如何在不包含文件名的情况下提取路径名

Question

有以下网址。

original = 'https://dev.s3.amazonaws.com/production/uploads/2017/11/filename.jpg'

我只想从中提取/production/uploads/2017/11 可以只提取文件名并提取包含文件名的路径，但是请告诉我是否有一种很好的方法来提取除文件名之外的路径。

original_image_name = original.split('/')[-1] # 'filename.jpg'

from urllib.parse import urlparse
original_image_url = urlparse(original)
original_image_path = original_image_url.path # '/production/uploads/2017/11/filename.jpg'

Answer 1

快速的方法是使用regular expressions，就像这样：

(http[s]?:\/\/)?([^\/\s]+)(.*)

观看现场演示：https://regex101.com/r/2jsyiI/2

Answer 2

urllib有足够的工具：

https://docs.python.org/3.6/library/urllib.parse.html#module-urllib.parse

Answer 3

您可以使用：

import os
os.path.dirname(original_image_path)

Answer 4

您将获得预期的输出

original = "https://dev.s3.amazonaws.com/production/uploads/2017/11/filename.jpg"

test = original.split("/")
test = test[3:]
test = test[:-1]
sd = '/'.join(test)
print sd

  

输出

production/uploads/2017/11