Python Regex to find a string in double quotes within a string

Here's all you need to do:

def doit(text):      
  import re
  matches = re.findall(r'"(.+?)"',text)
  # matches is now ['String 1', 'String 2', 'String3']
  return ",".join(matches)

doit('Regex should return "String 1" or "String 2" or "String3" ')

result:

'String 1,String 2,String3'

As pointed out by Li-aung Yip:

To elaborate, .+? is the "non-greedy" version of .+. It makes the regular expression match the smallest number of characters it can instead of the most characters it can. The greedy version, .+, will give String 1" or "String 2" or "String 3; the non-greedy version .+? gives String 1, String 2, String 3.

In addition, if you want to accept empty strings, change .+ to .*. Star * means zero or more while plus + means at least one.

Just try to fetch double quoted strings from the multiline string:

import re

s = """ 
"my name is daniel"  "mobile 8531111453733"[[[[[[--"i like pandas"
"location chennai"! -asfas"aadhaar du2mmy8969769##69869" 
@4343453 "pincode 642002""@mango,@apple,@berry" 
"""
print(re.findall(r'"(.*?)"', s))

The highly up-voted answer doesn't account for the possibility that the double-quoted string might contain one or more double-quote characters (properly escaped, of course). To handle this situation, the regex needs to accumulate characters one-by-one with a positive lookahead assertion stating that the current character is not a double-quote character that is not preceded by a backslash (which requires a negative lookbehind assertion):

"(?:(?:(?!(?<!\\)").)*)"

See Regex Demo

import re
import ast


def doit(text):
    matches=re.findall(r'"(?:(?:(?!(?<!\\)").)*)"',text)
    for match in matches:
        print(match, '=>', ast.literal_eval(match))


doit('Regex should return "String 1" or "String 2" or "String3" and "\\"double quoted string\\"" ')

Prints:

"String 1" => String 1
"String 2" => String 2
"String3" => String3
"\"double quoted string\"" => "double quoted string"