How to extract the substring between two markers?


Let’s say I have a string 'gfgfdAAA1234ZZZuijjk' and I want to extract just the '1234' part.

I only know what will be the few characters directly before AAA, and after ZZZ the part I am interested in 1234.

With sed it is possible to do something like this with a string:

echo "$STRING" | sed -e "s|.*AAA(.*)ZZZ.*|1|"

And this will give me 1234 as a result.

How to do the same thing in Python?

Asked By: ria



>>> s = 'gfgfdAAA1234ZZZuijjk'
>>> start = s.find('AAA') + 3
>>> end = s.find('ZZZ', start)
>>> s[start:end]

Then you can use regexps with the re module as well, if you want, but that’s not necessary in your case.

Answered By: Lennart Regebro
import re
print'AAA(.*?)ZZZ', 'gfgfdAAA1234ZZZuijjk').group(1)
Answered By: infrared

Using regular expressions – documentation for further reference

import re

text = 'gfgfdAAA1234ZZZuijjk'

m ='AAA(.+?)ZZZ', text)
if m:
    found =

# found: 1234


import re

text = 'gfgfdAAA1234ZZZuijjk'

    found ='AAA(.+?)ZZZ', text).group(1)
except AttributeError:
    # AAA, ZZZ not found in the original string
    found = '' # apply your error handling

# found: 1234
Answered By: eumiro

You can use re module for that:

>>> import re
>>> re.compile(".*AAA(.*)ZZZ.*").match("gfgfdAAA1234ZZZuijjk").groups()
Answered By: andreypopp

regular expression

import re"(?<=AAA).*?(?=ZZZ)", your_text).group(0)

The above as-is will fail with an AttributeError if there are no “AAA” and “ZZZ” in your_text

string methods


The above will return an empty string if either “AAA” or “ZZZ” don’t exist in your_text.

PS Python Challenge?

Answered By: tzot

Just in case somebody will have to do the same thing that I did. I had to extract everything inside parenthesis in a line. For example, if I have a line like ‘US president (Barack Obama) met with …’ and I want to get only ‘Barack Obama’ this is solution:

regex = '.*((.*?)).*'
matches =, line)
line = + 'n'

I.e. you need to block parenthesis with slash sign. Though it is a problem about more regular expressions that Python.

Also, in some cases you may see ‘r’ symbols before regex definition. If there is no r prefix, you need to use escape characters like in C. Here is more discussion on that.

Answered By: Denis Kutlubaev
>>> s = '/tmp/10508.constantstring'
>>> s.split('/tmp/')[1].split('constantstring')[0].strip('.')
Answered By: user1810100

With sed it is possible to do something like this with a string:

echo "$STRING" | sed -e "s|.*AAA(.*)ZZZ.*|1|"

And this will give me 1234 as a result.

You could do the same with re.sub function using the same regex.

>>> re.sub(r'.*AAA(.*)ZZZ.*', r'1', 'gfgfdAAA1234ZZZuijjk')

In basic sed, capturing group are represented by (..), but in python it was represented by (..).

Answered By: Avinash Raj

You can find first substring with this function in your code (by character index). Also, you can find what is after a substring.

def FindSubString(strText, strSubString, Offset=None):
        Start = strText.find(strSubString)
        if Start == -1:
            return -1 # Not Found
            if Offset == None:
                Result = strText[Start+len(strSubString):]
            elif Offset == 0:
                return Start
                AfterSubString = Start+len(strSubString)
                Result = strText[AfterSubString:AfterSubString + int(Offset)]
            return Result
        return -1

# Example:

Text = "Thanks for contributing an answer to Stack Overflow!"
subText = "to"

print("Start of first substring in a text:")
start = FindSubString(Text, subText, 0)
print(start); print("")

print("Exact substring in a text:")
print(Text[start:start+len(subText)]); print("")

print("What is after substring "%s"?" %(subText))
print(FindSubString(Text, subText))

# Your answer:

Text = "gfgfdAAA1234ZZZuijjk"
subText1 = "AAA"
subText2 = "ZZZ"

AfterText1 = FindSubString(Text, subText1, 0) + len(subText1)
BeforText2 = FindSubString(Text, subText2, 0) 

print("nYour answer:n%s" %(Text[AfterText1:BeforText2]))

One liners that return other string if there was no match.
Edit: improved version uses next function, replace "not-found" with something else if needed:

import re
res = next( ( for m in ["AAA(.*?)ZZZ", "gfgfdAAA1234ZZZuijjk" ),] if m), "not-found" )

My other method to do this, less optimal, uses regex 2nd time, still didn’t found a shorter way:

import re
res = ( ("AAA(.*?)ZZZ", "gfgfdAAA1234ZZZuijjk") or"()","") ).group(1) )
Answered By: MaxLZ

you can do using just one line of code

>>> import re

>>> re.findall(r'd{1,5}','gfgfdAAA1234ZZZuijjk')

>>> ['1234']

result will receive list…

Answered By: Mahesh Gupta

In python, extracting substring form string can be done using findall method in regular expression (re) module.

>>> import re
>>> s = 'gfgfdAAA1234ZZZuijjk'
>>> ss = re.findall('AAA(.+)ZZZ', s)
>>> print ss
Answered By: rashok

Surprised that nobody has mentioned this which is my quick version for one-off scripts:

>>> x = 'gfgfdAAA1234ZZZuijjk'
>>> x.split('AAA')[1].split('ZZZ')[0]
Answered By: Uncle Long Hair

Here’s a solution without regex that also accounts for scenarios where the first substring contains the second substring. This function will only find a substring if the second marker is after the first marker.

def find_substring(string, start, end):
    len_until_end_of_first_match = string.find(start) + len(start)
    after_start = string[len_until_end_of_first_match:]
    return string[string.find(start) + len(start):len_until_end_of_first_match + after_start.find(end)]
Answered By: Foobar
text = 'I want to find a string between two substrings'
left = 'find a '
right = 'between two'



Answered By: Fernando Wittmann

Another way of doing it is using lists (supposing the substring you are looking for is made of numbers, only) :

string = 'gfgfdAAA1234ZZZuijjk'
numbersList = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
output = []

for char in string:
    if char in numbersList: output.append(char)

print(f"output: {''.join(output)}")
### output: 1234
Answered By: Julio S.

Using PyParsing

import pyparsing as pp

word = pp.Word(pp.alphanums)

s = 'gfgfdAAA1234ZZZuijjk'
rule = pp.nestedExpr('AAA', 'ZZZ')
for match in rule.searchString(s):

which yields:


Answered By: Raphael

Typescript. Gets string in between two other strings.

Searches shortest string between prefixes and postfixes

prefixes – string / array of strings / null (means search from the start).

postfixes – string / array of strings / null (means search until the end).

public getStringInBetween(str: string, prefixes: string | string[] | null,
                          postfixes: string | string[] | null): string {

    if (typeof prefixes === 'string') {
        prefixes = [prefixes];

    if (typeof postfixes === 'string') {
        postfixes = [postfixes];

    if (!str || str.length < 1) {
        throw new Error(str + ' should contain ' + prefixes);

    let start = prefixes === null ? { pos: 0, sub: '' } : this.indexOf(str, prefixes);
    const end = postfixes === null ? { pos: str.length, sub: '' } : this.indexOf(str, postfixes, start.pos + start.sub.length);

    let value = str.substring(start.pos + start.sub.length, end.pos);
    if (!value || value.length < 1) {
        throw new Error(str + ' should contain string in between ' + prefixes + ' and ' + postfixes);

    while (true) {
        try {
            start = this.indexOf(value, prefixes);
        } catch (e) {
        value = value.substring(start.pos + start.sub.length);
        if (!value || value.length < 1) {
            throw new Error(str + ' should contain string in between ' + prefixes + ' and ' + postfixes);

    return value;
Answered By: Sergey Gurin

One liner with Python 3.8 if text is guaranteed to contain the substring:

Answered By: cookiemonster

also, you can find all combinations in the bellow function

s = 'Part 1. Part 2. Part 3 then more text'
def find_all_places(text,word):
    word_places = []
    while True:
        word_place = text.find(word,i)
        if i>=len(text):
        if word_place<0:
    return word_places
def find_all_combination(text,start,end):
    start_places = find_all_places(text,start)
    end_places = find_all_places(text,end)
    combination_list = []
    for start_place in start_places:
        for end_place in end_places:
            if start_place>=end_place:
    return combination_list


['Part 1. ', 'Part 1. Part 2. ', 'Part 2. ']
Answered By: yunus

In case you want to look for multiple occurences.

content ="Prefix_helloworld_Suffix_stuff_Prefix_42_Suffix_andsoon"
strings = []
for c in content.split('Prefix_'):
    spos = c.find('_Suffix')
    if spos!=-1:
        strings.append( c[:spos])
print( strings )

Or more quickly :

strings = [ c[:c.find('_Suffix')] for c in content.split('Prefix_') if c.find('_Suffix')!=-1 ]
Answered By: Adrien Mau
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