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Regex syntax to match equal sign in each word but do not take in consideration the first equal sign [closed]

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It might be confusing and I don't know if it even possible as I have some knowledge about regex using python but I couldn't solve the issue I have.
The thing is I have logs that I want to replace the equal sign in their URL to their url encoding (%3D) using Regex syntax.
For example I have logs of this:
request=www.google.com/fgdsg=gfsdg=gfdsd
request_access=https://regex101.com/eawf/?=dasf
All the equal sign after the first one that uses to assign value for variable I want to match them then replace them with %3D like this:
request=www.google.com/fgdsg%3Dgfsdg%3Dsgfdsd
request_access=https://regex101.com/eawf/?%3Ddasf
This is what I want to be.
The written text it only example I didn't wrote real logs.

You can make use of a negative lookbehind like so: (?<!request)= this will match all the = that do not come after request. A little
import re
sample = "request=www.google.com/fgdsg=gfsdg=gfdsd"
replaced = re.sub('(?<!request)=', '%3D', sample)
print(replaced)

First remove the part before (including) the first =, leaving only the request URL. Then you can simply substitute all occurrences of = with %3D in this string.
import re
log = "request=www.google.com/fgdsg=gfsdg=gfdsd"
request_url = re.sub(r'^\w+=', '', log, 1)
urlencoded_request_url = re.sub(r'=', '%3D', request_url)
print(urlencoded_request_url) # www.google.com/fgdsg%3Dgfsdg%3Dgfdsd

Replace a string in a URL with python [closed]

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I have a text containing a URL that needs to be reworked.
text='dfs:/?url=https://myserver/c12&ofg={"tes":{"id":1812}}'
I need to replace programmatically the id value (in this example 1812, which is unknown before the execution) with a fixed substring (e.g. 189). So the end result must be
'dfs:/?url=https://myserver/c12&ofg={"tes":{"id":189}}'
As I'm programming in Python, I guess that I should use the regular expression (module re) to automatically replace that value between "id": and }} but I couldn't find one that works for this use case.

I assume you are always generating the same URL with that pattern, and the value to 'change' is always in {"id":X}. One way to solve this particular problem is with a positive lookbehind + re.sub replacement.
import re
pattern = re.compile(r"(?<=\"id\":)\d+")
string = "dfs:/?url=https://myserver/c12&ofg={\"tes\":{\"id\":1812}}"
print(pattern.sub("desired_value", string))
Generated output will contain desired_value in place of the 1812. A good explanation of what is happening is done in regex101 but a quick rep of what is happening in the pattern:
Matches any digit one or more times ONLY if behind has "id":, without consuming characters

what about simply splitting the string twice? eg.
my_string = 'dfs:/?url=https://myserver/c12&ofg={"tes":{"id":1812}}'
substring = my_string.split('"id":',1)[1]
substring = substring.split('}}')[0]
print(my_string.replace(substring, "189"))

Python: how to extract the variables between 2 constant substring [closed]

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I am trying to extract the variables between 2 constant substring in a string. For example,
I wish to extract the variable Apple, Orange, Watermelon, Kiwi....13cups, 14cups...19cups. I am using the re expression to get to the first step of taking the variable between $ sign but I do not get anything results.
Anyone can advise on the correct expression or if there is a better way to extract it ?
Thanks.
import re
file = '$n$n$n$xa0$n$nSHOWALL$nSHOWALL%GROWTH$n$n$xa0$n$xa0$n$n$n$nApple$na$nOrange$n$nWatermelon$nKiwi$n$nBanana$nJackfruit$n$nGuava$na$nGrape$n$nPlum$na$nOrange$n$nCoconut$nWatermelon$n$n12cups$n13cups$n$n14cups$na$n15cups$n$n16cups$na$n17cups$n$n18cups$n19cups$n'
found = re.findall(r'(?=$(.*?)$)',file)
print(found)

Given that the rule(s) for identifying the required character sequences is ambiguous, I contend that RE is impractical. No doubt it could be done but here's a quick'n'dirty approach to the problem:-
data = '$n$n$n$xa0$n$nSHOWALL$nSHOWALL%GROWTH$n$n$xa0$n$xa0$n$n$n$nApple$na$nOrange$n$nWatermelon$nKiwi$n$nBanana$nJackfruit$n$nGuava$na$nGrape$n$nPlum$na$nOrange$n$nCoconut$nWatermelon$n$n12cups$n13cups$n$n14cups$na$n15cups$n$n16cups$na$n17cups$n$n18cups$n19cups$n'
for token in data.split('$n'):
if token not in ('SHOWALL%GROWTH', 'SHOWALL', '$xa0', 'a', ''):
print(token)

negative lookbehind not working as expected [closed]

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I have strings of this form:
FPLBX(2x3)ZE(53x13)(4x7)ZGQO
I want to find the blocks in parenthesis but only when they're not preceded by another group.
The other way around works perfectly fine but I can't make it work with preceding.
current regex:
(\(\d*x\d*\))(?<!\))

You simply need to put the so-called negative lookbehind assertion, i.e. the (?<!\))-part, in front of your search re:
>>> import re
>>> txt = "FPLBX(2x3)ZE(53x13)(4x7)ZGQO"
>>> re.findall(r"(?<!\))(\(\d*x\d*\))", txt)
['(2x3)', '(53x13)']

Python Regular Expression for pattern containing multiple lines [closed]

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I want to extract all the text printed after "AAAAAAAAAAAAAAAAAA"
Give me some text!
AAAAAAAAAAAAAAAAAA
S
p
p
p
Epppp
The following does not work:
import re
m = re.findall(r'AAAAAAAAAAAAAAAAAA(.*)', result)
print m[0]
Also, can I specify a variable in a regular expression instead of a hard coded string: "AAAAAAAAAAAAAAAAAA"?
Reason being, the text: "AAAAAAAAAAAAAAAAAA" is a variable and changes. So, I would like to look for a specific variable value in the pattern and then extract all the text after it.

Use re.S or re.DOTALL (they are synonyms) to have findall match across lines. Or, in your case, search is probably more appropriate since you only want one match. Also, to have it work for a non-hard-coded string, simply use string formatting or string concatenation. To avoid having unescaped regex characters in the string, run it through re.escape.
import re
result = """Give me some text!
AAAAAAAAAAAAAAAAAA
S
p
p
p
Epppp"""
s = 'AAAAAAAAAAAAAAAAAA'
# With formatting
m = re.search(r'{}(.*)'.format(re.escape(s)), result, re.S)
# With concatenation
m = re.search(re.escape(s) + r'(.*)', result, re.S)
print m.group(1)