Is it possible to just execute a shell script but without opening the terminal? I have a simple toggle script which should just be executed and not shown and currently I always get a black window with a loading animation which is annoying and not necessary. The desktop file looks like this: Code: [Desktop Entry] Encoding=UTF-8 Version=0.1 Name=FMTX Toggle Terminal=false Type=Application Comment=FMTX Toggle Exec=/home/user/apps/fmtransmtoggle.sh Icon=terminal #X-Window-Icon=timerwidget X-Window-Icon= X-HildonDesk-ShowInToolbar=false X-Osso-Type=application/x-executable and the script like this: Code: #!/bin/sh fmtx_client|grep disabled > /dev/null if [ $? -eq 0 ]; then fmtx_client -p 1 > /dev/null fmtx_client -f 106900 > /dev/null fmtx_client -s "N900 " > /dev/null fmtx_client -t "My N900 radio channel" > /dev/null echo 118 > /sys/class/i2c-adapter/i2c-2/2-0063/power_level echo "RF Enabled @106.90 with power `cat /sys/class/i2c-adapter/i2c-2/2-0063/power_level`" else fmtx_client -p 0 echo "RF Disabled" fi exit 0 Is it also possible to have only a shortcut on the desktop but not in the menu? Thanks
[Desktop Entry] Encoding=UTF-8 Version=0.1 Name=FMTX Toggle Terminal=false Type=Application Comment=FMTX Toggle Exec=/home/user/apps/fmtransmtoggle.sh Icon=terminal #X-Window-Icon=timerwidget X-Window-Icon= X-HildonDesk-ShowInToolbar=false X-Osso-Type=application/x-executable
#!/bin/sh fmtx_client|grep disabled > /dev/null if [ $? -eq 0 ]; then fmtx_client -p 1 > /dev/null fmtx_client -f 106900 > /dev/null fmtx_client -s "N900 " > /dev/null fmtx_client -t "My N900 radio channel" > /dev/null echo 118 > /sys/class/i2c-adapter/i2c-2/2-0063/power_level echo "RF Enabled @106.90 with power `cat /sys/class/i2c-adapter/i2c-2/2-0063/power_level`" else fmtx_client -p 0 echo "RF Disabled" fi exit 0