExternalCareerSite/job/Enterprise-Business-Planning-Analyst_3103928-1'
from urllib.parse import urlparse
from pathlib import PurePath
x = PurePath(urlparse(url).path).name
print(x)
# Enterprise-Business-Planning-Analyst_3103928-1
To print the text Enterprise-Business-Planning-Analyst_3103928-1 you can split() with respect to the / character:
url = 'https://hp.wd5.myworkdayjobs.com/en-US/ExternalCareerSite/job/Enterprise-Business-Planning-Analyst_3103928-1'
print(url.split("/")[-1])
# Enterprise-Business-Planning-Analyst_3103928-1
To print the text 3103928 you can replace the _ character with - and you can split() with respect to the - character:
url = 'https://hp.wd5.myworkdayjobs.com/en-US/ExternalCareerSite/job/Enterprise-Business-Planning-Analyst_3103928-1'
print(url.replace("_", "-").split("-")[-2])
# 3103928
Im passing through some urls and I'd like to strip a part of it which dynamically changes so I don't know it firsthand.
An example url is:
https://...?pid=2&gid=lostchapter&lang=en_GB&practice=1&channel=desktop&demo=2
And I'd like to strip the gid=lostchapter part without any of the rest.
How do I do that?
You can use urllib to convert the query string into a Python dict and access the desired item:
In [1]: from urllib import parse
In [2]: s = "https://...?pid=2&gid=lostchapter&lang=en_GB&practice=1&channel=desktop&demo=2"
In [3]: q = parse.parse_qs(parse.urlsplit(s).query)
In [4]: q
Out[4]:
{'pid': ['2'],
'gid': ['lostchapter'],
'lang': ['en_GB'],
'practice': ['1'],
'channel': ['desktop'],
'demo': ['2']}
In [5]: q["gid"]
Out[5]: ['lostchapter']
Here is the simple way to strip them
urls = "https://...?pid=2&gid=lostchapter&lang=en_GB&practice=1&channel=desktop&demo=2"
# Import the `urlparse` and `urlunparse` methods
from urllib.parse import urlparse, urlunparse
# Parse the URL
url = urlparse(urls)
# Convert the `urlparse` object back into a URL string
url = urlunparse(url)
# Strip the string
url = url.split("?")[1]
url = url.split("&")[1]
# Print the new URL
print(url) # Prints "gid=lostchapter"
Method 1: Using UrlParsers
from urllib.parse import urlparse
p = urlparse('https://.../?pid=2&gid=lostchapter&lang=en_GB&practice=1&channel=desktop&demo=2')
param: list[str] = [i for i in p.query.split('&') if i.startswith('gid=')]
Output: gid=lostchapter
Method 2: Using Regex
param: str = re.search(r'gid=.*&', 'https://.../?pid=2&gid=lostchapter&lang=en_GB&practice=1&channel=desktop&demo=2').group()[:-1]
you can change the regex pattern to appropriate pattern to match the expected outputs. currently it will extract any value.
We can try doing a regex replacement:
url = "https://...?pid=2&gid=lostchapter&lang=en_GB&practice=1&channel=desktop&demo=2"
output = re.sub(r'(?<=[?&])gid=lostchapter&?', '', url)
print(output) # https://...?pid=2&lang=en_GB&practice=1&channel=desktop&demo=2
For a more generic replacement, match on the following regex pattern:
(?<=[?&])gid=\w+&?
Using string slicing (I'm assuming there will be an '&' after gid=lostchapter)
url = r'https://...?pid=2&gid=lostchapter&lang=en_GB&practice=1&channel=desktop&demo=2'
start = url.find('gid')
end = start + url[url.find('gid'):].find('&')
url = url[start:] + url[:end-1]
print(url)
output
gid=lostchapter
What I'm trying to do here is:
find index of occurrence of "gid"
find the first "&" after "gid" is found
concatenate the parts of the url after"gid" and before "&"
I'm coding a python script to check a bunch of URL's and get their ID text, the URL's follow this sequence:
http://XXXXXXX.XXX/index.php?id=YY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX
http://XXXXXXX.XXX/index.php?id=YYY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX
http://XXXXXXX.XXX/index.php?id=YYYY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX
Up to
http://XXXXXXX.XXX/index.php?id=YYYYYYY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX
What I'm trying to do is get only the numbers after the id= and before the &
I've tried to use the regex (\D+)(\d+) but I'm also getting the auth numbers too.
Any suggestion on how to get only the id sequence?
Another way is to use split:
string = 'http://XXXXXXX.XXX/index.php?id=YY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX'
string.split('id=')[1].split('&auth=')[0]
Output:
YY
These are URL addresses, so I would just use url parser in that case.
Look at urllib.parse
Use urlparse to get query parameters, and then parse_qs to get query dict.
import urllib.parse as p
url = "http://XXXXXXX.XXX/index.php?id=YY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX"
query = p.urlparse(url).query
params = p.parse_qs(query)
print(params['id'])
You can include the start and stop tokens in the regex:
pattern = r'id=(\d+)(?:&|$)'
You can try this regex
import re
urls = ["http://XXXXXXX.XXX/index.php?id=YY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX", "http://XXXXXXX.XXX/index.php?id=YYY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX", "http://XXXXXXX.XXX/index.php?id=YYYY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX"]
for url in urls:
id_value = re.search(r"id=(.*)(?=&)", url).group(1)
print(id_value)
that will get you the id value from the URL
YY
YYY
YYYY
variables = """http://XXXXXXX.XXX/index.php?id=YY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX
http://XXXXXXX.XXX/index.php?id=YYY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX
http://XXXXXXX.XXX/index.php?id=YYYY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX""".splitlines()
for v in variables:
p1 = v.split("id=")[1]
p2 = p1.split("&")[0]
print(p2)
outoput:
YY
YYY
YYYY
If you prefer regex
import re
variables = """http://XXXXXXX.XXX/index.php?id=YY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX
http://XXXXXXX.XXX/index.php?id=YYY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX
http://XXXXXXX.XXX/index.php?id=YYYY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX"""
pattern = "id=(.*)\\&"
x = re.findall(pattern, variables)
print(x)
output:
['YY', 'YYY', 'YYYY']
I don't know if you mean with only numbers after id= and before & you mean that there could be letters and numbers between those letters, so I though to this
import re
variables = """http://XXXXXXX.XXX/index.php?id=5Y44Y&auth=XXXYYYXXXYYYXXXYYYXXXYYYX
http://XXXXXXX.XXX/index.php?id=Y2242YY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX
http://XXXXXXX.XXX/index.php?id=5YY453YY&auth=XXXYYYXXXYYYXXXYYYXXXYYYX"""
pattern = "id=(.*)\\&"
x = re.findall(pattern, variables)
print(x)
x2 = []
for p in x:
x2.append(re.sub("\\D", "", p))
print(x2)
Output:
['5Y44Y', 'Y2242YY', '5YY453YY']
['544', '2242', '5453']
Use the regex id=[0-9]+:
pattern = "id=[0-9]+"
id = re.findall(pattern, url)[0].split("id=")[1]
If you do it this way, there is no need for &auth to follow the id, which makes it very versatile. However, the &auth won't make the code stop working. It works for the edge cases, as well as the simple ones.
import re
import urllib
import HTMLParser
urlRegex = re.compile(r'(.+)&data=')
match=urlRegex.search('https://na01.safelinks.protection.outlook.com/?url=https%3A%2F%2Foffice.memoriesflower.com%2FPermission%2F%2525%2524%255E%2526%2526*%2523%2523%255E%2524%2525%255E%2526%255E*%2526%2523%255E%2525%2525%2526%2540%255E*%2523%2526%255E%2525%2523%2526%2540%2525*%255E%2540%255E%2523%2525%255E%2540%2526%2525*%255E%2540%2Foffice.php&data=01%7C01%7Cdavid.levin%40mheducation.com%7C0ac9a3770fe64fbb21fb08d50764c401%7Cf919b1efc0c347358fca0928ec39d8d5%7C0&sdata=PEoDOerQnha%2FACafNx8JAep8O9MdllcKCsHET2Ye%2B4%3D&reserved=0')
x = match.group()
urlRegex_1 = re.compile(r'url=(.+)&data=')
match_1 = urlRegex_1.search(x)
print match1.group(1)
htmlencodedurl = urllib.unquote(urllib.unquote(match1.group(1)))
actual_url = HTMLParser.HTMLParser().unescape(htmlencodedurl)
So the 'actual_url' displays this:
'https://office.memoriesflower.com/Permission/%$^&&##^$%^&^&#^%%&#^*#&^%'
I need it to display this:
https://office.memoriesflower.com/Permission/office.php
The following is cleaner as it uses the urlparse to extract the query string and then uses path operations to remove the unwanted component:
import posixpath as path
from urlparse import urlparse, parse_qs, urlunparse
url = 'https://na01.safelinks.protection.outlook.com/?url=https%3A%2F%2Foffice.memoriesflower.com%2FPermission%2F%2525%2524%255E%2526%2526*%2523%2523%255E%2524%2525%255E%2526%255E*%2526%2523%255E%2525%2525%2526%2540%255E*%2523%2526%255E%2525%2523%2526%2540%2525*%255E%2540%255E%2523%2525%255E%2540%2526%2525*%255E%2540%2Foffice.php&data=01%7C01%7Cdavid.levin%40mheducation.com%7C0ac9a3770fe64fbb21fb08d50764c401%7Cf919b1efc0c347358fca0928ec39d8d5%7C0&sdata=PEoDOerQnha%2FACafNx8JAep8O9MdllcKCsHET2Ye%2B4%3D&reserved=0'
target = parse_qs(urlparse(url).query)['url'][0]
p = urlparse(target)
q = p._replace(path=path.join(path.dirname(path.dirname(p.path)), path.basename(p.path)))
print urlunparse(q)
prints https://office.memoriesflower.com/Permission/office.php
I found this having a similar problem. Here's the code I used to resolve the issue. It's not particularly elegant, but you may be able to tweak it for your needs.
self.urls = (re.findall("safelinks\.protection\.outlook\.com/\?url=.*?sdata=", self.body.lower(), re.M))
if len(self.urls) > 0:
for i, v in enumerate(self.urls):
self.urls[i] = v[38:-11]
This works by getting the value in an ugly format and then stripping off the excess pieces of each item as a string. I believe the proper way to do this is with grouping, but this worked well enough for my needs.
In this snippet of code I am trying to obtain the links to images posted in a groupchat by a certain user:
import groupy
from groupy import Bot, Group, Member
prog_group = Group.list().first
prog_members = prog_group.members()
prog_messages = prog_group.messages()
rojer = str(prog_members[4])
rojer_messages = ['none']
rojer_pics = []
links = open('rojer_pics.txt', 'w')
print(prog_group)
for message in prog_messages:
if message.name == rojer:
rojer_messages.append(message)
if message.attachments:
links.write(str(message) + '\n')
links.close()
The issue is that in the links file it prints the entire message: ("Rojer Doewns: Heres a special one +https://i.groupme.com/406x1199.png.7679b4f1ee964656bde93448ff9cee12')>"
What I am wanting to do, is to get rid of characters that aren't part of the URL so it is written like so:
"https://i.groupme.com/406x1199.png.7679b4f1ee964656bde93448ff9cee12"
are there any methods in python that can manipulate a string like so?
I just used string.split() and split it into 3 parts by the parentheses:
for message in prog_messages:
if message.name == rojer:
rojer_messages.append(message)
if message.attachments:
link = str(message).split("'")
rojer_pics.append(link[1])
links.write(str(link[1]) + '\n')
This can done using string indices and the string method .find():
>>> url = "(\"Rojer Doewns: Heres a special one +https://i.groupme.com/406x1199.png.7679b4f1ee964656bde93448ff9cee12')"
>>> url = url[url.find('+')+1:-2]
>>> url
'https://i.groupme.com/406x1199.png.7679b4f1ee964656bde93448ff9cee12'
>>>
>>> string = '("Rojer Doewns: Heres a special one +https://i.groupme.com/406x1199.png.7679b4f1ee964656bde93448ff9cee12\')>"'
>>> string.split('+')[1][:-4]
'https://i.groupme.com/406x1199.png.7679b4f1ee964656bde93448ff9cee12'