Compatible Support Forums: How do I find if what version of Win2k I have?

Jump to content

Page 1 of 1
  • You cannot start a new topic
  • You cannot reply to this topic

How do I find if what version of Win2k I have?

#1 User is offline   RazoR 

  • newbie
  • Group: Members
  • Posts: 7
  • Joined: 12-January 00

Posted 12 January 2000 - 06:54 AM

Hi,
I just want to know what version of windows 2000 I have, such as final retail, beta, rc(What are RC's?)
My System panel says: Microsoft Windows 2000 5.00.2195

Is this the retail final version?
0

#2 User is offline   Jerry - 

  • member
  • Group: Members
  • Posts: 114
  • Joined: 03-November 99

Posted 12 January 2000 - 10:28 AM

RC's = Release Candidates...
In the command prompt, type winver. If there is no expiration date listed you have the final release code.
0

#3 User is offline   RazoR 

  • newbie
  • Group: Members
  • Posts: 7
  • Joined: 12-January 00

Posted 13 January 2000 - 05:47 AM

Ok cool, It doesn't say an expiration date, if anyone has the beta, can you goto command prompt and type WINVER and take a screenshot of it?
0

#4 User is offline   RazoR 

  • newbie
  • Group: Members
  • Posts: 7
  • Joined: 12-January 00

Posted 13 January 2000 - 05:51 AM

Btw, can someone goto safemode with the retail? Mine says Build 2195 FREE, what does that mean?
0

#5 User is offline   Seldzar 

  • addict
  • Group: Members
  • Posts: 468
  • Joined: 09-September 99

Posted 13 January 2000 - 07:59 AM

It means it's the 120 day trial version.
0

#6 User is offline   RazoR 

  • newbie
  • Group: Members
  • Posts: 7
  • Joined: 12-January 00

Posted 13 January 2000 - 11:53 PM

But when I type winver in the run command it doesnt say and experation day, like Jerry said there would be if you had a trial.
And I had to enter a CD-Key when I first installed it.

[This message has been edited by RazoR (edited 13 January 2000).]
0

Share this topic:


Page 1 of 1
  • You cannot start a new topic
  • You cannot reply to this topic

1 User(s) are reading this topic
0 members, 1 guests, 0 anonymous users