How do I catch "split" exceptions in python?

Question

I am trying to parse a list of email addresses to remove the username and '@' symbol only leaving the domain name.

Example: [email protected] Desired output: gmail.com

I have accomplished this with the following code:

for row in cr: emailaddy = row[0] (emailuser, domain) = row[0].split('@') print domain

but my issue is when I encounter a improperly formatted email address. For example if the row contains "aaaaaaaaa" (instead of a valid email address) the program crashes with the error

(emailuser, domain) = row[0].split('@') ValueError: need more than 1 value to unpack.

(as you would expect) Rather than check all the email addresses for their validity, I would rather just not update grab the domain and move on to the next record. How can I properly handle this error and just move on?

So for the list of:

[email protected] [email protected] youououou [email protected]

I would like the output to be:

gmail.com hotmail.com yahoo.com

Thanks!

Just to note, about the only safe way to validate an email is to send an email to the address asking the account holder to verify that they've received it. — Open AI - Opting Out
– Open AI - Opting Out, Commented Feb 28, 2012 at 10:48
Note that several @'s are perfectly fine by the email spec. So it's much more sensible to do rfind("@") or something.. — Voo
– Voo, Commented Feb 28, 2012 at 16:36

Felix Yan · Accepted Answer · 2012-02-28 05:56:18Z

28

You want something like this?

try: (emailuser, domain) = row[0].split('@') except ValueError: continue

answered Feb 28, 2012 at 5:56

Felix Yan

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Comments

RanRag · Accepted Answer · 2012-02-28 05:58:00Z

You can just filter out the address which does not contain @.

>>> [mail.split('@')[1] for mail in mylist if '@' in mail] ['gmail.com', 'hotmail.com', 'yahoo.com'] >>>

glglgl · Accepted Answer · 2012-02-28 05:59:01Z

What about

splitaddr = row[0].split('@') if len(splitaddr) == 2: domain = splitaddr[1] else: domain = ''

This even handles cases like aaa@bbb@ccc and makes it invalid ('').

NPatel · Accepted Answer · 2012-02-28 05:59:09Z

Try this

In [28]: b = ['[email protected]', '[email protected]', 'youououou', '[email protected]'] In [29]: [x.split('@')[1] for x in b if '@' in x] Out[29]: ['gmail.com', 'hotmail.com', 'yahoo.com']

the wolf · Accepted Answer · 2012-02-28 16:31:29Z

This does what you want:

import re l=["[email protected]","[email protected]", "youououou","[email protected]","amy@[email protected]"] for e in l: if '@' in e: l2=e.split('@') print l2[-1] else: print

Output:

gmail.com hotmail.com yahoo.com youso.com

It handles the case where an email might have more than one '@' and just takes the RH of that.

tobyodavies · Accepted Answer · 2012-02-28 05:58:37Z

1

if '@' in row[0]: user, domain = row[0].split('@') print domain

answered Feb 28, 2012 at 5:58

tobyodavies

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Comments

Ture Friese · Accepted Answer · 2021-05-27 19:12:31Z

Maybe the best solution is to avoid exception handling all together. You can do this by using the builtin function partition(). It is similar to split() but does not raise ValueError when the seperator is not found.
Read more:
https://docs.python.org/3/library/stdtypes.html#str.partition

Walk · Accepted Answer · 2019-03-19 06:18:28Z

We can consider the string not having '@' symbol, as a simple username:

try: (emailuser, domain) = row[0].split('@') print "Email User" + emailuser print "Email Domain" + domain except ValueError: emailuser = row[0] print "Email User Only" + emailuser O/P: Email User : abc Email Domain : gmail.com Email User : xyz Email Domain : gmail.com Email User Only : usernameonly

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