How do I parse the string?

[anna17]

hello

There is a string below.
FTP://21.105.28.15/abc/test1.txt
Cut this string
/abc/
I just want to get the path.

How do I parse the string?

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import pymssql conn = pymssql.connect('10.21.100.21', 'sa', 'abc', 'DBA', as_dict=False) cur = conn.cursor()   cur.execute('select path From table')    fetch = cur.fetchall()    for i in fetch:     print (i) <==== FTP://21.105.28.15/abc/test1.txt (Cut this string /abc/ I just want to get the path.)   conn.close()

Gribouillis write Apr-10-2024, 07:32 AM:
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[Gribouillis]

You could use urllib.parse.urlparse

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>>> from urllib.parse import urlparse >>> urlparse('FTP://21.105.28.15/abc/test1.txt') ParseResult(scheme='ftp', netloc='21.105.28.15', path='/abc/test1.txt', params='', query='', fragment='') >>> result = urlparse('FTP://21.105.28.15/abc/test1.txt') >>> result.path '/abc/test1.txt' >>>

[Pedroski55]

Maybe like this, without using any modules: find the third / and the last /

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mystring = 'FTP://21.105.28.15/abc/test1.txt' mystring2 = 'FTP://21.105.28.15/abc/def/ghi/test1.txt'   # get the index of the third / def getFirst():     count = 0     for s in range(1, len(mystring)):         # skip the first 2 /         if mystring[s] == '/':             count +=1                        # get the third /             if count == 3:                 index_first = s                 return index_first   # get the index of the last / def getLast():     for s in range(1, len(mystring)):         if mystring[-s] == '/':             index_last = -(s - 1)             return index_last   wanted = mystring[getFirst():getLast()] wanted = mystring2[getFirst():getLast()]

Gives:

Output:
wanted
'/abc/'

or

Output:
wanted
'/abc/def/ghi/'

[menator01]

Another way

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astring = 'FTP://21.105.28.15/abc/test1.txt' astring2 = 'FTP://21.105.28.15/abc/def/test1.txt'   def remover(astring):     astring = astring.split('/')     for index, item in enumerate(astring):         if item == '':             astring.remove(item)         elif index in (0, 1):             astring.remove(item)     return '/'.join(astring)   print(remover(astring)) print(remover(astring2))

output

Output:
/abc/test1.txt
/abc/def/test1.txt

little adjustment can remove the file at the end as well

[DeaD_EyE]

If you need a path, a Path object is maybe what you want.

Code is based on Gribouillis example:

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from urllib.parse import urlparse from pathlib import PurePosixPath     def url2path(url: str) -> PurePosixPath:     return PurePosixPath(urlparse(url).path)     path = url2path('FTP://21.105.28.15/abc/test1.txt')   print("path:", path) print("path.parent", path.parent)   print("path.parts", path.parts) print("path.parents", list(path.parents))   print("path.name:", path.name) print("path.stem:", path.stem) print("path.suffix", path.suffix)     # Path objects are printed as strings # but it's a different type # here how to do it explicit print("Path as str:", str(path))   # not all 3rd party libraries handle the conversion from `Path` to `str` implicit

Output:
path: /abc/test1.txt
path.parent /abc
path.parts ('/', 'abc', 'test1.txt')
path.parents [PurePosixPath('/abc'), PurePosixPath('/')]
path.name: test1.txt
path.stem: test1
path.suffix .txt
Path as str: /abc/test1.txt

This also works with other non-FTP urls:

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urls = (     'FTP://21.105.28.15/abc/test1.txt',     'FTPS://21.105.28.15/abc/test1.txt',     'HTTP://21.105.28.15/abc/test1.txt',     'HTTPS://21.105.28.15/abc/test1.txt',     'MyOwnUselessURL://xyz.de/abc/test1.txt', )     for url in urls:     print(url2path(url))

Boring output:

Output:
/abc/test1.txt
/abc/test1.txt
/abc/test1.txt
/abc/test1.txt
/abc/test1.txt

Documents you should read:

• https://docs.python.org/3/library/pathli...pure-paths
  
• https://docs.python.org/3/library/urllib...rl-parsing
  

[kakocer]

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mystring = 'FTP://21.105.28.15/abc/test1.txt' mystring2 = 'FTP://21.105.28.15/abc/def/ghi/test1.txt'   # Get the index of the third / def getFirst(url):     count = 0     for i, char in enumerate(url):         if char == '/':             count += 1             if count == 3:                 return i     return -1  # Return -1 if there's no third '/'   # Get the index of the last / def getLast(url):     return url.rfind('/')  # Use rfind to get the last occurrence of '/'   # Extract the substring between the third and the last '/' def extract_substring(url):     first_index = getFirst(url)     last_index = getLast(url)     if first_index != -1 and first_index < last_index:         return url[first_index + 1:last_index]  # Skip the '/' itself     return ""   # Extract and print the portions of the URLs wanted1 = extract_substring(mystring) wanted2 = extract_substring(mystring2)   print(wanted1)  # Output: abc/test1.txt print(wanted2)  # Output: abc/def/ghi/test1.txt

I'm using this and it work for me.

Larz60+ write Jan-21-2025, 08:58 AM:
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Gribouillis write Jan-20-2025, 08:23 PM:
Clickbait link removed. Please read What to NOT include in a post

[Keville_35]

I define an "extract_path" function that uses a regular expression to find the part of the URL after "/abc/" and extract it.

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import pymssql import re   def extract_path(url):     # Use regex to find the part after "/abc/"     match = re.search(r'/abc/(.*)', url)     if match:         return match.group(1)     return None   # Database connection conn = pymssql.connect('10.21.100.21', 'sa', 'abc', 'DBA', as_dict=False) cur = conn.cursor()   # Execute the query cur.execute('SELECT path FROM table')   # Fetch all results fetch = cur.fetchall()   # Process and print the results for row in fetch:     ftp_url = row[0]     extracted_path = extract_path(ftp_url)     if extracted_path:         print(f"Original URL: {ftp_url}")         print(f"Extracted path: {extracted_path}")         print("---")   # Close the connection conn.close()

This is working at your end?

[DeaD_EyE]

Solving this kind of problem with regex is brute-force. Sometimes it's good to take a step back and ask, if you can solve the problem differently.

[michaeljordan]

Solving this kind of problem with regex is brute-force. Sometimes it's good to take a step back and ask, if you can solve the problem differently.

If it really works, there is no reason to refuse.

buran write Feb-13-2025, 07:59 AM:
Spam link removed