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Not able get desired output after string parsing through regex

input =
6:/BENM/Gravity Exports/REM//INV: 3267/FEB20:65:ghgh
6:/BENM/Tabuler Trading/REM//IMP/2020-341
original_regex = 6:[A-Za-z0-9 \/\.\-:] - bt this is taking full string 6:/BENM/Gravity Exports/REM//INV: 3267/FEB20:65:ghgh
modified_regex_pattern = 6:[A-Za-z0-9 \/\.\-:]{1,}[\/-:]
In the first string i want output till
6:/BENM/Gravity Exports/REM//INV: 3267/FEB20
but its giving till :65:
Can anyone suggest better way to write this.
Example as below
https://regex101.com/r/pAduvy/1

You could for example use a capturing group with an optional part at the end to match the :digits:a-z part.
(6:[A-Za-z0-9 \/.:-]+?)(?::\d+:[a-z]+)?$
( Capture group 1
6:[A-Za-z0-9 \/.:-]+? Match any of the listed in the character class as least as possible
) Close group 1
(?::\d+:[a-z]+)? optionally match the part at the end that you don't want to include
$ End of string
Regex demo
Note Not sure if intended, but the last part of your pattern [\/-:] denotes a range from ASCII range 47 - 58.
Or a more precise pattern to get the match only
6:/\w+/\w+ \w+/[A-Z]+//[A-Z]+(?:: \d+)?/[A-Z]*\d+(?:-\d+)?
6:/\w+/\w+ Match 6 and 2 times / followed by 1+ word chars and a space
\w+/[A-Z]+//[A-Z]+ Match 1+ word chars, / and uppercase chars, // and again uppercase chars
(?:: \d+)? Optionally match a space and 1+ digits
/[A-Z]*\d+ Match /, optional uppercase chars and 1+ digits
(?:-\d+)? Optionally match - and 1+ digits
Regex demo

Finding exact values associated with given word using regex in python

am trying to find values associated with a particular word using regex but not getting expected results.
I wrote a pattern that is working fine for standard input only and I want to so the same for all sorts of inputs.
What I have now:
string = r'''results on 12/28/2012: WBC=8.110*3, RBC=3.3010*6, Hgb=11.3gm/dL'''
Pattern which I wrote:
re.findall(r'{}=(.*)'.format(detected_word), search_query)[0].split(',')[0]
detected_word is variable where am detecting left side part of equals sign like (WBC, RBC,...) using another technique.
In this above case, it's working fine, but if I change the sentence pattern like below am unable to find a generic pattern.
string = r'''results on 12/28/2012: WBC=8.110*3, RBC=3.3010*6 and Hgb=11.3gm/dL'''
string = r'''results for WBC, RBC and Hgb are 8.110*3, 3.3010*6 and 11.3gm/dL'''
no matter of string format I can able to detect WBC, RBC, and Hgb these words but detecting the value for an associated word is worrying me
Could anyone please help me with this?
Thanks in advance

Here is an idea: use two separate patterns for the strings you provided as sample input, the first one will extract values coming after expected word= and the other will extract them from clauses of expected word1 + optional expected word2 + optional expected word3 + "to be" verb + value1, optional value2 and optional value3.
Pattern 1:
\b(WBC|RBC|Hgb)=(\S*)\b
See the regex demo.
\b(WBC|RBC|Hgb) - a whole word WBC, RBC or Hgb
= - a = char
(\S*)\b - Group 2: 0 or more non-whitespaces, that stops at last word boundary position
Pattern 2:
\b(WBC|RBC|Hgb)(?:(?:\s+and)?(?:\s*,)?\s+(WBC|RBC|Hgb))?(?:(?:\s+and)?(?:\s*,)?\s*(WBC|RBC|Hgb))?\s+(?:is|(?:a|we)re|was|will\s+be)(?:\s*,)?\s*(\d\S*)\b(?:(?:\s+and)?(?:\s*,)?\s*(\d\S*)\b)?(?:(?:\s+and)?(?:\s*,)?\s*(\d\S*)\b)?
See regex demo.
\b(WBC|RBC|Hgb) - Group 1 capturing the searched word
(?:(?:\s+and)?(?:\s*,)?\s*(WBC|RBC|Hgb))? - an optional pattern:
(?:\s+and)? - an optional sequence of 1+ whitespaces and then and
(?:\s*,)? - an optional sequence of 0+ whitespaces and then a comma
\s*(WBC|RBC|Hgb) - 0+ whitespaces and Group 2 capturing the searched word
(?:(?:\s+and)?(?:\s*,)?\s*(WBC|RBC|Hgb))? - same as above, captures the 3rd optional searched word into Group 3
\s+ - 1+ whitespaces
(?:is|(?:a|we)re|was|will\s+be) - a VERB, you may add more if you expect them to be at this position, or plainly try a \S+ or \w+ pattern instead
(?:\s*,)?\s* - an optional 0+ whitespaces and a comma sequence, then 0+ whitespaces
(\d\S*)\b - Group 4 (pair it with Group 1 value): a digit and then 0+ non-whitespace chars limited by a word boundary
(?:(?:\s+and)?(?:\s*,)?\s*(\d\S*)\b)? - an optional group matching
(?:\s+and)? - an optional sequence of 1+ whitespaces and and
(?:\s*,)?\s* - an optional 0+ whitespaces and a comma, then 0+ whitespaces
(\d\S*)\b - Group 5 (pair it with Group 2 value): a digit and then 0+ non-whitespace chars limited by a word boundary
(?:(?:\s+and)?(?:\s*,)?\s*(\d\S*)\b)? - same as above, with a capture group 6 that must be paired with Group 3.

Regex include only one digit between chars

I have to parse a PDF document and I'm using PyPDF2 with re(regex).
The file includes several lines like the one below:
18-02-202010:44:48PEDMILANO OVEST- BINASCOA1,40
I need to extract from this line the text( bold ) between the time and the amount:
PEDMILANO OVEST- BINASCOA
The following code is working but sometimes this code doesn't find anything since can be a number between these chars, for example, 18-02-202010:44:48PEDMILANO OVE3ST- BINASCOA1,40.
regex = re.compile(r'\d\d-\d\d-\d\d\d\d\d\d:\d\d:\d\d\D+\d+,\d\d')
Is there a way to include a number in this regular expression?

The following should simplify the current regex:
import re
s = '18-02-202010:44:48PEDMILANO OVE3ST- BINASCOA1,40'
re.search(r'\:\d+([A-Z].*?)(?=\d+\,\d+$)', s).group(1)
# 'PEDMILANO OVE3ST- BINASCOA'
See demo
\d+([A-Z].*?)(?=\d+\,\d+$)
\: matches the character : literally (case sensitive)
\d+: matches a digit (equal to [0-9])
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
1st Capturing Group ([A-Z].*?)
Match a single character present in the list below [A-Z]
A-Z a single character in the range between A (index 65) and Z (index 90) (case sensitive)
.*? matches any character (except for line terminators)
*? Quantifier — Matches between zero and unlimited times, as few times as possible, expanding as needed (lazy)
Positive Lookahead (?=\d+\,\d+$)
Assert that the Regex below matches
\d+ matches a digit (equal to [0-9])
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
\, matches the character , literally (case sensitive)
\d+ matches a digit (equal to [0-9])
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
$ asserts position at the end of a line

I suggest using
import re
text = "18-02-202010:44:48PEDMILANO OVEST- BINASCOA1,40"
print( re.sub(r'^\d{2}-\d{2}-\d{5,6}:\d{2}:\d{2}(.*?)\d+(?:,\d+)?$', r'\1', text) )
It can also be written as
re.sub(r'^\d{2}-\d{2}-\d{5,6}:\d{2}:\d{2}|\d+(?:,\d+)?$', '', text)
Or, if you prefer matching and capturing:
m = re.search(r'^\d{2}-\d{2}-\d{5,6}:\d{2}:\d{2}(.*?)\d+(?:,\d+)?$', text)
if m:
print( m.group(1) )
See an online Python demo. With this solution, your data may start with any char, and will contain any char (excluding line break chars, since your data is on single lines).
Regex details
^ - start of string
\d{2}-\d{2}-\d{5,6}:\d{2}:\d{2} - datetime string: two digits, -, two digits, -, five or six digits, :, two digits, : two digits
(.*?) - Group 1: any zero or more chars other than line break chars, as few as possible
\d+(?:,\d+)? - an int/float value pattern: 1+ digits followed with an optional sequence of , and 1+ digits
$ - end of string.
See the regex demo.

regular expression of python

I am struggling when writing regular expression in python.
For instance I get the following right
"GET /images/launch-logo.gif HTTP/1.0" 220 1839
is matched by
"(\S+) (\S+)\s*(\S*)" (\d{3}) (\S+)
however I still need to include the following cases all together
"GET /history/history.html hqpao/hqpao_home.html
HTTP/1.0" 200 1502
"GET /shuttle/missions/missions.html Shuttle Launches from
Kennedy Space Center HTTP/1.0"200 8677
"GET /finger #net.com HTTP/1.0"404 -
obviously I should change the bold part of the expression
"(\S+) (\S+)\s*(\S*)" (\d{3}) (\S+)
But how should I change it. I have one approach in mind which is change the bold part to
[\s |(\s*)(\S+) |(\S+)(12) |(\S+)]
where the 2nd, 3rd , 4th expression is the (1), (2), (3) extra cases I need to deal with.
But my expression do not work. What do I misunderstand about regular expression as I simply deal with it case by case.

This Might be a bit messy but it works:
\"(\S+) (\S+[\s\w\.\#]*)\s*(\S*)\"\s?(\d{3})\s(\S+)*
You can play with it on Regexr. Regexr Shared Link

You may use
^"([^\s"]+)\s+([^\s"]+)(?:\s+([^"]+?))?\s+([A-Z]+/\d[\d.]*)"\s*(\d{3})\s*(\S+)$
See the regex demo
Details
^ - start of a line (use re.M if you are reading the whole file into a variable, f.read())
" - a double quotation mark
([^\s"]+) - Group 1: one or more chars other than whitespace and a double quotation mark
\s+ - 1+ whitespaces
([^\s"]+) - Group 2: one or more chars other than whitespace and a double quotation mark
(?:\s+([^"]+?))? - an optional non-capturing group matching
\s+ - 1+ whitespaces
([^"]+?) - Group 3: any 1 or more chars other than ", as few as possible
\s+ - 1+ whitespaces
([A-Z]+/\d[\d.]*) - Group 4: 1+ uppercase letters, / and then 1 digit followed with any 0+ digits or . chars
" - a double quotation mark
\s+ - 0+ whitespaces
(\d{3}) - Group 5: three digits
\s* - 0+ whitespaces
(\S+) - 1 or more non-whitespace chars
$ - end of string.

Regex return match and extended matches

Can regex return matches and extended matches. What I mean is one regex expression that can return different number of found elements depending on the structure. My text is:
AB : CDE / 123.456.1; 1
AC : DEF / 3.1.2
My return (match) should be:
'AB', 'CDE', '123.456.1', '1'
'AC', 'DEF','3.1.2'
So if there is a value after a semicolon then the regex should match and return that as well. But if is not there it should still match the part and return the rest.
My code is:
import re
s = '''AB : CDE / 123.456.1; 1
AC : DEF / 3.1.2'''
match1 = re.search(r'((?:AB|AC))\s*:\s*(\w+)\s*\/\s*([\w.]+)\s*(;\s*\d+)', s)
print(match1[0])
match2 = re.search(r'((?:AB|AC))\s*:\s*(\w+)\s*\/\s*([\w.]+)\s*', s)
print(match2[0])
Where match1 only matches the first occurrance and match2 only the second. What would be the regex to work in both cases?

The r'((?:AB|AC))\s*:\s*(\w+)\s*\/\s*([\w.]+)\s*(;\s*\d+)' pattern contains an obligatory (;\s*\d+) pattern at the end. You need to make it optional and you may do it by adding a ? quantifier after it, so as to match 1 or 0 occurrences of the subpattern.
With other minor enhancements, you may use
r'A[BC]\s*:\s*\w+\s*/\s*[\w.]+\s*(?:;\s*\d+)?'
Note all capturing groups are removed, and non-capturing ones are introduced since you only get the whole match value in the end.
Details
A[BC] - AB or AC
\s*:\s* - a colon enclosed with 0+ whitespace chars
\w+ - or more word chars
\s*/\s* - a / enclosed with 0+ whitespace chars
[\w.]+ - 1 or more word or . chars
\s* - 0+ whitespaces
(?:;\s*\d+)? - an optional sequence of
; - a ;
\s* - 0+ whitespaces
\d+ - 1+ digits