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Invalid email addresses defined

bspink
Explorer
0 Likes
1,512

2021-06-01-17-44-54.pngHi Everyone,

I am trying to send a broadcast message to all Learners and am being met with the below message. I have checked the user emails in several ways including filtering the users by "Wrong Email" and also exporting the user list to Excel to review using formulas.

I found only one email address with an extra space at the end whilst using the filter and corrected it, but still get the same message.

Does anyone know another way to find the invalid emails?

Thanks,

Byron

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Answers (1)

Answers (1)

DirkManuel
Active Contributor

Did you check the Reply Email Address in the Server Settings, in case it is objecting to the from email and not the to email? Are any other automated emails going through OK? Or can you send a broadcast message to a single user you KNOW is correct (like yourself).If not, it may be down to the SMTP server.

bspink
Explorer
0 Likes

Hi Dirk,

Thanks for responding.

Yes I have been able to send myself a message and also the Standard Authors. It's one or some email(s) in the learner population that have the issue.

The reply email address has been left blank so it the hana.ondemand sender.

Thanks,

Byron

DirkManuel
Active Contributor
0 Likes

Hmmm, yeah, then it has to be one of those email addresses. You said you checked them using a formula in Excel. Not doubting your capabilities there, but maybe check the formula again - what exactly are you checking for? Do you have the ability to run them through a Regular Expression, as that may be a bit tighter in that you can specify only A-Z,0-9, etc. - just in case there is some invalid character in there. (I saw something recently on spam where they were using a character that looked like a regular alphanumeric but was really a Cyrillic character or something...just a thought.) [a-zA-Z0-9+_.-]+@[a-zA-Z0-9.-]+$ would work.

bspink
Explorer
0 Likes

Another great shout Dirk. I'm merely a semi-competent Excel user so the formula was straight forward too. I was simply checking for the "@" in the right place and a "." after it. The formula I was using was "=ISNUMBER(MATCH("*@*.?*",A2,0))" borrowed from a forum. I do not have the ability to use regular expressions but will investigate and hopefully learn.

I will post back here if I find the answer.

Any other helpful tips are gratefully received.