How can I retrieve the page title of a webpage using Python?
How can I retrieve the page title of a webpage (title html tag) using Python?
Asked by: Daryl172 | Posted: 28-01-2022
Answer 1
Here's a simplified version of @Vinko Vrsalovic's answer:
import urllib2
from BeautifulSoup import BeautifulSoup
soup = BeautifulSoup(urllib2.urlopen("https://www.google.com"))
print soup.title.string
NOTE:
soup.title finds the first title element anywhere in the html document
title.string assumes it has only one child node, and that child node is a string
For beautifulsoup 4.x, use different import:
from bs4 import BeautifulSoup
Answered by: Melissa190 | Posted: 01-03-2022
Answer 2
I'll always use lxml for such tasks. You could use beautifulsoup as well.
import lxml.html
t = lxml.html.parse(url)
print(t.find(".//title").text)
EDIT based on comment:
from urllib2 import urlopen
from lxml.html import parse
url = "https://www.google.com"
page = urlopen(url)
p = parse(page)
print(p.find(".//title").text)
Answered by: Charlie208 | Posted: 01-03-2022
Answer 3
No need to import other libraries. Request has this functionality in-built.
>> hearders = {'headers':'Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:51.0) Gecko/20100101 Firefox/51.0'}
>>> n = requests.get('http://www.imdb.com/title/tt0108778/', headers=hearders)
>>> al = n.text
>>> al[al.find('<title>') + 7 : al.find('</title>')]
u'Friends (TV Series 1994\u20132004) - IMDb'
Answered by: Kristian532 | Posted: 01-03-2022
Answer 4
The mechanize Browser object has a title() method. So the code from this post can be rewritten as:
from mechanize import Browser
br = Browser()
br.open("http://www.google.com/")
print br.title()
Answered by: Julia694 | Posted: 01-03-2022
Answer 5
This is probably overkill for such a simple task, but if you plan to do more than that, then it's saner to start from these tools (mechanize, BeautifulSoup) because they are much easier to use than the alternatives (urllib to get content and regexen or some other parser to parse html)
Links: BeautifulSoup mechanize
#!/usr/bin/env python
#coding:utf-8
from bs4 import BeautifulSoup
from mechanize import Browser
#This retrieves the webpage content
br = Browser()
res = br.open("https://www.google.com/")
data = res.get_data()
#This parses the content
soup = BeautifulSoup(data)
title = soup.find('title')
#This outputs the content :)
print title.renderContents()
Answered by: Steven651 | Posted: 01-03-2022
Answer 6
Using HTMLParser:
from urllib.request import urlopen
from html.parser import HTMLParser
class TitleParser(HTMLParser):
def __init__(self):
HTMLParser.__init__(self)
self.match = False
self.title = ''
def handle_starttag(self, tag, attributes):
self.match = tag == 'title'
def handle_data(self, data):
if self.match:
self.title = data
self.match = False
url = "http://example.com/"
html_string = str(urlopen(url).read())
parser = TitleParser()
parser.feed(html_string)
print(parser.title) # prints: Example Domain
Answered by: Ted424 | Posted: 01-03-2022
Answer 7
Use soup.select_one to target title tag
import requests
from bs4 import BeautifulSoup as bs
r = requests.get('url')
soup = bs(r.content, 'lxml')
print(soup.select_one('title').text)
Answered by: Joyce533 | Posted: 01-03-2022
Answer 8
Using regular expressions
import re
match = re.search('<title>(.*?)</title>', raw_html)
title = match.group(1) if match else 'No title'
Answered by: Fenton408 | Posted: 01-03-2022
Answer 9
soup.title.string
actually returns a unicode string.
To convert that into normal string, you need to do
string=string.encode('ascii','ignore')
Answer 10
Here is a fault tolerant HTMLParser
implementation.
You can throw pretty much anything at get_title()
without it breaking, If anything unexpected happens
get_title()
will return None
.
When Parser()
downloads the page it encodes it to ASCII
regardless of the charset used in the page ignoring any errors.
It would be trivial to change to_ascii()
to convert the data into UTF-8
or any other encoding. Just add an encoding argument and rename the function to something like to_encoding()
.
By default HTMLParser()
will break on broken html, it will even break on trivial things like mismatched tags. To prevent this behavior I replaced HTMLParser()
's error method with a function that will ignore the errors.
#-*-coding:utf8;-*-
#qpy:3
#qpy:console
'''
Extract the title from a web page using
the standard lib.
'''
from html.parser import HTMLParser
from urllib.request import urlopen
import urllib
def error_callback(*_, **__):
pass
def is_string(data):
return isinstance(data, str)
def is_bytes(data):
return isinstance(data, bytes)
def to_ascii(data):
if is_string(data):
data = data.encode('ascii', errors='ignore')
elif is_bytes(data):
data = data.decode('ascii', errors='ignore')
else:
data = str(data).encode('ascii', errors='ignore')
return data
class Parser(HTMLParser):
def __init__(self, url):
self.title = None
self.rec = False
HTMLParser.__init__(self)
try:
self.feed(to_ascii(urlopen(url).read()))
except urllib.error.HTTPError:
return
except urllib.error.URLError:
return
except ValueError:
return
self.rec = False
self.error = error_callback
def handle_starttag(self, tag, attrs):
if tag == 'title':
self.rec = True
def handle_data(self, data):
if self.rec:
self.title = data
def handle_endtag(self, tag):
if tag == 'title':
self.rec = False
def get_title(url):
return Parser(url).title
print(get_title('http://www.google.com'))
Answered by: Adelaide360 | Posted: 01-03-2022
Answer 11
In Python3, we can call method urlopen
from urllib.request
and BeautifulSoup
from bs4
library to fetch the page title.
from urllib.request import urlopen
from bs4 import BeautifulSoup
html = urlopen("https://www.google.com")
soup = BeautifulSoup(html, 'lxml')
print(soup.title.string)
Here we are using the most efficient parser 'lxml'.
Answered by: Joyce440 | Posted: 01-03-2022Answer 12
Using lxml...
Getting it from page meta tagged according to the Facebook opengraph protocol:
import lxml.html.parse
html_doc = lxml.html.parse(some_url)
t = html_doc.xpath('//meta[@property="og:title"]/@content')[0]
or using .xpath with lxml:
t = html_doc.xpath(".//title")[0].text
Answered by: Sienna336 | Posted: 01-03-2022
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