Meta Stack Exchange is where users like you discuss bugs, features, and support issues that affect the software powering all 158 Stack Exchange communities.

What is meta?
Here's how it works:
  1. Any Stack Exchange user can ask a question
  2. The community provides support, votes on ideas, and reports bugs
  3. Your voice helps shape the way Stack Exchange operates

How do I ask a question on metastackoverflow without an openid?

It says, "you must type your name or openid" and where do I type a nickname if I have no openid?

On Stack Overflow, at least, it asks for "openid or name email homepage", but here I see only an openid textbox.

share|improve this question

migrated from stackoverflow.com Dec 3 '09 at 17:06

This question came from our site for professional and enthusiast programmers.

    
Belons on meta . . . sorry, I couldn't resist :) – Binary Worrier Dec 3 '09 at 17:05
    
See my bug report related to this as well... it's really really similar, but not quite a dupe. – Pops Sep 24 '10 at 4:32

You can't.

share|improve this answer
    
Are those... invisible emoticons? – mmyers Dec 3 '09 at 17:08
    
They're advanced space fillers. – JSONBog Dec 3 '09 at 17:09
    
That's what Jeff said, but your post generates a lot more XML tags, which requires more CPU cycles, and thus takes longer to display and consumes more power. You're too frippery. <em><code> </code></em> You can't. <em><code> </code></em> – Adam Davis Dec 3 '09 at 17:52
4  
Welbog, stop stealing our CPU cycles, you magnificent bastard! – Jeff Atwood Dec 3 '09 at 18:29

Meta requires OpenID?

Yes.

How do I ask a question on metastackoverflow without an openid?

  • Use a throw away openid account - they are trivially easy to get from lots of providers. (best solution)
  • Ask on SO and have the question migrated (frowned upon).
  • Get one of the tens of thousands of SO users to install a keylogger, and use their openid (possibly illegal).
share|improve this answer
1  
+1 both helpful and funny – C. Ross Dec 3 '09 at 18:25

Meta requires an openid; anonymous posting is disabled here.

share|improve this answer
    
That's what I said, but you used a lot more letters. You're too verbose, Jeff. – JSONBog Dec 3 '09 at 17:16
14  
The funny thing is that he actually managed it to ask an anonymous question here ;) So the correct answer is: ask on SO and get migrated. – tanascius Dec 3 '09 at 17:24
2  
lol, you're right! I hadn't considered that "loophole"... – Jeff Atwood Dec 3 '09 at 18:28

If Meta requires OpenID, then FIX YOUR STUFF!

1) Attempting to ask or answer a question here without being logged in produces the following error message:

"•name and email, or your OpenID, are missing"

That implies that a name and email address would be sufficient.

2) Click around enough (the "register" and "log in" links are confusing and the whole thing needs re-thinking, so I can't remember for certain, but I think it was on the "forgot your account information" page, which I can't find again at this moment), and you'll be informed, and I quote (yes, I copied the text when I had that page open in my browser):

"...does not require an email address to participate..."

I guess that's an ambiguous exerpt, but the text on that page led me to believe I didn't need an OpenID associated with an email address.

I'm a geek too, but I prefer NOT to use OpenID, and the hassle of using SO without one is so great, that I try to avoid SO for the most part, which is a bummer, because everything else you've done is AWESOME

share|improve this answer
    
so use Facebook login, that's OAuth 2.0 instead. Congrats, you're not using OpenID! :) – Jeff Atwood Sep 24 '10 at 4:31
    
Let's not get started on Facebook. ;-) Point is, SO Meta's own error text and other information give the exact opposite information - that OpenID is NOT required to use Meta. It'd be a lot more clear if this were in the FAQ or on the front page - I had to read a few posts all the way through before I could figure it out, and I have more PATIENCE. – user151637 Sep 24 '10 at 4:41

You must log in to answer this question.