问题：如何获取URL中最后一个斜杠之后的所有内容？
如何提取Python中URL中最后一个斜杠之后的内容？例如，这些URL应该返回以下内容：
URL: http://www.test.com/TEST1
returns: TEST1
URL: http://www.test.com/page/TEST2
returns: TEST2
URL: http://www.test.com/page/page/12345
returns: 12345
我已经尝试过urlparse，但这给了我完整的路径文件名，例如page/page/12345。
How can I extract whatever follows the last slash in a URL in Python? For example, these URLs should return the following:
URL: http://www.test.com/TEST1
returns: TEST1
URL: http://www.test.com/page/TEST2
returns: TEST2
URL: http://www.test.com/page/page/12345
returns: 12345
I’ve tried urlparse, but that gives me the full path filename, such as page/page/12345.
回答 0
您不需要花哨的东西，只需在标准库中查看字符串方法，就可以轻松地在“ filename”部分和其余部分之间拆分url：
url.rsplit('/', 1)
因此，您可以简单地通过以下方式获得您感兴趣的部分：
url.rsplit('/', 1)[-1]
You don’t need fancy things, just see the string methods in the standard library and you can easily split your url between ‘filename’ part and the rest:
url.rsplit('/', 1)
So you can get the part you’re interested in simply with:
url.rsplit('/', 1)[-1]
回答 1
另一种（惯用的）方式：
URL.split("/")[-1]
One more (idio(ma)tic) way:
URL.split("/")[-1]
回答 2
rsplit 应该完成任务：
In [1]: 'http://www.test.com/page/TEST2'.rsplit('/', 1)[1]
Out[1]: 'TEST2'
rsplit should be up to the task:
In [1]: 'http://www.test.com/page/TEST2'.rsplit('/', 1)[1]
Out[1]: 'TEST2'
回答 3
您可以这样：
head, tail = os.path.split(url)
其中tail是您的文件名。
You can do like this:
head, tail = os.path.split(url)
Where tail will be your file name.
回答 4
如果需要，可以使用urlparse（例如，摆脱任何查询字符串参数）。
import urllib.parse
urls = [
'http://www.test.com/TEST1',
'http://www.test.com/page/TEST2',
'http://www.test.com/page/page/12345',
'http://www.test.com/page/page/12345?abc=123'
]
for i in urls:
url_parts = urllib.parse.urlparse(i)
path_parts = url_parts[2].rpartition('/')
print('URL: {}\nreturns: {}\n'.format(i, path_parts[2]))
输出：
URL: http://www.test.com/TEST1
returns: TEST1
URL: http://www.test.com/page/TEST2
returns: TEST2
URL: http://www.test.com/page/page/12345
returns: 12345
URL: http://www.test.com/page/page/12345?abc=123
returns: 12345
urlparse is fine to use if you want to (say, to get rid of any query string parameters).
import urllib.parse
urls = [
'http://www.test.com/TEST1',
'http://www.test.com/page/TEST2',
'http://www.test.com/page/page/12345',
'http://www.test.com/page/page/12345?abc=123'
]
for i in urls:
url_parts = urllib.parse.urlparse(i)
path_parts = url_parts[2].rpartition('/')
print('URL: {}\nreturns: {}\n'.format(i, path_parts[2]))
Output:
URL: http://www.test.com/TEST1
returns: TEST1
URL: http://www.test.com/page/TEST2
returns: TEST2
URL: http://www.test.com/page/page/12345
returns: 12345
URL: http://www.test.com/page/page/12345?abc=123
returns: 12345
回答 5
os.path.basename(os.path.normpath('/folderA/folderB/folderC/folderD/'))
>>> folderD
os.path.basename(os.path.normpath('/folderA/folderB/folderC/folderD/'))
>>> folderD
回答 6
这是更通用的正则表达式方法：
re.sub(r'^.+/([^/]+)$', r'\1', url)
Here’s a more general, regex way of doing this:
re.sub(r'^.+/([^/]+)$', r'\1', url)
回答 7
extracted_url = url[url.rfind("/")+1:];
extracted_url = url[url.rfind("/")+1:];
回答 8
partition并且rpartition对于此类事情也很方便：
url.rpartition('/')[2]
First extract the path element from the URL:
from urllib.parse import urlparse
parsed= urlparse('https://www.dummy.example/this/is/PATH?q=/a/b&r=5#asx')
and then you can extract the last segment with string functions:
parsed.path.rpartition('/')[2]
(example resulting to 'PATH')
回答 9
分割网址并弹出最后一个元素
url.split('/').pop()
Split the url and pop the last element
url.split('/').pop()
回答 10
url ='http://www.test.com/page/TEST2'.split('/')[4]
print url
输出：TEST2。
url ='http://www.test.com/page/TEST2'.split('/')[4]
print url
Output: TEST2.
