How can I retrieve the page title of a webpage using Python?

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

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)

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()

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()

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)

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'.

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