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

Posted on 2000-05-07
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Last Modified: 2008-03-06
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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Question by:bhajanpreets
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Expert Comment

by:el_ouhd
ID: 2785920
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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Expert Comment

by:terajiv
ID: 2787151
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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by:bhajanpreets
ID: 2830049
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Author Comment

by:bhajanpreets
ID: 2830077
thanks to all
trrajiv was right!
in my own m/c i had to use InetAddress.getLocalHost();
else
ip address
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by:terajiv
ID: 2830131
This question no longer is pending deletion
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Accepted Solution

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terajiv earned 150 total points
ID: 2830132
So Im Claiming for the Points...

I hope u can give me Immediatly..
Thanx

Rajiv
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Author Comment

by:bhajanpreets
ID: 2930132
Answer accepted
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