如何在Python中验证URL？（是否畸形）

我已经从用户那里__了ABC0，我必须用获取的HTML进行回复。

如何检查URL是否格式错误？

例如：

url = 'google' # Malformed
url = 'google.com' # Malformed
url = 'http://google.com' # Valid
url = 'http://google' # Malformed
276606 次浏览
小开
最佳答案

django url validation regex (source):

import re
regex = re.compile(
r'^(?:http|ftp)s?://' # http:// or https://
r'(?:(?:[A-Z0-9](?:[A-Z0-9-]{0,61}[A-Z0-9])?\.)+(?:[A-Z]{2,6}\.?|[A-Z0-9-]{2,}\.?)|' #domain...
r'localhost|' #localhost...
r'\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})' # ...or ip
r'(?::\d+)?' # optional port
r'(?:/?|[/?]\S+)$', re.IGNORECASE)


print(re.match(regex, "http://www.example.com") is not None) # True
print(re.match(regex, "example.com") is not None)            # False
小开

Actually, I think this is the best way.

from django.core.validators import URLValidator
from django.core.exceptions import ValidationError


val = URLValidator(verify_exists=False)
try:
val('http://www.google.com')
except ValidationError, e:
print e

If you set verify_exists to True, it will actually verify that the URL exists, otherwise it will just check if it's formed correctly.

edit: ah yeah, this question is a duplicate of this: How can I check if a URL exists with Django’s validators?

小开

note - lepl is no longer supported, sorry (you're welcome to use it, and i think the code below works, but it's not going to get updates).

rfc 3696 http://www.faqs.org/rfcs/rfc3696.html defines how to do this (for http urls and email). i implemented its recommendations in python using lepl (a parser library). see http://acooke.org/lepl/rfc3696.html

to use:

> easy_install lepl
...
> python
...
>>> from lepl.apps.rfc3696 import HttpUrl
>>> validator = HttpUrl()
>>> validator('google')
False
>>> validator('http://google')
False
>>> validator('http://google.com')
True
小开

Use the validators package:

>>> import validators
>>> validators.url("http://google.com")
True
>>> validators.url("http://google")
ValidationFailure(func=url, args={'value': 'http://google', 'require_tld': True})
>>> if not validators.url("http://google"):
...     print "not valid"
...
not valid
>>>

Install it from PyPI with pip (pip install validators).

小开

I landed on this page trying to figure out a sane way to validate strings as "valid" urls. I share here my solution using python3. No extra libraries required.

See https://docs.python.org/2/library/urlparse.html if you are using python2.

See https://docs.python.org/3.0/library/urllib.parse.html if you are using python3 as I am.

import urllib
from pprint import pprint


invalid_url = 'dkakasdkjdjakdjadjfalskdjfalk'
valid_url = 'https://stackoverflow.com'
tokens = [urllib.parse.urlparse(url) for url in (invalid_url, valid_url)]


for token in tokens:
pprint(token)


min_attributes = ('scheme', 'netloc')  # add attrs to your liking
for token in tokens:
if not all([getattr(token, attr) for attr in min_attributes]):
error = "'{url}' string has no scheme or netloc.".format(url=token.geturl())
print(error)
else:
print("'{url}' is probably a valid url.".format(url=token.geturl()))

ParseResult(scheme='', netloc='', path='dkakasdkjdjakdjadjfalskdjfalk', params='', query='', fragment='')

ParseResult(scheme='https', netloc='stackoverflow.com', path='', params='', query='', fragment='')

'dkakasdkjdjakdjadjfalskdjfalk' string has no scheme or netloc.

'https://stackoverflow.com' is probably a valid url.

Here is a more concise function:

from urllib.parse import urlparse


min_attributes = ('scheme', 'netloc')




def is_valid(url, qualifying=min_attributes):
tokens = urlparse(url)
return all([getattr(tokens, qualifying_attr)
for qualifying_attr in qualifying])
小开

A True or False version, based on @DMfll answer:

try:
# python2
from urlparse import urlparse
except:
# python3
from urllib.parse import urlparse


a = 'http://www.cwi.nl:80/%7Eguido/Python.html'
b = '/data/Python.html'
c = 532
d = u'dkakasdkjdjakdjadjfalskdjfalk'
e = 'https://stackoverflow.com'


def uri_validator(x):
try:
result = urlparse(x)
return all([result.scheme, result.netloc])
except:
return False


print(uri_validator(a))
print(uri_validator(b))
print(uri_validator(c))
print(uri_validator(d))
print(uri_validator(e))

Gives:

True
False
False
False
True
小开

EDIT

As pointed out by @Kwame , the below code does validate the url even if the .com or .co etc are not present.

also pointed out by @Blaise, URLs like https://www.google is a valid URL and you need to do a DNS check for checking if it resolves or not, separately.

This is simple and works:

So min_attr contains the basic set of strings that needs to be present to define the validity of a URL, i.e http:// part and google.com part.

urlparse.scheme stores http:// and

urlparse.netloc store the domain name google.com

from urlparse import urlparse
def url_check(url):


min_attr = ('scheme' , 'netloc')
try:
result = urlparse(url)
if all([result.scheme, result.netloc]):
return True
else:
return False
except:
return False

all() returns true if all the variables inside it return true. So if result.scheme and result.netloc is present i.e. has some value then the URL is valid and hence returns True.

小开

Nowadays, I use the following, based on the Padam's answer:

$ python --version
Python 3.6.5

And this is how it looks:

from urllib.parse import urlparse


def is_url(url):
try:
result = urlparse(url)
return all([result.scheme, result.netloc])
except ValueError:
return False

Just use is_url("http://www.asdf.com").

Hope it helps!

小开

Validate URL with urllib and Django-like regex

The Django URL validation regex was actually pretty good but I needed to tweak it a little bit for my use case. Feel free to adapt it to yours!

Python 3.7

import re
import urllib


# Check https://regex101.com/r/A326u1/5 for reference
DOMAIN_FORMAT = re.compile(
r"(?:^(\w{1,255}):(.{1,255})@|^)" # http basic authentication [optional]
r"(?:(?:(?=\S{0,253}(?:$|:))" # check full domain length to be less than or equal to 253 (starting after http basic auth, stopping before port)
r"((?:[a-z0-9](?:[a-z0-9-]{0,61}[a-z0-9])?\.)+" # check for at least one subdomain (maximum length per subdomain: 63 characters), dashes in between allowed
r"(?:[a-z0-9]{1,63})))" # check for top level domain, no dashes allowed
r"|localhost)" # accept also "localhost" only
r"(:\d{1,5})?", # port [optional]
re.IGNORECASE
)
SCHEME_FORMAT = re.compile(
r"^(http|hxxp|ftp|fxp)s?$", # scheme: http(s) or ftp(s)
re.IGNORECASE
)


def validate_url(url: str):
url = url.strip()


if not url:
raise Exception("No URL specified")


if len(url) > 2048:
raise Exception("URL exceeds its maximum length of 2048 characters (given length={})".format(len(url)))


result = urllib.parse.urlparse(url)
scheme = result.scheme
domain = result.netloc


if not scheme:
raise Exception("No URL scheme specified")


if not re.fullmatch(SCHEME_FORMAT, scheme):
raise Exception("URL scheme must either be http(s) or ftp(s) (given scheme={})".format(scheme))


if not domain:
raise Exception("No URL domain specified")


if not re.fullmatch(DOMAIN_FORMAT, domain):
raise Exception("URL domain malformed (domain={})".format(domain))


return url

Explanation

  • The code only validates the scheme and netloc part of a given URL. (To do this properly, I split the URL with urllib.parse.urlparse() in the two according parts which are then matched with the corresponding regex terms.)

  • The netloc part stops before the first occurrence of a slash /, so port numbers are still part of the netloc, e.g.: 

    https://www.google.com:80/search?q=python
^^^^^   ^^^^^^^^^^^^^^^^^
|             |
|             +-- netloc (aka "domain" in my code)
+-- scheme

  • IPv4 addresses are also validated

IPv6 Support

If you want the URL validator to also work with IPv6 addresses, do the following:

  • Add is_valid_ipv6(ip) from Markus Jarderot's answer, which has a really good IPv6 validator regex
  • Add and not is_valid_ipv6(domain) to the last if

Examples

Here are some examples of the regex for the netloc (aka domain) part in action:

小开

All of the above solutions recognize a string like "http://www.google.com/path,www.yahoo.com/path" as valid. This solution always works as it should

import re


# URL-link validation
ip_middle_octet = u"(?:\.(?:1?\d{1,2}|2[0-4]\d|25[0-5]))"
ip_last_octet = u"(?:\.(?:[1-9]\d?|1\d\d|2[0-4]\d|25[0-4]))"


URL_PATTERN = re.compile(
u"^"
# protocol identifier
u"(?:(?:https?|ftp|rtsp|rtp|mmp)://)"
# user:pass authentication
u"(?:\S+(?::\S*)?@)?"
u"(?:"
u"(?P<private_ip>"
# IP address exclusion
# private & local networks
u"(?:localhost)|"
u"(?:(?:10|127)" + ip_middle_octet + u"{2}" + ip_last_octet + u")|"
u"(?:(?:169\.254|192\.168)" + ip_middle_octet + ip_last_octet + u")|"
u"(?:172\.(?:1[6-9]|2\d|3[0-1])" + ip_middle_octet + ip_last_octet + u"))"
u"|"
# IP address dotted notation octets
# excludes loopback network 0.0.0.0
# excludes reserved space >= 224.0.0.0
# excludes network & broadcast addresses
# (first & last IP address of each class)
u"(?P<public_ip>"
u"(?:[1-9]\d?|1\d\d|2[01]\d|22[0-3])"
u"" + ip_middle_octet + u"{2}"
u"" + ip_last_octet + u")"
u"|"
# host name
u"(?:(?:[a-z\u00a1-\uffff0-9_-]-?)*[a-z\u00a1-\uffff0-9_-]+)"
# domain name
u"(?:\.(?:[a-z\u00a1-\uffff0-9_-]-?)*[a-z\u00a1-\uffff0-9_-]+)*"
# TLD identifier
u"(?:\.(?:[a-z\u00a1-\uffff]{2,}))"
u")"
# port number
u"(?::\d{2,5})?"
# resource path
u"(?:/\S*)?"
# query string
u"(?:\?\S*)?"
u"$",
re.UNICODE | re.IGNORECASE
)
def url_validate(url):
""" URL string validation
"""
return re.compile(URL_PATTERN).match(url)
小开

Not directly relevant, but often it's required to identify whether some token CAN be a url or not, not necessarily 100% correctly formed (ie, https part omitted and so on). I've read this post and did not find the solution, so I am posting my own here for the sake of completeness.

def get_domain_suffixes():
import requests
res=requests.get('https://publicsuffix.org/list/public_suffix_list.dat')
lst=set()
for line in res.text.split('\n'):
if not line.startswith('//'):
domains=line.split('.')
cand=domains[-1]
if cand:
lst.add('.'+cand)
return tuple(sorted(lst))


domain_suffixes=get_domain_suffixes()


def reminds_url(txt:str):
"""
>>> reminds_url('yandex.ru.com/somepath')
True
    

"""
ltext=txt.lower().split('/')[0]
return ltext.startswith(('http','www','ftp')) or ltext.endswith(domain_suffixes)
小开

Here's a regex solution since top voted regex doesn't work for weird cases like top-level domain. Some test cases down below.

regex = re.compile(
r"(\w+://)?"                # protocol                      (optional)
r"(\w+\.)?"                 # host                          (optional)
r"((\w+)\.(\w+))"           # domain
r"(\.\w+)*"                 # top-level domain              (optional, can have > 1)
r"([\w\-\._\~/]*)*(?<!\.)"  # path, params, anchors, etc.   (optional)
)

cases = [
"http://www.google.com",
"https://www.google.com",
"http://google.com",
"https://google.com",
"www.google.com",
"google.com",
"http://www.google.com/~as_db3.2123/134-1a",
"https://www.google.com/~as_db3.2123/134-1a",
"http://google.com/~as_db3.2123/134-1a",
"https://google.com/~as_db3.2123/134-1a",
"www.google.com/~as_db3.2123/134-1a",
"google.com/~as_db3.2123/134-1a",
# .co.uk top level
"http://www.google.co.uk",
"https://www.google.co.uk",
"http://google.co.uk",
"https://google.co.uk",
"www.google.co.uk",
"google.co.uk",
"http://www.google.co.uk/~as_db3.2123/134-1a",
"https://www.google.co.uk/~as_db3.2123/134-1a",
"http://google.co.uk/~as_db3.2123/134-1a",
"https://google.co.uk/~as_db3.2123/134-1a",
"www.google.co.uk/~as_db3.2123/134-1a",
"google.co.uk/~as_db3.2123/134-1a",
"https://...",
"https://..",
"https://.",
"https://.google.com",
"https://..google.com",
"https://...google.com",
"https://.google..com",
"https://.google...com"
"https://...google..com",
"https://...google...com",
".google.com",
".google.co."
"https://google.co."
]
for c in cases:
print(c, regex.match(c).span()[1] - regex.match(c).span()[0] == len(c))
小开

Function based on Dominic Tarro answer:

import re
def is_url(x):
return bool(re.match(
r"(https?|ftp)://" # protocol
r"(\w+(\-\w+)*\.)?" # host (optional)
r"((\w+(\-\w+)*)\.(\w+))" # domain
r"(\.\w+)*" # top-level domain (optional, can have > 1)
r"([\w\-\._\~/]*)*(?<!\.)" # path, params, anchors, etc. (optional)
, x))

提交答案