blocks|key|3114518|text|你应该能够用urlparse+(文档：python2，python3)做到这一点：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3114519|from+urllib.parse+import+urlparse
#+from+urlparse+import+urlparse++#+Python+2
parsed_uri+=+urlparse('http://stackoverflow.com/questions/1234567/blah-blah-blah-blah'+)
result+=+'{uri.scheme}://{uri.netloc}/'.format(uri=parsed_uri)
print(result)

#+gives
'http://stackoverflow.com/'|code-block|syntax|javascript|3114520|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/urlparse.html|1|https://docs.python.org/3/library/urllib.parse.html^0|6|8|J|7|0|R|7|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]|$9|10|A|11|1|12]]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]|S|$5|N|O|P|E|$Q|T]]]]

You should be able to do it with <code>urlparse</code> (docs: <a href="https://docs.python.org/2/library/urlparse.html" rel="noreferrer">python2</a>, <a href="https://docs.python.org/3/library/urllib.parse.html" rel="noreferrer">python3</a>):

<pre><code>from urllib.parse import urlparse
# from urlparse import urlparse # Python 2
parsed_uri = urlparse('http://stackoverflow.com/questions/1234567/blah-blah-blah-blah' )
result = '{uri.scheme}://{uri.netloc}/'.format(uri=parsed_uri)
print(result)

# gives
'http://stackoverflow.com/'
</code></pre>

blocks|key|2900306|text|https://github.com/john-kurkowski/tldextract|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2900307|这是urlparse更详细的版本。它会为你检测域和子域。|2900308|从他们的文档中：|2900309|>>>+import+tldextract
>>>+tldextract.extract('http://forums.news.cnn.com/')
ExtractResult(subdomain='forums.news',+domain='cnn',+suffix='com')
>>>+tldextract.extract('http://forums.bbc.co.uk/')+#+United+Kingdom
ExtractResult(subdomain='forums',+domain='bbc',+suffix='co.uk')
>>>+tldextract.extract('http://www.worldbank.org.kg/')+#+Kyrgyzstan
ExtractResult(subdomain='www',+domain='worldbank',+suffix='org.kg')|code-block|syntax|javascript|2900310|ExtractResult是一个命名元组，所以很容易访问你想要的部分。|style|CODE|2900311|>>>+ext+=+tldextract.extract('http://forums.bbc.co.uk')
>>>+ext.domain
'bbc'
>>>+'.'.join(ext[:2])+#+rejoin+subdomain+and+domain
'forums.bbc'|2900312|entityMap|0|LINK|mutability|MUTABLE|url^0|0|18|0|0|0|0|0|0|D|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]|$1|D|3|E|5|6|7|13|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|14|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|15|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|16|8|@$A|17|B|18|O|P]]|9|@]|C|$]]|$1|Q|3|R|5|J|7|19|8|@]|9|@]|C|$K|L]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|4]]]]

<a href="https://github.com/john-kurkowski/tldextract" rel="noreferrer">https://github.com/john-kurkowski/tldextract</a>

This is a more verbose version of urlparse. It detects domains and subdomains for you.

From their documentation:

<pre><code>&gt;&gt;&gt; import tldextract
&gt;&gt;&gt; tldextract.extract('http://forums.news.cnn.com/')
ExtractResult(subdomain='forums.news', domain='cnn', suffix='com')
&gt;&gt;&gt; tldextract.extract('http://forums.bbc.co.uk/') # United Kingdom
ExtractResult(subdomain='forums', domain='bbc', suffix='co.uk')
&gt;&gt;&gt; tldextract.extract('http://www.worldbank.org.kg/') # Kyrgyzstan
ExtractResult(subdomain='www', domain='worldbank', suffix='org.kg')
</code></pre>

<code>ExtractResult</code> is a namedtuple, so it's simple to access the parts you want.

<pre><code>&gt;&gt;&gt; ext = tldextract.extract('http://forums.bbc.co.uk')
&gt;&gt;&gt; ext.domain
'bbc'
&gt;&gt;&gt; '.'.join(ext[:2]) # rejoin subdomain and domain
'forums.bbc'
</code></pre>

blocks|key|2900314|text|这有点笨拙，但在两个方向上都使用了urlparse：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2900315|import+urlparse
def+uri2schemehostname(uri):
++++urlparse.urlunparse(urlparse.urlparse(uri)[:2]+%2B+("",)+*+4)|code-block|syntax|javascript|2900316|奇数("",)+*+4位是因为urlparse期望一个恰好是len(urlparse.ParseResult._fields)+=6序列|2900317|entityMap^0|H|8|0|0|2|9|U|X|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

This is a bit obtuse, but uses <code>urlparse</code> in both directions:

<pre><code>import urlparse
def uri2schemehostname(uri):
 urlparse.urlunparse(urlparse.urlparse(uri)[:2] + ("",) * 4)
</code></pre>

that odd <code>("",) * 4</code> bit is because urlparse expects a sequence of exactly <code>len(urlparse.ParseResult._fields)</code> = 6

blocks|key|3114598|text|纯字符串操作有什么问题吗：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3114599|url+=+'http://stackoverflow.com/questions/9626535/get-domain-name-from-url'
parts+=+url.split('//',+1)
print+parts[0]%2B'//'%2Bparts[1].split('/',+1)[0]
>>>+http://stackoverflow.com|code-block|syntax|javascript|3114600|如果您更喜欢附加一个尾随斜杠，可以像这样扩展此脚本：|3114601|parts+=+url.split('//',+1)
base+=+parts[0]%2B'//'%2Bparts[1].split('/',+1)[0]
print+base+%2B+(len(url)+>+len(base)+and+url[len(base)]=='/'and'/'+or+'')|3114602|这可能会被优化一点...|3114603|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Is there anything wrong with pure string operations:

<pre><code>url = 'http://stackoverflow.com/questions/9626535/get-domain-name-from-url'
parts = url.split('//', 1)
print parts[0]+'//'+parts[1].split('/', 1)[0]
&gt;&gt;&gt; http://stackoverflow.com
</code></pre>

If you prefer having a trailing slash appended, extend this script a bit like so:

<pre><code>parts = url.split('//', 1)
base = parts[0]+'//'+parts[1].split('/', 1)[0]
print base + (len(url) &gt; len(base) and url[len(base)]=='/'and'/' or '')
</code></pre>

That can probably be optimized a bit ...

blocks|key|235796|text|使用urlsplit的Python3|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|235797|from+urllib.parse+import+urlsplit
url+=+"http://stackoverflow.com/questions/9626535/get-domain-name-from-url"
base_url+=+"{0.scheme}://{0.netloc}/".format(urlsplit(url))
print(base_url)
#+http://stackoverflow.com/|code-block|syntax|javascript|235798|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/3/library/urllib.parse#urllib.parse.urlsplit^0|2|8|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

<h3>Python3 using <a href="http://docs.python.org/3/library/urllib.parse#urllib.parse.urlsplit">urlsplit</a>:</h3>

<pre><code>from urllib.parse import urlsplit
url = "http://stackoverflow.com/questions/9626535/get-domain-name-from-url"
base_url = "{0.scheme}://{0.netloc}/".format(urlsplit(url))
print(base_url)
# http://stackoverflow.com/
</code></pre>

blocks|key|235800|text|>>>+import+urlparse
>>>+url+=+'http://stackoverflow.com/questions/1234567/blah-blah-blah-blah'
>>>+urlparse.urljoin(url,+'/')
'http://stackoverflow.com/'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|235801|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>&gt;&gt;&gt; import urlparse
&gt;&gt;&gt; url = 'http://stackoverflow.com/questions/1234567/blah-blah-blah-blah'
&gt;&gt;&gt; urlparse.urljoin(url, '/')
'http://stackoverflow.com/'
</code></pre>

blocks|key|3114651|text|纯字符串操作:)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3114652|>>>+url+=+"http://stackoverflow.com/questions/9626535/get-domain-name-from-url"
>>>+url.split("//")[-1].split("/")[0].split('?')[0]
'stackoverflow.com'
>>>+url+=+"stackoverflow.com/questions/9626535/get-domain-name-from-url"
>>>+url.split("//")[-1].split("/")[0].split('?')[0]
'stackoverflow.com'
>>>+url+=+"http://foo.bar?haha/whatever"
>>>+url.split("//")[-1].split("/")[0].split('?')[0]
'foo.bar'|code-block|syntax|javascript|3114653|就这些了，伙计们。|3114654|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Pure string operations :):

<pre><code>&gt;&gt;&gt; url = "http://stackoverflow.com/questions/9626535/get-domain-name-from-url"
&gt;&gt;&gt; url.split("//")[-1].split("/")[0].split('?')[0]
'stackoverflow.com'
&gt;&gt;&gt; url = "stackoverflow.com/questions/9626535/get-domain-name-from-url"
&gt;&gt;&gt; url.split("//")[-1].split("/")[0].split('?')[0]
'stackoverflow.com'
&gt;&gt;&gt; url = "http://foo.bar?haha/whatever"
&gt;&gt;&gt; url.split("//")[-1].split("/")[0].split('?')[0]
'foo.bar'
</code></pre>

That's all, folks.

blocks|key|235814|text|下面是一个略微改进的版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|235815|urls+=+[
++++"http://stackoverflow.com:8080/some/folder?test=/questions/9626535/get-domain-name-from-url",
++++"Stackoverflow.com:8080/some/folder?test=/questions/9626535/get-domain-name-from-url",
++++"http://stackoverflow.com/some/folder?test=/questions/9626535/get-domain-name-from-url",
++++"https://StackOverflow.com:8080?test=/questions/9626535/get-domain-name-from-url",
++++"stackoverflow.com?test=questions&v=get-domain-name-from-url"]
for+url+in+urls:
++++spltAr+=+url.split("://");
++++i+=+(0,1)[len(spltAr)>1];
++++dm+=+spltAr[i].split("?")[0].split('/')[0].split(':')[0].lower();
++++print+dm|code-block|syntax|javascript|235816|输出|235817|stackoverflow.com
stackoverflow.com
stackoverflow.com
stackoverflow.com
stackoverflow.com|235818|小提琴：https://pyfiddle.io/fiddle/23e4976e-88d2-4757-993e-532aa41b7bf0/?i=true|offset|length|235819|entityMap|0|LINK|mutability|MUTABLE|url|https://pyfiddle.io/fiddle/23e4976e-88d2-4757-993e-532aa41b7bf0/?i=true^0|0|0|0|0|4|1Z|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@$M|11|N|12|1|13]]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

Here is a slightly improved version:

<pre><code>urls = [
 "http://stackoverflow.com:8080/some/folder?test=/questions/9626535/get-domain-name-from-url",
 "Stackoverflow.com:8080/some/folder?test=/questions/9626535/get-domain-name-from-url",
 "http://stackoverflow.com/some/folder?test=/questions/9626535/get-domain-name-from-url",
 "https://StackOverflow.com:8080?test=/questions/9626535/get-domain-name-from-url",
 "stackoverflow.com?test=questions&amp;v=get-domain-name-from-url"]
for url in urls:
 spltAr = url.split("://");
 i = (0,1)[len(spltAr)&gt;1];
 dm = spltAr[i].split("?")[0].split('/')[0].split(':')[0].lower();
 print dm
</code></pre>

Output

<pre><code>stackoverflow.com
stackoverflow.com
stackoverflow.com
stackoverflow.com
stackoverflow.com
</code></pre>

Fiddle: <a href="https://pyfiddle.io/fiddle/23e4976e-88d2-4757-993e-532aa41b7bf0/?i=true" rel="noreferrer">https://pyfiddle.io/fiddle/23e4976e-88d2-4757-993e-532aa41b7bf0/?i=true</a>

blocks|key|3114714|text|如果你认为你的url是有效的，那么这将一直起作用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3114715|domain+=+"http://google.com".split("://")[1].split("/")[0]+|code-block|syntax|javascript|3114716|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

if you think your url is valid then this will work all the time

<pre><code>domain = "http://google.com".split("://")[1].split("/")[0] 
</code></pre>

blocks|key|238736|text|如果它包含的斜杠少于3个，那么你就得到了它，如果不是，那么我们可以找到它之间的出现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|238737|import+re

link+=+http://forum.unisoftdev.com/something

slash_count+=+len(re.findall("/",+link))
print+slash_count+#+output:+3

if+slash_count+>+2:
+++regex+=+r'\:\/\/(.*?)\/'
+++pattern++=+re.compile(regex)
+++path+=+re.findall(pattern,+url)

+++print+path|code-block|syntax|javascript|238738|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If it contains less than 3 slashes thus you've it got and if not then we can find the occurrence between it:

<pre><code>import re

link = http://forum.unisoftdev.com/something

slash_count = len(re.findall("/", link))
print slash_count # output: 3

if slash_count &gt; 2:
 regex = r'\:\/\/(.*?)\/'
 pattern = re.compile(regex)
 path = re.findall(pattern, url)

 print path
</code></pre>

blocks|key|2900514|text|我知道这是一个古老的问题，但我今天也遇到了这个问题。用一行代码解决了这个问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2900515|import+re
result+=+re.sub(r'(.*://)?([%5E/?]%2B).*',+'\g<1>\g<2>',+url)|code-block|syntax|javascript|2900516|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I know it's an old question, but I too encountered it today.
Solved this with an one-liner:

<pre><code>import re
result = re.sub(r'(.*://)?([^/?]+).*', '\g&lt;1&gt;\g&lt;2&gt;', url)
</code></pre>

blocks|key|233231|text|获取域/主机名和源*|type|unstyled|depth|inlineStyleRanges|entityRanges|data|233232|url+=+'https://stackoverflow.com/questions/9626535/get-protocol-host-name-from-url'
hostname+=+url.split('/')[2]+#+stackoverflow.com
origin+=+'/'.join(url.split('/')[:3])+#+https://stackoverflow.com|code-block|syntax|javascript|233233|*在XMLHttpRequest标头中使用Origin|offset|length|style|CODE|233234|entityMap^0|0|0|2|E|L|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

to get domain/hostname and Origin*

<pre><code>url = 'https://stackoverflow.com/questions/9626535/get-protocol-host-name-from-url'
hostname = url.split('/')[2] # stackoverflow.com
origin = '/'.join(url.split('/')[:3]) # https://stackoverflow.com
</code></pre>

*<code>Origin</code> is used in <code>XMLHttpRequest</code> headers

blocks|key|3114777|text|它可以通过re.search()来解决|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3114778|import+re
url+=+'https://docs.google.com/spreadsheet/ccc?key=blah-blah-blah-blah#gid=1'
result+=+re.search(r'%5Ehttp[s]*:\/\/[\w\.]*',+url).group()
print(result)

#result
'https://docs.google.com'|code-block|syntax|javascript|3114779|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

It could be solved by re.search()

<pre><code>import re
url = 'https://docs.google.com/spreadsheet/ccc?key=blah-blah-blah-blah#gid=1'
result = re.search(r'^http[s]*:\/\/[\w\.]*', url).group()
print(result)

#result
'https://docs.google.com'
</code></pre>

blocks|key|233247|text|标准库函数urllib.parse.urlsplit()就是您所需要的。以下是Python3的一个示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|233248|>>>+import+urllib.parse
>>>+o+=+urllib.parse.urlsplit('https://user:pass@www.example.com:8080/dir/page.html?q1=test&q2=a2#anchor1')
>>>+o.scheme
'https'
>>>+o.netloc
'user:pass@www.example.com:8080'
>>>+o.hostname
'www.example.com'
>>>+o.port
8080
>>>+o.path
'/dir/page.html'
>>>+o.query
'q1=test&q2=a2'
>>>+o.fragment
'anchor1'
>>>+o.username
'user'
>>>+o.password
'pass'|code-block|syntax|javascript|233249|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/urllib.parse.html#urllib.parse.urlparse^0|5|N|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

The standard library function <a href="https://docs.python.org/3/library/urllib.parse.html#urllib.parse.urlparse" rel="noreferrer">urllib.parse.urlsplit()</a> is all you need. Here is an example for Python3:

<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; import urllib.parse
&gt;&gt;&gt; o = urllib.parse.urlsplit('https://user:pass@www.example.com:8080/dir/page.html?q1=test&amp;q2=a2#anchor1')
&gt;&gt;&gt; o.scheme
'https'
&gt;&gt;&gt; o.netloc
'user:pass@www.example.com:8080'
&gt;&gt;&gt; o.hostname
'www.example.com'
&gt;&gt;&gt; o.port
8080
&gt;&gt;&gt; o.path
'/dir/page.html'
&gt;&gt;&gt; o.query
'q1=test&amp;q2=a2'
&gt;&gt;&gt; o.fragment
'anchor1'
&gt;&gt;&gt; o.username
'user'
&gt;&gt;&gt; o.password
'pass'
</code></pre>

blocks|key|2900627|text|您可以简单地使用urljoin和相对根'/‘作为第二个参数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2900628|import+urllib.parse


url+=+'https://stackoverflow.com/questions/9626535/get-protocol-host-name-from-url'
root_url+=+urllib.parse.urljoin(url,+'/')
print(root_url)|code-block|syntax|javascript|2900629|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can simply use urljoin with relative root '/' as second argument:
<pre><code>import urllib.parse


url = 'https://stackoverflow.com/questions/9626535/get-protocol-host-name-from-url'
root_url = urllib.parse.urljoin(url, '/')
print(root_url)
</code></pre>

blocks|key|237071|text|这是获取任何域的根URL的简单方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|237072|from+urllib.parse+import+urlparse

url+=+urlparse('https://stackoverflow.com/questions/9626535/')
root_url+=+url.scheme+%2B+'://'+%2B+url.hostname
print(root_url)+#+https://stackoverflow.com|code-block|syntax|javascript|237073|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is the simple way to get the root URL of any domain.
<pre><code>from urllib.parse import urlparse

url = urlparse('https://stackoverflow.com/questions/9626535/')
root_url = url.scheme + '://' + url.hostname
print(root_url) # https://stackoverflow.com
</code></pre>

In my Django app, I need to get the host name from the referrer in <code>request.META.get('HTTP_REFERER')</code> along with its protocol so that from URLs like:

<ul>
<li><a href="https://docs.google.com/spreadsheet/ccc?key=blah-blah-blah-blah#gid=1" rel="noreferrer">https://docs.google.com/spreadsheet/ccc?key=blah-blah-blah-blah#gid=1</a></li>
<li><a href="https://stackoverflow.com/questions/1234567/blah-blah-blah-blah">https://stackoverflow.com/questions/1234567/blah-blah-blah-blah</a></li>
<li><a href="http://www.example.com" rel="noreferrer">http://www.example.com</a></li>
<li><a href="https://www.other-domain.com/whatever/blah/blah/?v1=0&amp;v2=blah+blah" rel="noreferrer">https://www.other-domain.com/whatever/blah/blah/?v1=0&amp;v2=blah+blah</a> ...</li>
</ul>

I should get:

<ul>
<li><a href="https://docs.google.com/" rel="noreferrer">https://docs.google.com/</a></li>
<li><a href="https://stackoverflow.com/">https://stackoverflow.com/</a></li>
<li><a href="http://www.example.com" rel="noreferrer">http://www.example.com</a></li>
<li><a href="https://www.other-domain.com/" rel="noreferrer">https://www.other-domain.com/</a></li>
</ul>

I looked over other related questions and found about urlparse, but that didn't do the trick since

<pre><code>&gt;&gt;&gt; urlparse(request.META.get('HTTP_REFERER')).hostname
'docs.google.com'
</code></pre>

Get protocol + host name from URL

Django

Python

在我的Django应用程序中，我需要从request.META.get('HTTP_REFERER')中的referrer及其协议中获取主机名，以便从URL中获取，例如： ...我应该得到：我查看了其他相关问题，找到了关于urlparse的信息，但这并没有起到作用，因为>>> urlparse(request.META...

问从URL获取协议+主机名
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