How to replace next word after a specific word in python

Question

I have a string like below. Here, I want to replace just next immediate word after a particular word like, '%(db_user)s' and '%(db_passsword)s' but the words I can search for in the string are --db-user and --db-passwords becuase the above will be substituted by values.

Input:

 "cd scripts &&   bash setup.sh  --client-name %(client_name)s --is-db-auth-enabled %(is_db_auth_enabled)s --db-user '%(db_user)s' --db-password '%(db_password)s' "

Output:

 "cd scripts &&   bash setup.sh  --client-name %(client_name)s --is-db-auth-enabled %(is_db_auth_enabled)s --db-user '***' --db-password '****' "

So please help me with a function where I will provide an array of words and a string, which will replace next words to those supplied words.

Answer 1

This will help -

import re

def char_index(sentence, word_index): 
    sentence = re.split('(\s)',sentence) #Parentheses keep split characters 
    return len(''.join(sentence[:word_index*2]))

def print_secure_message(msg):
    secure_words = ['--db-user', '--db-password']
    # Removing extra white spaces within string
    msg = re.sub(' +', ' ', msg)
    cpy_msg = msg.split(" ")
    for word in secure_words:
        # Getting index of the word's first characters
        t = re.search(word, msg)
        # Getting index of the next word of the searched word's
        word_index = cpy_msg.index(word)+2;
        index= char_index(msg, word_index)
        print(t.end(), word_index, index)
        msg = msg[0:t.end() + 1] + "'****'" + msg[index - 1:]
    print(''.join(msg))