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Use re to detect string with any numbers, symbols, or letters

I have a python program and am trying to do a re.search to find a specific pattern in text. The issue I am facing is that the middle search for "[a-zA-Z0-9/" ]+" does not find any number/symbol/or letter and I have to specify each type of symbol I want it to pick up on.
re.search(r'[0-9] [a-zA-Z0-9/" ]+ [0-9]', text)
I am trying to detect strings in text.

I guess you are looking for non space, so each time you may not specify each time of symbol in the character class.
x = re.search(r'[0-9] \S+ [0-9]', text)
Samples are provided in the below link. Try this, if it helps you.
https://www.w3schools.com/python/python_regex.asp

Extracting a word between two path separators that comes after a specific word

I have the following path stored as a python string 'C:\ABC\DEF\GHI\App\Module\feature\src' and I would like to extract the word Module that is located between words \App\ and \feature\ in the path name. Note that there are file separators '\' in between which ought not to be extracted, but only the string Module has to be extracted.
I had the few ideas on how to do it:
Write a RegEx that matches a string between \App\ and \feature\
Write a RegEx that matches a string after \App\ --> App\\[A-Za-z0-9]*\\, and then split that matched string in order to find the Module.
I think the 1st solution is better, but that unfortunately it goes over my RegEx knowledge and I am not sure how to do it.
I would much appreciate any help.
Thank you in advance!

The regex you want is:
(?<=\\App\\).*?(?=\\feature\\)
Explanation of the regex:
(?<=behind)rest matches all instances of rest if there is behind immediately before it. It's called a positive lookbehind
rest(?=ahead) matches all instances of rest where there is ahead immediately after it. This is a positive lookahead.
\ is a reserved character in regex patterns, so to use them as part of the pattern itself, we have to escape it; hence, \\
.* matches any character, zero or more times.
? specifies that the match is not greedy (so we are implicitly assuming here that \feature\ only shows up once after \App\).
The pattern in general also assumes that there are no \ characters between \App\ and \feature\.
The full code would be something like:
str = 'C:\\ABC\\DEF\\GHI\\App\\Module\\feature\\src'
start = '\\App\\'
end = '\\feature\\'
pattern = rf"(?<=\{start}\).*?(?=\{end}\)"
print(pattern) # (?<=\\App\\).*?(?=\\feature\\)
print(re.search(pattern, str)[0]) # Module
A link on regex lookarounds that may be helpful: https://www.regular-expressions.info/lookaround.html

We can do that by str.find somethings like
str = 'C:\\ABC\\DEF\\GHI\\App\\Module\\feature\\src'
import re
start = '\\App\\'
end = '\\feature\\'
print( (str[str.find(start)+len(start):str.rfind(end)]))
print("\n")
output
Module

Your are looking for groups. With some small modificatians you can extract only the part between App and Feature.
(?:App\\\\)([A-Za-z0-9]*)(?:\\\\feature)
The brackets ( ) define a Match group which you can get by match.group(1). Using (?:foo) defines a non-matching group, e.g. one that is not included in your result. Try the expression here: https://regex101.com/r/24mkLO/1

Exact search of a string that has parenthesis using regex

I am new to regexes.
I have the following string : \n(941)\n364\nShackle\n(941)\nRivet\n105\nTop
Out of this string, I want to extract Rivet and I already have (941) as a string in a variable.
My thought process was like this:
Find all the (941)s
filter the results by checking if the string after (941) is followed by \n, followed by a word, and ending with \n
I made a regex for the 2nd part: \n[\w\s\'\d\-\/\.]+$\n.
The problem I am facing is that because of the parenthesis in (941) the regex is taking 941 as a group. In the 3rd step the regex may be wrong, which I can fix later, but 1st I needed help in finding the 2nd (941) so then I can apply the 3rd step on that.
PS.
I know I can use python string methods like find and then loop over the searches, but I wanted to see if this can be done directly using regex only.
I have tried the following regex: (?:...), (941){1} and the make regex literal character \ like this \(941\) with no useful results. Maybe I am using them wrong.
Just wanted to know if it is possible to be done using regex. Though it might be useful for others too or a good share for future viewers.
Thanks!

Assuming:
You want to avoid matching only digits;
Want to match a substring made of word-characters (thus including possible digits);
Try to escape the variable and use it in the regular expression through f-string:
import re
s = '\n(941)\n364\nShackle\n(941)\nRivet\n105\nTop'
var1 = '(941)'
var2 = re.escape(var1)
m = re.findall(fr'{var2}\n(?!\d+\n)(\w+)', s)[0]
print(m)
Prints:
Rivet

If you have text in a variable that should be matched exactly, use re.escape() to escape it when substituting into the regexp.
s = '\n(941)\n364\nShackle\n(941)\nRivet\n105\nTop'
num = '(941)'
re.findall(rf'(?<=\n{re.escape(num)}\n)[\w\s\'\d\-\/\.]+(?=\n)', s)
This puts (941)\n in a lookbehind, so it's not included in the match. This avoids a problem with the \n at the end of one match overlapping with the \n at the beginning of the next.

Trying to find the regex for this particular case? Also can I parse this without creating groups?

text to capture looks like this..
Policy Number ABCD000012345 other text follows in same line....
My regex looks like this
regex value='(?i)(?:[P|p]olicy\s[N|n]o[|:|;|,][\n\r\s\t]*[\na-z\sA-Z:,;\r\d\t]*[S|s]e\s*[H|h]abla\s*[^\n]*[\n\s\r\t]*|(?i)[P|p]olicy[\s\n\t\r]*[N|n]umber[\s\n\r\t]*)(?P<policy_number>[^\n]*)'
this particular case matches with the second or case.. however it is also capturing everything after the policy number. What can be the stopping condition for it to just grab the number. I know something is wrong but can't find a way out.
(?i)[P|p]olicy[\s\n\t\r]*[N|n]umber[\s\n\r\t]*)
current output
ABCD000012345othertextfollowsinsameline....
expected output
ABCD000012345

You may use a more simple regex, just finding from the beginning "[P|p]olicy\s*[N|n]umber\s*\b([A-Z]{4}\d+)\b.*" and use the word boundary \b
pattern = re.compile(r"[P|p]olicy\s*[N|n]umber\s*\b([A-Z0-9]+)\b.*")
line = "Policy Number ABCD000012345 other text follows in same line...."
matches = pattern.match(line)
id_res = matches.group(1)
print(id_res) # ABCD000012345
And if there's always 2 words before you can use (?:\w+\s+){2}\b([A-Z0-9]+)\b.*
Also \s is for [\r\n\t\f\v ] so no need to repeat them, your [\n\r\s\t] is just \s

you don't need the upper and lower case p and n specified since you're already specifying case insensitive.
Also \s already covers \n, \t and \r.
(?i)policy\s+number\s+([A-Z]{4}\d+)\b
for verification purpose: Regex
Another Solution:
^[\s\w]+\b([A-Z]{4}\d+)\b
for verification purpose: Regex
I like this better, in case your text changes from policy number

How do I create a regex with regular variable and some fixed text in Python?

In code i only want to fetch variable name from a c file which is used in if condition.
Following is code snippet of regex:
fieldMatch = re.findall(itemFieldList[i]+"=", codeline, re.IGNORECASE);
here i can find variable itemFieldList[i] from file.
But when i try to add if as shown below nothing is extracted as output even though variable exist in c code in if condition .
fieldMatch = re.findall(("^(\w+)if+[(](\w+)("+itemFieldList[i]+")="), codeline, re.IGNORECASE|re.MULTILINE);
Can anyone suggest how can we create regex to fetch mentioned scenario.
Sample Input :
IF(WORK.env_flow_ind=="R")
OR
IF( WORK.qa_flow_ind=="Q" OR WORK.env_flow_ind=="R")
here itemFieldList[i] = WORK.env_flow_ind

I don't have enough reputation to make this a comment, which it should be and I can't say that I fully understand the question. But to point out a few things:
it's about adding variables to your regex then you should be using string templates to make it more understandable for us and your future self.
"^{}".format(variable)
Doing that will allow you to create a dynamic regex that searches for what you want.
Secondly, I don't think that is your problem. I think that your regex is malformed. I don't know what exactly you are trying to search for but I recommend reading the python regex documentation and testing your regex on a resource like regex101 to make sure that you're capturing what you intend to. From what I can see you are a bit confused about groups. When you put parenthesis around a pattern you are identifying it as a group. You were on the right track trying to exclude the parenthesis in your search by surrounding it with square brackets but it's simpler and cleaner to escape them.
if you are trying to capture this statement:
if(someCondition == fries)
and you want to extract the keyword fries the valid syntax for that pattern is:
(?=if\((?:[\w=\s])+(fries)\))
Since you want this to be dynamic you would replace the string fries with your string template, and you'll get code that ends up something like this:
p = re.compile("(?=if\((?:[\w=\s])+({})\))".format(search), re.IGNORECASE)
p.findall(string)
Regex101 does a better job of breaking down my regex than I ever will:
Link cuz i have no rep

You can build the regex pattern as:
pattern = r"\bif\b\s*\(.*?\b" + re.escape(variablename) + r"\b"
This will look for the word “if” in lowercase, then optionally any spaces, then an opening parenthesis, then optionally any characters, and then your search term, its beginning and its end at word boundaries.
So if variablename is "WORK.env_flow_ind", then re.findall(pattern, textfile) will match the following lines:
if(blabla & WORK.env_flow_ind == "a")
if (WORK.env_flow_id == "b")
if(WORK.env_flow_id == "b")
if( WORK.env_flow_id == "b")
and these won't match:
if (WORK.env_bla == "c")
if (WORK.env_flow_id2 == "d")