Hi!
I'm trying to delegate my free subdomain (onehalf3544.strangled.net) to ns*.he.net nameservers:
raspberry-pi% dig ns onehalf3544.strangled.net @ns1.afraid.org
; <<>> DiG 9.8.1-P1 <<>> ns onehalf3544.strangled.net @ns1.afraid.org
;; global options: +cmd
;; Got answer:
;; ->>HEADER<<- opcode: QUERY, status: NOERROR, id: 25376
;; flags: qr rd; QUERY: 1, ANSWER: 0, AUTHORITY: 4, ADDITIONAL: 0
;; WARNING: recursion requested but not available
;; QUESTION SECTION:
;onehalf3544.strangled.net. IN NS
;; AUTHORITY SECTION:
onehalf3544.strangled.net. 3600 IN NS ns3.he.net.
onehalf3544.strangled.net. 3600 IN NS ns2.he.net.
onehalf3544.strangled.net. 3600 IN NS ns1.he.net.
onehalf3544.strangled.net. 3600 IN NS ns4.he.net.
;; Query time: 166 msec
;; SERVER: 2607:f0d0:1102:d5::2#53(2607:f0d0:1102:d5::2)
;; WHEN: Tue Oct 23 19:51:03 2012
;; MSG SIZE rcvd: 118
But all I get when trying to add this domain via dns.he.net web interface is this:
Zone failed validation test. ERROR: Delegation was not found. Please delegate to ns1, ns2, ns3, ns4 and ns5.he.net then retry. (strangled.net / onehalf3544.strangled.net).
Answers in this topic (http://www.tunnelbroker.net/forums/index.php?topic=1050.0) suggest that dns.he.net checks existence of whois record, which my subdomain does not have. OTOH, co.cc domains do not have(?) such records either and they work (as mentioned in that forum topic).
Ok, I've mailed dnsadmin@he.net and the problem was solved - I have successfully added my domain to the zone.
Don't know why this did not work at first, however.
Hi onehalf3544,
I´m having exactly the same problem with my domain obtained through afraid.org.
When I try to register my jumpingcrab.com domain, I get the same error :(
Any suggestion?
Quote from: caza on November 08, 2012, 04:54:11 AM
Hi onehalf3544,
I´m having exactly the same problem with my domain obtained through afraid.org.
When I try to register my jumpingcrab.com domain, I get the same error :(
Any suggestion?
Mail dnsadmin@he.net and ask. I
Quote
Don't know why this did not work at first, however.