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Sockets and Internet Protocols
this below gives 192.168.10.2 well in which case this gives 127.0.0.1
Leonidas Savvides
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Posts: 403
posted 13 years ago
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DatagramPacket receivePacket = new DatagramPacket( data, data.length ); [b]receivePacket.getAddress()[/b] //this below gives [b]192.168.10.2[/b] well, App run localhost and this is a local Lan ip, well in which case this gives [b]127.0.0.1[/b] ??? displayMessage( "\nPacket received:" + "\nFrom host: " + receivePacket.getAddress() + "\nHost port: " + receivePacket.getPort() + "\nLength: " + receivePacket.getLength() + "\nContaining:\n\t" + new String( receivePacket.getData(), 0, receivePacket.getLength() ) );
Nitesh Kant
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posted 13 years ago
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Your question is really not clear. It will be nice if you can explain more about your problem.
apigee
, a better way to API!
Leonidas Savvides
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posted 13 years ago
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receivePacket.getAddress()
//this below gives
192.168.10.2
well, App run localhost and this is a local Lan ip, well in which case this gives
127.0.0.1
and which case give as given as tried:
192.168.10.2
? ???
Carol Enderlin
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posted 13 years ago
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I see question marks, but do not see a question.
Leonidas Savvides
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posted 13 years ago
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when we get 127.0.0.1 and when 192.168.10.2 (my PC address in LAN) ?
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