please i need help i tried using the re mod to print the text in the var excluding the title but my code printed an empty list

Question

import re
word = """
title="hello how are you doing"
"""
find_word = re.findall(r"title\=([a-z])", str(word))

print(find_word)

Answer 1

Ruba Kh It will work as long as there are no more double quotes after the matching pattern in the docstring. For example, if the docstring were

word = """
title="hello how are you doing"
subject="something"
"""

the whole substring from title to the final " is matched.

Answer 2

Ruba Kh You could use the lazy quantifier instead though - *? (just thought of that) to make

find_word = re.findall(r"title\=\"(.*?)\"", str(word))

Answer 3

The string says 'title="hello...' whereas the regex pattern misses out the double quotation marks after the '='.

Change to
find_word = re.findall(r"title\=\"([a-z])", str(word)) # ['h']

Answer 4

Ruba Kh 's answer will certainly work for your example where all characters are either a lower case letter or a space, but you may want to capture more complicated strings than those, in which there is a simple solution. After the 'title="', you want to match any character that is not a '"'. This is done easily using the ^ metacharacter which matches anything except the ones you include in the character class.

find_word = re.findall(r"title\=\"([^\"]*)", str(word))

Answer 5

Victorr Ruba Kh 's answer worked fine, you probably missed the space in the brackets [a-z ].

Answer 6

Maybe you want something like that

find_word = re.findall(r"title\=\"([a-z ]*)\"", str(word))

Answer 7

You are absolutely right I was just adjusting on the example I did not try to generalize the answer

Answer 8

Russ, Thanks so much sir i really appreciate you're d best!!!

Answer 9

Ruba Kh sir yours worked but it only printed hello thanks though really appreciate you guys for the help.

Answer 10

Sir it worked buh it still got a problem
it only printed the first letter "h"
it didn't print the whole sentence "hello how are you doing"

Answer 11

It can be done also like that right?
Russ

find_word = re.findall(r"title\=\"(.*)\"", str(word))

Answer 12

Russ you are right sir it worked when i added the space THANKS

Answer 13

Russ Ah thanks alot now I understand