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Regex to add quotes around hyphenated strings

I want to add quotes around all hyphenated words in a string.
With an example string, the desired function add_quotes() should perform like this:
>>> s = '{name = first-name}'
>>> add_quotes(s)
{name = "first-name"}
I know how to find all occurances of hyphenated works using this Regex selector, but don't know how to add quotes around each of those occurances in the original string.
>>> import re
>>> s = '{name = first-name}'
>>> re.findall(r'\w+(?:-\w+)+', s)
['first-name']

Regex can be used to do this with Python Module re from the standard library.
import re
def add_quotes(s):
return re.sub(r'\w+(?:-\w+)+', r'"\g<0>"', s)
s = '{name = first-name}'
add_quotes(s) # returns '{name = "first-name"}'
where the occurances of hyphenated words are found using this selector.

Getting word from string

How can i get word example from such string:
str = "http://test-example:123/wd/hub"
I write something like that
print(str[10:str.rfind(':')])
but it doesn't work right, if string will be like
"http://tests-example:123/wd/hub"

You can use this regex to capture the value preceded by - and followed by : using lookarounds
(?<=-).+(?=:)
Regex Demo
Python code,
import re
str = "http://test-example:123/wd/hub"
print(re.search(r'(?<=-).+(?=:)', str).group())
Outputs,
example
Non-regex way to get the same is using these two splits,
str = "http://test-example:123/wd/hub"
print(str.split(':')[1].split('-')[1])
Prints,
example

You can use following non-regex because you know example is a 7 letter word:
s.split('-')[1][:7]
For any arbitrary word, that would change to:
s.split('-')[1].split(':')[0]

many ways
using splitting:
example_str = str.split('-')[-1].split(':')[0]
This is fragile, and could break if there are more hyphens or colons in the string.
using regex:
import re
pattern = re.compile(r'-(.*):')
example_str = pattern.search(str).group(1)
This still expects a particular format, but is more easily adaptable (if you know how to write regexes).

I am not sure why do you want to get a particular word from a string. I guess you wanted to see if this word is available in given string.
if that is the case, below code can be used.
import re
str1 = "http://tests-example:123/wd/hub"
matched = re.findall('example',str1)

Split on the -, and then on :
s = "http://test-example:123/wd/hub"
print(s.split('-')[1].split(':')[0])
#example

using re
import re
text = "http://test-example:123/wd/hub"
m = re.search('(?<=-).+(?=:)', text)
if m:
print(m.group())

Python strings has built-in function find:
a="http://test-example:123/wd/hub"
b="http://test-exaaaample:123/wd/hub"
print(a.find('example'))
print(b.find('example'))
will return:
12
-1
It is the index of found substring. If it equals to -1, the substring is not found in string. You can also use in keyword:
'example' in 'http://test-example:123/wd/hub'
True

regex to extract data between quotes

As title says string is '="24digit number"' and I want to extract number between "" (example: ="000021484123647598423458" should get me '000021484123647598423458').
There are answers that answer how to get data between " but in my case I also need to confirm that =" exist without capturing (there are also other "\d{24}" strings, but they are for other stuff) it.
I couldn't modify these answers to get what I need.
My latest regex was ((?<=\")\d{24}(?=\")) and string is ="000021484123647598423458".
UPDATE: I think I will settle with pattern r'^(?:\=\")(\d{24})(?:\")' because I just want to capture digit characters.
word = '="000021484123647598423458"'
pattern = r'^(?:\=\")(\d{24})(?:\")'
match = re.findall(pattern, word)[0]
Thank you all for suggestions.

You could have it like:
=(['"])(\d{24})\1
See a demo on regex101.com.
In Python:
import re
string = '="000021484123647598423458"'
rx = re.compile(r'''=(['"])(\d{24})\1''')
print(rx.search(string).group(2))
# 000021484123647598423458

Any one of the following works:
>>> st = '="000021484123647598423458"'
>>> import re
>>> re.findall(r'".*\d+.*"',st)
['"000021484123647598423458"']
or
>>> re.findall(r'".*\d{24}.*"',st)
['"000021484123647598423458"']
or
>>> re.findall(r'"\d{24}"',st)
['"000021484123647598423458"']

python String Pattern Matching

two string expression： #RequestMapping(value = "/list/base/info") or #RequestMapping("/list/base/info")
How do I get uri /list/base/info value by String Pattern Matching?

In this case you can try getting it by split:
expression.split("\"")[1]

import re
re.match(
'#RequestMapping\((value\s*=\s*)?"([^"]+)"\)',
'#RequestMapping(value = "/list/base/info")'
).group(2)
That would output:
'/list/base/info'

Regex to retrieve the last few characters of a string

Regex to retrieve the last portion of a string:
https://play.google.com/store/apps/details?id=com.lima.doodlejump
I'm looking to retrieve the string followed by id=
The following regex didn't seem to work in python
sampleURL = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
re.search("id=(.*?)", sampleURL).group(1)
The above should give me an output:
com.lima.doodlejump
Is my search group right?

Your regular expression
(.*?)
will not work because, it will match between zero and unlimited times, as few times as possible (becasue of the ?). So, you have the following choices of RegEx
(.*) # Matches the rest of the string
(.*?)$ # Matches till the end of the string
But, you don't need RegEx at all here, simply split the string like this
data = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
print data.split("id=", 1)[-1]
Output
com.lima.doodlejump
If you really have to use RegEx, you can do like this
data = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
import re
print re.search("id=(.*)", data).group(1)
Output
com.lima.doodlejump

I'm surprised that nobody has mentioned urlparse yet...
>>> s = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
>>> urlparse.urlparse(s)
ParseResult(scheme='https', netloc='play.google.com', path='/store/apps/details', params='', query='id=com.lima.doodlejump', fragment='')
>>> urlparse.parse_qs(urlparse.urlparse(s).query)
{'id': ['com.lima.doodlejump']}
>>> urlparse.parse_qs(urlparse.urlparse(s).query)['id']
['com.lima.doodlejump']
>>> urlparse.parse_qs(urlparse.urlparse(s).query)['id'][0]
'com.lima.doodlejump'
The HUGE advantage here is that if the url query string gets more components then it could easily break the other solutions which rely on a simple str.split. It won't confuse urlparse however :).

Just split it in the place you want:
id = url.split('id=')[1]
If you print id, you'll get:
com.lima.doodlejump
Regex isn't needed here :)
However, in case there are multiple id=s in your string, and you only wanted the last one:
id = url.split('id=')[-1]
Hope this helps!

This works:
>>> import re
>>> sampleURL = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
>>> re.search("id=(.+)", sampleURL).group(1)
'com.lima.doodlejump'
>>>
Instead of capturing non-greedily for zero or more characters, this code captures greedily for one or more.