Python 使用urllib更改URL的主机名
我想更改URL的主机名Python 使用urllib更改URL的主机名,python,python-3.x,url,urllib,urlparse,Python,Python 3.x,Url,Urllib,Urlparse,我想更改URL的主机名 >>> import urllib >>> url = "https://foo.bar.com:9300/hello" >>> parsed = urllib.parse.urlparse(url) >>> parsed ParseResult(scheme='https', netloc='foo.bar.com:9300', path='/hello', params
>>> import urllib
>>> url = "https://foo.bar.com:9300/hello"
>>> parsed = urllib.parse.urlparse(url)
>>> parsed
ParseResult(scheme='https', netloc='foo.bar.com:9300', path='/hello', params='', query='', fragment='')
由于解析的是一个命名的元组,因此可以替换方案:
>>> parsed_replaced = parsed._replace(scheme='http')
>>> urllib.parse.urlunparse(parsed_replaced)
'http://foo.bar.com:9300/hello'
解析的对象也有一个主机名属性:
>>> parsed.hostname
'foo.bar.com'
但它不是namedtuple中的字段之一,因此不能像scheme那样替换它
有没有办法只替换URL中的主机名?您正在查找的netloc
url = 'https://foo.bar.com:9300/hello'
parsed = urllib.parse.urlparse(url)
parsed_replaced = parsed._replace(netloc='spam.eggs.com:9300')
urllib.parse.urlunparse(parsed_replaced)
'https://spam.eggs.com:9300/hello'
您正在查找netloc
url = 'https://foo.bar.com:9300/hello'
parsed = urllib.parse.urlparse(url)
parsed_replaced = parsed._replace(netloc='spam.eggs.com:9300')
urllib.parse.urlunparse(parsed_replaced)
'https://spam.eggs.com:9300/hello'
import urllib.parse
url=”https://foo.bar.com:9300/hello"
parsed=urllib.parse.urlparse(url)
hostname=parsed.hostname
new\u hostname=“my.new.hostname”
parsed_replaced=parsed._replace(netloc=parsed.netloc.replace(主机名,新主机名))
打印(已解析或已替换)
导入urllib.parse
url=”https://foo.bar.com:9300/hello"
parsed=urllib.parse.urlparse(url)
hostname=parsed.hostname
new\u hostname=“my.new.hostname”
parsed_replaced=parsed._replace(netloc=parsed.netloc.replace(主机名,新主机名))
打印(已解析或已替换)