[ra.ravi.rav]

I have the follwing code, which is a simple mp3 player, I am running it on Nokia 7610 (S60 2ndEd FP1):

import appuifw, e32, audio, sys

song = None
curr_pos = 0

def play():
    s = appuifw.query(u"Path", "text")
    print s
    song = open(s)
    song.play()

def pause():
    curr_pos = song.current_position()
    song.stop

def stop():
    song.stop()

def exit():
    exit = appuifw.query(u"You want to exit?", "query")
    if exit:
        sys.exit(0)
    else:
        pass


if __name__ == '__main__':
    appuifw.app.menu = [(u"Play", play), (u"Pause", pause), (u"Stop", stop)]
    appuifw.app.exit_key_handler = exit
    appuifw.app.title = u"FooMP3"
    app_lock = e32.Ao_lock()
    app_lock.wait()

On choosing play menu, I enter the filename as "E:\\Sounds\\song.mp3" and get the error as:

Traceback (most recent call last):
  File "E:\System\Apps\Python\my\mp3player.py", line 9, in play
    song = audio.Sound.open(s)
  File "E:\system\libs\audio.py", line 44, in open
    raise SymbianError,(callback_error[0],
SymbianError: [Errno -28] Error opening file: KErrBadName

However if I try to open thr file usin normal open() call it works soomthly as:

str = appuifw.query(u"Enter file path", "text", "E:\\Sounds\\song.mp3")
f = open(str)
print dir(f)

What may be the possible reason?

[ra.ravi.rav]

I got the solution:

enetring "E:\\Sounds\\song.mp3" translates as "E:\\\\Sounds\\\\song.mp3" by appuifw.query() function.

[ck.umraliya]

hi ra.ravi.rav

Instead of entering the full path each time,
you can find out all mp3 files on the phone.
Go through this article.

Regards,
Chintan

[ra.ravi.rav]

Thanks for your reply. Adding autosearch was indeed the next job on the list.

[croozeus]

Originally Posted by ra.ravi.rav

I got the solution:

enetring "E:\\Sounds\\song.mp3" translates as "E:\\\\Sounds\\\\song.mp3" by appuifw.query() function.

Rather enter "E:\Sounds\song.mp3" in the appuifw.query which would be manipulated to the correct path i.e. "E:\\Sounds\\song.mp3"

Best Regards,
Croozeus