Python如何实现转换URL详解

<scheme>://<netloc>/<path>?<query_params>#<fragment>

URL = namedtuple("URL", ["schema", "netloc", "path", "params", "fragment"])
def url_parse1(url):
assert url.startswith("http")
# 初始化每部分为空
schema = netloc = params = fragment = path = None
# 从 :// 切分 url，前面部分是shema
i = url.find('://')
if i > 0:
schema = url[:i]
url = url[i + 3:]
# 获取netloc
for c in "/?#": # 三个分隔符的顺利很重要
a = url.find(c)
if a > 0: # 只要有三个字符中的任意字符，立即切分，前部分就是netloc，剩下的部分进行后续处理
netloc, url = url[0:a], url[a:]
break
else:
netloc, url = url, '' # 如果三个分隔符都不在url中，那么这是一个只包含
# 同样的方式获取path
for c in "?#":
a = url.find(c)
if a > 0:
path, url = url[0:a], url[a:]
break
else:
path, url = url or None, ''
if "#" in url:
url, fragment = url.split("#", 1)
if '?' in url:
url, params = url.split('?', 1)
return URL(schema=schema, netloc=netloc, path=path, params=_params_parse(params), fragment=fragment)
def _params_parse(params):
if not params:
return None
pairs = [s for s in params.split('&')]
param_dict = dict()
for pair in pairs:
k, v = pair.split('=', 1)
param_dict[k] = v
return param_dict

def url_parse2(url):
rex = r'^(http[s]?):\/\/([^\/\s]+)([\/\w\-\.]+[^#?\s]*)?(\?([^#]*))?(#(.*))?$'
schema = netloc = params = fragment = path = ''
pattern = re.compile(rex)
match = pattern.match(url)
if match:
schema = match.group(1)
netloc = match.group(2)
path = match.group(3)
params = match.group(5)
fragment = match.group(7)
return URL(schema=schema, netloc=netloc, path=path, params=_params_parse(params), fragment=fragment)