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v. v. urgent.....ravindra/anthony please clarify chat query

i changed the code as told by ravindra
and opend the applet by java web server as told by anthony
previosly the error was
com.ms.security.SecurityExceptionEx[newguiapp.makeconnection]: cannot access "127.0.0.1":1032
and now the error is
com.ms.security.SecurityExceptionEx[newguiapp.makeconnection]: unable to
check numeric address "12.0.0.1"

as i told.. my server as well as client machine address is 132.147.2.7 so where does this 120.0.0.1 come from.

the things you suggested worked fine but now this error???
please suggest
thanks
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bhajanpreets
Asked:
bhajanpreets
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1 Solution
 
el_ouhdCommented:
I presume that both 12.0.0.1 and 120.0.0.1 should be 127.0.0.1.  This is the "loopback" address for IP requests and it is (essentially) an alias for the local machine ("localhost.")  If you really are seeing one of the first two sets of numbers, your networking configuration is probably wrong in some place.

Are you using "localhost" or some similar alias anywhere in your code?

R.
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terajivCommented:
Start Server at ur System and try to access it from another system using Internet giving the IP address... That time u will get the Reqd. Output. Now what is Happening is Ur Server as well as Client are same. So U either have to use Localhost or 127.0.0.1
If u have LAN then try it out on another system. or else ask one of ur friend to check ur server...
That will be the Best Solution...

Ur problem will be solved...

Check it out...

Rajiv
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bhajanpreetsAuthor Commented:
This question has a deletion request Pending
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bhajanpreetsAuthor Commented:
thanks to all
trrajiv was right!
in my own m/c i had to use InetAddress.getLocalHost();
else
ip address
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terajivCommented:
This question no longer is pending deletion
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terajivCommented:
So Im Claiming for the Points...

I hope u can give me Immediatly..
Thanx

Rajiv
0
 
bhajanpreetsAuthor Commented:
Answer accepted
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