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LoneWolf121188
Aug 9, 2007, 01:54 PM
I've installed SSH and the binkit and all that, but now I need a little help. I followed the directions here (http://www.uninnovate.com/2007/08/01/myphone-share-your-iphones-music-collection-via-wifi/) and it asked me to put MyPhone.py in /usr/local/bin. I didn't have a /local under bin, so I created it, created /bin, and put MyPhone.py in there. Now its asking me to "Log into your iPhone via ssh and run MyPhone with this command: nohup python /usr/local/bin/MyPhone.py&". I don't understand what this is saying. Where do I put in this command? iphoneinterface doesn't recognize it.

FYI: iPhone version 1.01, Win XP, SSH client WinSCP.

Thanks!



The General
Aug 9, 2007, 04:13 PM
Login with SSH and run that command.

Where does it say to use iPhoneInterface to do that? :)

Also, PuTTY is what I'd recommend instead of WinSCP.

LoneWolf121188
Aug 10, 2007, 01:42 PM
I just tried running that command from the Terminal and I get this: "46
-sh: /usr/bin/nohup: Permission denied". Haven't tried it in XP yet. Help?

Yes, I ran jailbreak (well, I used iFuntastic 2.5, but same thing).

In XP using WinSCP, I tried nohup python /usr/local/bin/MyPhone.py& and I got a message saying "Connection has been unexpectedly closed. Server sent command exit status 2." Putty gave me "53 -sh: /usr/bin/nohup: Permission denied".