I am trying to get the name of the resource, I will share with you the

CodePudding user response：

You can use:

^. \/(. )\.. $

• ^. - From the start, match as many characters as possible
• \/ - Match a literal /.
• (. ) - Match one or more characters and capture them in a group
• \. - Match a literal .
• . $ - Match one or more characters at the end of the string (the extension)

Live demo here.

CodePudding user response：

There are many ways:

Couple below using python:

#using regexp:

>>> file_name='https://res-3.cloudinary.com/ngxcoder/image/upload/f_auto,q_auto/v1/blog-images/5oonz9.jpg'
>>> regexpr = r".*/([^\/] ).jpg$"
>>> re.match(regexpr, file_name).group(1)
'5oonz9'
>>>

#to get any file name:

>>> regexpr = r".*/([^\/] )$"
>>> re.match(regexpr, file_name).group(1)
'5oonz9.jpg'

#if interested, here is one using split & take last

>>> (file_name.split("/")[-1]).split(".")[0]
'5oonz9'
>>>

CodePudding user response：

You don't need a capture group, just a match:

(?<=\/)[^\/.] (?=\.[^\/.] $)

Demo

We can write the expression in free-spacing mode to make it self-documenting:

(?<=      # begin a negative lookbehind
  \/      # match '/'
)         # end negative lookbehind
[^\]      # match one or more characters other than '/'
(?=       # begin a positive lookahead
  \.      # match '.'
  [^\/]   # match one or more characters other than '/'
  $       # match end of string
)         # end the positive lookahead

You should not use a regex for this, however, as Python provides os.path:

import os

str = 'https://res-3.cloudinary.com/ngxcoder/image/'\
      'upload/f_auto,q_auto/v1/blog-images/5oonz9.jpg'
base = os.path.basename(str)
print(os.path.splitext(base)[0])
  #=> "5oonz9"

Here base #=> "5oonz9.jpg".

See it in action

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