I have a string like this:
string = r'''<img height="233" src="monline/" title="email example" width="500" ..
title="second example title" width="600"...
title="one more title"...> '''
I am trying to get anything that appears as title (title="Anything here") I have already tried this but it does not work correctly.
re.findall(r'title=\"(.*)\"',string)
I think your Regex is too Greedy. You can try something like this
re.findall(r'title=\"(?P<title>[\w\s]+)\"', string)
As @Austin and @Plato77 said in the comments, there is a better way to parse HTML in python. See other SO Answers for more context. There are a few common tools for this like:
If you would like to read more on performance testing of different python HTML parsers you can learn more here
As @Austin and @Plato77 said in the comments, there is a better way to parse HTML in python. I stand by this too, but if you want to get it done through regex this may help
c = re.finditer(r'title=[\"]([a-zA-Z0-9\s]+)[\" ]', string)
for i in c:
print(i.group(1))
The problem here is that the next " symbol is parsed as a character and is considered part of the (.*) of your RE. For your usecase, you can use only letters and numbers.
