CODEWITHSUNDEEP
pythonregexpython-2.7non-greedy
PIMg021
PIMg021

Reputation: 83

REGEX extracting specific part non greedy

I'm new to Python 2.7. Using regular expressions, I'm trying to extract from a text file just the emails from input lines. I am using the non-greedy method as the emails are repeated 2 times in the same line. Here is my code:

import re
f_hand = open('mail.txt')
for line in f_hand:
    line.rstrip()
    if re.findall('\S+@\S+?',line): print re.findall('\S+@\S+?',line)

however this is what i"m getting instead of just the email address:

['href="mailto:[email protected]">sercetary@a']

What shall I use in re.findall to get just the email out?

Upvotes: 0

Views: 259

Answers (5)

Glen Ragan
Glen Ragan

Reputation: 41

\S accepts many characters that aren't valid in an e-mail address. Try a regular expression of

[a-zA-Z0-9-_.]+@[a-zA-Z0-9-_.]+\\.[a-zA-Z0-9-_.]+

(presuming you are not trying to support Unicode -- it seems that you aren't since your input is a "text file").

This will require a "." in the server portion of the e-mail address, and your match will stop on the first character that is not valid within the e-mail address.

Upvotes: 1

work.bin
work.bin

Reputation: 1108

This is the format of an email address - https://www.rfc-editor.org/rfc/rfc5322#section-3.4.1.

Keeping that in mind the regex that you need is - r"([a-zA-Z0-9_.+-]+@[a-zA-Z0-9-]+\.[a-zA-Z0-9-.]+)". (This works without having to depend on the text surrounding an email address.)

The following lines of code -

html_str = r'<a href="mailto:[email protected]">[email protected]</a>'
email_regex = r"([a-zA-Z0-9_.+-]+@[a-zA-Z0-9-]+\.[a-zA-Z0-9-.]+)"
print re.findall(email_regex, html_str)

yields -

['[email protected]', '[email protected]']

P.S. - I got the regex for email addresses by googling for "email address regex" and clicking on the first site - http://emailregex.com/

Upvotes: 0

saurabh baid
saurabh baid

Reputation: 1877

try this re.findall('mailto:(\S+@\S+?\.\S+)\"',str))

It should give you something like ['[email protected]']

Upvotes: 1

Casimir et Hippolyte
Casimir et Hippolyte

Reputation: 89557

If you parse a simple file with anchors for email addresses and always the same syntax (like double quotes to enclose attributes), you can use:

for line in f_hand: 
    print re.findall(r'href="mailto:([^"@]+@[^"]+)">\1</a>', line)

(re.findall returns only the capture group. \1 stands for the content of the first capture group.)

If the file is a more complicated html file, use a parser, extract the links and filter them.
Or eventually use XPath, something like:
substring-after(//a/@href[starts-with(., "mailto:")], "mailto:")

Upvotes: 1

Laurel
Laurel

Reputation: 6173

\S means not a space. " and > are not spaces.

You should use mailto:([^@]+@[^"]+) as the regex (quoted form: 'mailto:([^@]+@[^"]+)'). This will put the email address in the first capture group.

Upvotes: 1

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