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

Question

A code in python using regex that can perform something like this

Input: Regex should return \"String 1\" or \"String 2\" or \"String3\" 
Output: String 1,St         


        

Answer 1

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"

Answer 2

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

import re

str=""" 
"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'["](.*?)["]',str))

Answer 3

import re
r=r"'(\\'|[^'])*(?!<\\)'|\"(\\\"|[^\"])*(?!<\\)\""

texts=[r'"aerrrt"',
r'"a\"e'+"'"+'rrt"',
r'"a""""arrtt"""""',
r'"aerrrt',
r'"a\"errt'+"'",
r"'aerrrt'",
r"'a\'e"+'"'+"rrt'",
r"'a''''arrtt'''''",
r"'aerrrt",
r"'a\'errt"+'"',
      "''",'""',""]

for text in texts:
     print (text,"-->",re.fullmatch(r,text))

results:

"aerrrt" --> <_sre.SRE_Match object; span=(0, 8), match='"aerrrt"'>
"a\"e'rrt" --> <_sre.SRE_Match object; span=(0, 10), match='"a\\"e\'rrt"'>
"a""""arrtt""""" --> None
"aerrrt --> None
"a\"errt' --> None
'aerrrt' --> <_sre.SRE_Match object; span=(0, 8), match="'aerrrt'">
'a\'e"rrt' --> <_sre.SRE_Match object; span=(0, 10), match='\'a\\\'e"rrt\''>
'a''''arrtt''''' --> None
'aerrrt --> None
'a\'errt" --> None
'' --> <_sre.SRE_Match object; span=(0, 2), match="''">
"" --> <_sre.SRE_Match object; span=(0, 2), match='""'>
 --> None

Answer 4

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: (I nearly quote)

.+? 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 .+? 'String 1,String 2,String3'

In addition (Johan speaking again), if you want to accept empty strings, change .+ to .*. Star means zero or more - plus means at least one.