Python URL をパース¶
>>> from urllib.parse import urlparse >>> url = "http://basicwerl.com/funny-face.jpg?size=27000x3" >>> u = urlparse(url) >>> u ParseResult(scheme='http', netloc='basicwerl.com', path='/funny-face.jpg', params='', query='size=27000x3', fragment='') >>> url = "../img/funny-face.jpg" >>> u = urlparse(url) >>> u ParseResult(scheme='', netloc='', path='../img/funny-face.jpg', params='', query='', fragment='') >>> url = "http://basicwerk.com/memoize2/#python.BeautifulSoup.modify_attrs.txt" >>> u = urlparse(url) >>> u ParseResult(scheme='http', netloc='basicwerk.com', path='/memoize2/', params='', query='', fragment='python.BeautifulSoup.modify_attrs.txt') >>> u.netloc 'basicwerk.com' 参考: Python 3.5.2 21.8. urllib.parse — URL を解析して構成要素にする
Last modified: 2016-10-05 | ||
|
||
|
||
© Shin Nakamura/BasicWerk 2008 - 2024 |