How to match to capture group 1 with regex

Question

My goal is to capture the date from the following string:

Reviewed 2 days ago <a class="viaMobile" href="/apps" target="_blank" onclick="ta.util.cookie.setPIDCookie(24487)">via mobile </a>

To do this I’m using the regex:title="(*?)"
Which returns Match (group 0): title="November 9, 2017" Group 1: November 9, 2017

I need my match returned by regex to be just the date, what is currently group 1. Is there a simple way to do this? I am new to regex but I could find direction on this online.

Note: I’m not writing regex for the structure of a date because some strings have multiple dates and I only want the date in title. Thanks!

Asked By: hackerman

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Source

Answer 1

You can use re.findall:

import re
s = """
  <span class="ui_bubble_rating bubble_50"></span><span class="ratingDate relativeDate" title="November 9, 2017">Reviewed 2 days ago </span><a class="viaMobile" href="/apps" target="_blank" onclick="ta.util.cookie.setPIDCookie(24487)"><span class="ui_icon mobile-phone"></span>via mobile </a>
  """
date = re.findall('title="(.*?)"', s)[0]

Output:

'November 9, 2017'

Answered By: Ajax1234

Answer 2

You can use negative lookahead / lookbehind instead of capture groups

(?<=title=").+?(?=")

This will ensure it starts with title, without actually selecting it

Answered By: user184994

Answer 3

You can also use

title="K.*?(?=")

This will look for the value between title=" and "

Answered By: Sanskar

How to match to capture group 1 with regex

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