Develop Reference

Match text between parenthesis that end with .md

I need to get the text inside the parenthesis where the text ends with .md using a regex (if you know another way you can say it) in python.
Original string:
[Romanian (Romania)](books/free-programming-books-ro.md)
Expected result:
books/free-programming-books-ro.md

This should work:
import re
s = '[Romanian (Romania)](books/free-programming-books-ro.md)'
result = re.findall(r'[^\(]+\.md(?=\))',s)
['books/free-programming-books-ro.md']

Regex filter containing word at beginning but not containing another word

suppose i have the following string
GPH_EPL_GK_FIN
i want a regex that ill be using in python that looks for such string from a csv file (not relevant to this question) for records that start with GPH but DONT contain EPL
i know carrot ^ is used for searching at beginning
so i have something like this
^GPH_.*
i want to include the NOT contain part as well, how do i chain the regex?
i.e.
(^GPH_.*)(?!EPL)
i would like to take this a step further eventually and any records that are returned without EPL, i.e.
GPH_ABC_JKL_OPQ
to include AFTER GPH_ the EPL part
i.e. desired result
GPH_EPL_ABC_JKL_OPQ

To cover both requirements:
compose a pattern to match lines that start with GPH but DONT contain EPL
insert EPL_ part into matched line to a particular position
import re
# sample string containing lines
s = '''GPH_EPL_GK_FIN
GPH_ABC_JKL_OPQ'''
pat = re.compile(r'^(GPH_)(?!.*EPL.*)')
for line in s.splitlines():
print(pat.sub('\\1EPL_', line))
The output:
GPH_EPL_GK_FIN
GPH_EPL_ABC_JKL_OPQ

This here would do, I think:
^GPH_(?!EPL).*
This will return any string that start with GPH and does not have EPL after GPH_.

I'm just guessing that one option would be,
(?<=^GPH_(?!EPL))
and re.sub with,
EPL_
Test
import re
print(re.sub(r"(?<=^GPH_(?!EPL))", "EPL_", "GPH_ABC_JKL_OPQ"))
Output
GPH_EPL_ABC_JKL_OPQ

Simply use this:
https://regex101.com/r/GwBsg2/2
pattern: ^(?!^(?:[^_\n]+_)*EPL_?(?:[^_\n]+_?)*)(.*)GPH
substitute: \1GPH_EPL
flags: gm

The way to unescape escaped regex pattern Python

I'm trying to unescape the escaped regex pattern to apply it to a string.
It's actually dynamic I don't exactly know what it would look like, but throughout my testing I encountered one problem, the string with escaped regex pattern looks like this:
\\d{4}
I've written a simple regex which replaces every single combination of backslash and a character with just a character
And I'm applying it this way:
sub(r"\\(.)", "\\1", escaped_pattern)
But what it gives me afterwards is d{4} not \d{4} as I expect.
I've tried using raw strings for repl, escape\unescape it, it still doesnt return what I expect it to return. Would appreciate any help.
EDIT
escaped_pattern = settings.reg_exp
regexp = sub(r"\\(.)", "\\1", escaped_pattern)
search(regexp, string_to_regexp).group()[0]

Based on you update I'm pretty sure that you would get exactly your desired output if you just stopped trying to unescape it.
import re
s1 = "1234astring"
matches = re.search("\\d{4}", s1)
matches.group(0)
"1234"
matches.group()[0]
"1"

Try r"\\\\(.)" in search pattern and '\\\1' in substitution pattern.
works OK here: https://regex101.com/r/M3ikqj/1

python regular expression not matching file contents with re.match and re.MULTILINE flag

I'm reading in a file and storing its contents as a multiline string. Then I loop through some values I get from a django query to run regexes based on the query results values. My regex seems like it should be working, and works if I copy the values returned by the query, but for some reason isn't matching when all the parts are working together that ends like this
My code is:
with open("/path_to_my_file") as myfile:
data=myfile.read()
#read saved settings then write/overwrite them into the config
items = MyModel.objects.filter(some_id="s100009")
for item in items:
regexString = "^\s*"+item.feature_key+":"
print regexString #to verify its what I want it to be, ie debug
pq = re.compile(regexString, re.M)
if pq.match(data):
#do stuff
So basically my problem is that the regex isn't matching. When I copy the file contents into a big old string, and copy the value(s) printed by the print regexString line, it does match, so I'm thinking theres some esoteric python/django thing going on (or maybe not so esoteric as python isnt my first language).
And for examples sake, the output of print regexString is :
^\s*productDetailOn:
File contents:
productDetailOn:true,
allOff:false,
trendingWidgetOn:true,
trendingWallOn:true,
searchResultOn:false,
bannersOn:true,
homeWidgetOn:true,
}
Running Python 2.7. Also, dumped the types of both item.feature and data, and both were unicode. Not sure if that matters? Anyway, I'm starting to hit my head off the desk after working this for a couple hours, so any help is appreciated. Cheers!

According to documentation, re.match never allows searching at the beginning of a line:
Note that even in MULTILINE mode, re.match() will only match at the beginning of the string and not at the beginning of each line.
You need to use a re.search:
regexString = r"^\s*"+item.feature_key+":"
pq = re.compile(regexString, re.M)
if pq.search(data):
A small note on the raw string (r"^\s+"): in this case, it is equivalent to "\s+" because there is no \s escape sequence (like \r or \n), thus, Python treats it as a raw string literal. Still, it is safer to always declare regex patterns with raw string literals in Python (and with corresponding notations in other languages, too).

How to regex split, but keep the split string?

I have the following URL pattern:
http://www.hulu.jp/watch/589851/supernatural-dub-hollywood-babylon/en
I would like to get everything up until and inclusive of /watch/\d+/.
So far I have:
>>> re.split(r'watch/\d+/', 'http://www.hulu.jp/watch/589851/supernatural-dub-hollywood-babylon/en')
['http://www.hulu.jp/', 'supernatural-dub-hollywood-babylon/en']
But this does not include the split string (the string which appears between the domain and the path). The end answer I want to achieve is:
http://www.hulu.jp/watch/589851

You need to use capture group :
>>> re.split(r'(watch/\d+/)', 'http://www.hulu.jp/watch/589851/supernatural-dub-hollywood-babylon/en')
['http://www.hulu.jp/', 'watch/589851/', 'supernatural-dub-hollywood-babylon/en']

As mentioned in the other answer, you need to use groups to capture the "glue" between the split strings.
I wonder though, is what you want here a split() or a search()? It looks (from the sample) that you're trying to extract from a URL everything from the first occurrence of /watch/XXX/ where XXX is 1 or more digits, to the end of the string. If that's the case, then a match/search might be more suitable, as with a split if the search regex can match multiple times you'll split into multiple groups. Ex:
re.split(r'(watch/\d+/)', 'http://www.hulu.jp/watch/589851/supernatural-dub-hollywood-babylon/watch/2342/fdsaafsdf')
['http://www.hulu.jp/', 'watch/589851/', 'supernatural-dub-hollywood-babylon/', 'watch/2342/', 'fdsaafsdf']
Which doesn't look like what you want. Instead perhaps:
result = re.search(r'(watch/\d+/)(.*)', 'http://www.hulu.jp/watch/589851/supernatural-dub-hollywood-babylon/watch/2342/fdsaafsdf')
result.groups() if result else []
which gives:
('watch/589851/', 'supernatural-dub-hollywood-babylon/watch/2342/fdsaafsdf')
You could also use this approach combined with named groups to get extra fancy:
result = re.search(r'(?P<watchId>watch/\d+/)(?P<path>.*)', 'http://www.hulu.jp/watch/589851/supernatural-dub-hollywood-babylon/watch/2342/fdsaafsdf')
result.groupdict() if result else {}
giving:
{'path': 'supernatural-dub-hollywood-babylon/watch/2342/fdsaafsdf', 'watchId': 'watch/589851/'}
If you're set on the split() approach, you can also set the maxsplit parameter to ensure it's only split once:
re.split(r'(watch/\d+/)', 'http://www.hulu.jp/watch/589851/supernatural-dub-hollywood-babylon/watch/2342/fdsaafsdf', maxsplit=1)
giving:
['http://www.hulu.jp/', 'watch/589851/', 'supernatural-dub-hollywood-babylon/watch/2342/fdsaafsdf']
Personally though, I find that when parsing URL's into constituent parts the search() with named groups approach works extremely well as it allows you to name the various parts in the regex itself, and via groupdict() get a nice dictionary you can use for working with those parts.

You've surely seen the Stack Overflow don't-parse-HTML-with-regex post, yes?
You can't parse [X]HTML with regex. Because HTML can't be parsed by regex. Regex is not a tool that can be used to correctly parse HTML. As I have answered in HTML-and-regex questions here so many times before, the use of regex will not allow you to consume HTML.
Well, regex can parse URLs, but trying to do so when there's a plethora of better tools is foolish.
This is what a regex for URLs looks like:
^(?:(?:https?|ftp):\/\/)(?:\S+(?::\S*)?#)?(?:(?!10(?:\.\d{1,3}){3})(?!127(?:\.\d{1,3}){3})(?!169\.254(?:\.\d{1,3}){2})(?!192\.168(?:\.\d{1,3}){2})(?!172\.(?:1[6-9]|2\d|3[0-1])(?:\.\d{1,3}){2})(?:[1-9]\d?|1\d\d|2[01]\d|22[0-3])(?:\.(?:1?\d{1,2}|2[0-4]\d|25[0-5])){2}(?:\.(?:[1-9]\d?|1\d\d|2[0-4]\d|25[0-4]))|(?:(?:[a-z\x{00a1}-\x{ffff}0-9]+-?)*[a-z\x{00a1}-\x{ffff}0-9]+)(?:\.(?:[a-z\x{00a1}-\x{ffff}0-9]+-?)*[a-z\x{00a1}-\x{ffff}0-9]+)*(?:\.(?:[a-z\x{00a1}-\x{ffff}]{2,})))(?::\d{2,5})?(?:\/[^\s]*)?$ (+ caseless flag)
It's just a mess of characters, right? Exactly!
Don't parse URLs with regex... almost.
There is one simple thing:
A path-relative URL must be zero or more path segments separated from each other by a "/".
Splitting the URL should be as simple as url.split("/").
from urllib.parse import urlparse, urlunparse
myurl = "http://www.hulu.jp/watch/589851/supernatural-dub-hollywood-babylon/en"
# Run a parser over it
parts = urlparse(myurl)
# Crop the path to UP TO length 2
new_path = str("/".join(parts.path.split("/")[:3]))
# Unparse
urlunparse(parts._replace(path=new_path))
#>>> 'http://www.hulu.jp/watch/589851'

You can try following regex
.*\/watch\/\d+
Working Demo