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?

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

>>> 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 /

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

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:

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:

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]

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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Tags have been added. Please use BBCode tags on future posts

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.

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