Python 如何在请求中获取页面标题
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How to get page title in requests
提问by David542
What would be the simplest way to get the title of a page in Requests?
在请求中获取页面标题的最简单方法是什么？
r = requests.get('http://www.imdb.com/title/tt0108778/')
# ? r.title
Friends (TV Series 1994–2004) - IMDb
采纳答案by alecxe
You need an HTML parser to parse the HTML response and get the titletag's text:
您需要一个 HTML 解析器来解析 HTML 响应并获取title标签的文本：
Example using lxml.html:
使用示例lxml.html：
>>> import requests
>>> from lxml.html import fromstring
>>> r = requests.get('http://www.imdb.com/title/tt0108778/')
>>> tree = fromstring(r.content)
>>> tree.findtext('.//title')
u'Friends (TV Series 1994\u20132004) - IMDb'
There are certainly other options, like, for example, mechanizelibrary:
当然还有其他选项，例如mechanize库：
>>> import mechanize
>>> br = mechanize.Browser()
>>> br.open('http://www.imdb.com/title/tt0108778/')
>>> br.title()
'Friends (TV Series 1994\xe2\x80\x932004) - IMDb'
What option to choose depends on what are you going to do next: parse the page to get more data, or, may be, you want to interact with it: click buttons, submit forms, follow links etc.
选择哪个选项取决于您接下来要做什么：解析页面以获取更多数据，或者，您可能想要与之交互：单击按钮、提交表单、关注链接等。
Besides, you may want to use an API provided by IMDB, instead of going down to HTML parsing, see:
此外，您可能希望使用 提供的 API IMDB，而不是深入到 HTML 解析，请参阅：
Example usage of an IMDbPYpackage:
IMDbPY包的使用示例：
>>> from imdb import IMDb
>>> ia = IMDb()
>>> movie = ia.get_movie('0108778')
>>> movie['title']
u'Friends'
>>> movie['series years']
u'1994-2004'
回答by Greg
You could use beautifulsoup to parse the HTML.
您可以使用 beautifulsoup 来解析 HTML。
Install it using pip install beautifulsoup4
安装它使用 pip install beautifulsoup4
>>> import requests
>>> r = requests.get('http://www.imdb.com/title/tt0108778/')
>>> import bs4
>>> html = bs4.BeautifulSoup(r.text)
>>> html.title
<title>Friends (TV Series 1994–2004) - IMDb</title>
>>> html.title.text
u'Friends (TV Series 1994\u20132004) - IMDb'
回答by Rahul Chawla
No need to import other libraries. Request has this functionality in-built.
无需导入其他库。Request 内置了此功能。
>>> 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'
Update after ZN13'scomment
ZN13评论后更新
>>> import re
>>> import requests
>>> n = requests.get('https://www.libsdl.org/release/SDL-1.2.15/docs/html/guideinputkeyboard.html')
>>> al = n.text
>>> d = re.search('<\W*title\W*(*)</title', al, re.IGNORECASE)
>>> d.group(1)
u'Handling the Keyboard'
This will work for all cases whether extra non alphabetical characters are present with title tag or not.
这将适用于所有情况，无论标题标签是否存在额外的非字母字符。
回答by Vitaly Zdanevich
Regex with lookbehind and lookforward:
具有后视和前瞻的正则表达式：
re.search('(?<=<title>).+?(?=</title>)', mytext, re.DOTALL).group().strip()
re.DOTALLbecause title can have a new line character \n
re.DOTALL因为标题可以有一个换行符 \n
回答by u5602117
Pythonic HTML Parsing for Humans.
人类的 Pythonic HTML 解析。
from requests_html import HTMLSession
print(HTMLSession().get('http://www.imdb.com/title/tt0108778/').html.find('title', first=True).text)
