How to store matched part of regex in python?

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The string is

\xa0\n  xxxxxx\nBirchgrove 101,Durga Saffron Square,\nKariyammana Agrahara, Bellandur,    xxxxxx@xxxxxxx.com\n'

and it is in a list called shipto and it is at index 0 of this list.

the regex I am using is

shipto_re=re.compile(r"\\n  xxxxxx\\n(.*)(.*?)    xxxxxx")

The part of the string that I want is

Birchgrove 101,Durga Saffron Square,\nKariyammana Agrahara, Bellandur,

How do I iterate through the list shipto and store the required regex match in a string variable?

string-matching python loops regex
1 Answers

Your actual text is

 
  xxxxxx
Birchgrove 101,Durga Saffron Square,
Kariyammana Agrahara, Bellandur,    xxxxxx@xxxxxxx.com

So, the regex you need may look like

\n  xxxxxx\n(.*\n.*)    xxxxxx
\n {2}xxxxxx\n(.*\n.*) {4}xxxxxx

See the regex demo.

See a Python demo below:

import re
shipto=['\xa0\n  xxxxxx\nBirchgrove 101,Durga Saffron Square,\nKariyammana Agrahara, Bellandur,    xxxxxx@xxxxxxx.com\n']

Here, I printed the variable to see the literal text:

>>> print(shipto[0])
 
  xxxxxx
Birchgrove 101,Durga Saffron Square,
Kariyammana Agrahara, Bellandur,    xxxxxx@xxxxxxx.com

Next:

match = re.search(r'\n {2}xxxxxx\n(.*\n.*) {4}xxxxxx', shipto[0])
if match:
    print(match.group(1))

Output:

Birchgrove 101,Durga Saffron Square,
Kariyammana Agrahara, Bellandur,
user_38329700
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