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How to remove text before a particular character or string in multi-line text?

I want to remove all the text before and including */ in a string.
For example, consider:
string = ''' something
other things
etc. */ extra text.
'''
Here I want extra text. as the output.
I tried:
string = re.sub("^(.*)(?=*/)", "", string)
I also tried:
string = re.sub(re.compile(r"^.\*/", re.DOTALL), "", string)
But when I print string, it did not perform the operation I wanted and the whole string is printing.

I suppose you're fine without regular expressions:
string[string.index("*/ ")+3:]
And if you want to strip that newline:
string[string.index("*/ ")+3:].rstrip()

The problem with your first regex is that . does not match newlines as you noticed. With your second one, you were closer but forgot the * that time. This would work:
string = re.sub(re.compile(r"^.*\*/", re.DOTALL), "", string)
You can also just get the part of the string that comes after your "*/":
string = re.search(r"(\*/)(.*)", string, re.DOTALL).group(2)

Update: After doing some research, I found that the pattern (\n|.) to match everything including newlines is inefficient. I've updated the answer to use [\s\S] instead as shown on the answer I linked.
The problem is that . in python regex matches everything except newlines. For a regex solution, you can do the following:
import re
strng = ''' something
other things
etc. */ extra text.
'''
print(re.sub("[\s\S]+\*/", "", strng))
# extra text.
Add in a .strip() if you want to remove that remaining leading whitespace.

to keep text until that symbol you can do:
split_str = string.split(' ')
boundary = split_str.index('*/')
new = ' '.join(split_str[0:boundary])
print(new)
which gives you:
something
other things
etc.

string_list = string.split('*/')[1:]
string = '*/'.join(string_list)
print(string)
gives output as
' extra text. \n'

Convert string with escapes to one without

I'm working on a exercism.io exercise in Python where one of the tests requires that I convert an SGF value with escape characters into one without. I don't know why they leave newline characters intact, however.
input_val = "\\]b\nc\nd\t\te \n\\]"
output_val = "]b\nc\nd e \n]"
I tried some codecs and ats functions to no avail. Any suggestions? Thanks in advance.

The purpose of your exercise is unclear, but the solution is trivial:
input_val.replace("\\", "").replace("\t", " ")

You can use this code:
def no_escapes(text): # get text argument
# get a list of strings split with \ and join them together without it
text = text.split('\\')
text = [t.split('\t') for t in text]
text = [i for t in text for i in t]
return ''.join(text)
It will first turn "\\]b\nc\nd\t\te \n\\]" into ["]b\nc\nd\te \n"]. It'll then turn it into [["]b\nc\nd", "e \n"]]. Next, it'll flatten it out into ["]b\nc\nd", "e \n"] and it'll join them together without anything between the strings, so you'll end up with "]b\nc\nd e \n]"
Example:
>>> print(no_escapes('\\yeet\nlol\\'))
yeet
lol
And if you want it raw:
>>> string = no_escapes('\\yeet\nlol\\')
>>> print(f'{string!r}')
yeet\nlol

After looking at SGF text value rules here which says, 'all whitespaces except line breaks become spaces,' I came up with this solution. It oddly doesn't say '\\' characters should be erased, though. Not sure if there's a cleaner way to do this?
s = '\\]b\nc\nd\t\te \n\\]'
r = re.sub(r'[^\S\n]', ' ', s).replace(r'\\', '')
print(r)
# ']b\nc\nd e \n]'

Remove Characters from string with replace not working

I have a number of strings from which I am aiming to remove charactars using replace. However, this dosent seem to wake. To give a simplified example, this code:
row = "b'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38'"
row = row.replace("b'", "").replace("'", "").replace('b"', '').replace('"', '')
print(row.encode('ascii', errors='ignore'))
still ouputs this b'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38' wheras I would like it to output James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38. How can I do this?
Edit: Updataed the code with a better example.

You seem to be mistaking single quotes for double quotes. Simple replace 'b:
>>> row = "xyz'b"
>>> row.replace("'b", "")
'xyz'
As an alternative to str.replace, you can simple slice the string to remove the unwanted leading and trailing characters:
>>> row[2:-1]
'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38'

In your first .replace, change b' to 'b. Hence your code should be:
>>> row = "xyz'b"
>>> row = row.replace("'b", "").replace("'", "").replace('b"', '').replace('"', '')
# ^ changed here
>>> print(row.encode('ascii', errors='ignore'))
xyz
I am assuming rest of the conditions you have are the part of other task/matches that you didn't mentioned here.
If all you want is to take the string before first ', then you may just do:
row.split("'")[0]

You haven't listed this to remove 'b:
.replace("'b", '')

import ast
row = "b'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38'"
b_string = ast.literal_eval(row)
print(b_string)
u_string = b_string.decode('utf-8')
print(u_string)
out:
b_string:b'James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38'
u_string: James Bray,/citations?user=8IqSrdIAAAAJ&hl=en&oe=ASCII,1985,6020,188.12,42,1.31,76,2.38
The real question is how to convert a string to python object.
You get a string which contains an a binary string, to convert it to python's binary string object, you should use eval(). ast.literal_eval() is more safe way to do it.
Now you get a binary string, you can convert it to unicode string which do not start with "b" by using decode()

Delete all characters after a backslash in python?

I have a for loop that changes the string current_part.
Current_part should have a format of 1234 but sometimes it has the format of 1234/gg
Other formats exist but in all of them, anything after the backlash need to be deleted.
I found a similar example below so I tried it but it didn't work. How can I fix this? Thanks
current_part = re.sub(r"\B\\\w+", "", str(current_part))

No need for regexes here, why don't you simply go for current_part = current_part.split('/')[0] ?

Find the position of '/' and replace your string with all characters preceding '/'
st = "12345/gg"
n = st.find('/');
st = st[:n]
print(st)

You can split your string using string.split()
for example:
new_string = current_part.split("/")[0]

Using strip() to clean up a string

I am new to python and I have a string that looks like this
Temp = "', '/1412311.2121\n"
my desired output is just getting the numbers and decimal itself.. so im looking for
1412311.2121
as the output.. trying to get rid of the ', '/\n in the string.. I have tried Temp.strip("\n") and Temp.rstrip("\n") for trying to remove \n but i still seems to remain in my string. :/... Does anyone have any ideas? Thanks for your help.

Strings are immutable. string.strip() doesn't change string, it's a function that returns a value. You need to do:
Temp = Temp.strip()
Note also that calling strip() without any parameters causes it to remove all whitespace characters, including \n
As stalk said, you can achieve your desired result by calling strip("',/\n") on Temp.

If the data are like you show, numbers that are wrapped from right and left with non-number data, you can use a very simple regular expression:
g = re.search('[0-9.]+', s) # capture the inner number only
print g.group(0)

I would use a regular expression to do this:
In [8]: s = "', '/1412311.2121\n"
In [9]: re.findall(r'([+-]?\d+(?:\.\d+)?(?:[eE][+-]\d+)?)', s)
Out[9]: ['1412311.2121']
This returns a list of all floating-point numbers found in the string.