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Xfce + Debian Testing ... I have several custom Application Menus and only the first one can be open via a keyboard shortcut. How can I designate a keyboard shortcut for the other custom menus?
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Here is one option that might work for with using the applications menu. What this does, is set the custom-menu-file name via xfconf-query then pops-up the applications-menu using that custom-menu file, and then resets the menu to the default menu. What you see is the custom menu popping up that you can interact with, and if you click the menu button instead, the default menu will show up.
#!/bin/bash
### Set these values to suit
PLUGIN_NUM=1
DEFAULT_MENU=xfce-applications.menu
### Do not change anything below
# make sure custom menus are enabled
xfconf-query -c xfce4-panel -p /plugins/plugin-$PLUGIN_NUM/custom-menu -s true
# change to the correct custom menu
xfconf-query -c xfce4-panel -p /plugins/plugin-$PLUGIN_NUM/custom-menu-file -s ~/.config/menus/$1
# pause for the change to happen
sleep .5s
# popup the menu
xfce4-popup-applicationsmenu
# reset to the default
xfconf-query -c xfce4-panel -p /plugins/plugin-$PLUGIN_NUM/custom-menu-file -s ~/.config/menus/$DEFAULT_MENU
Make sure you set the value of PLUGIN_NUM (the number of the applicationsmenu plugin - "xfconf-query -c xfce4-panel -lv" will help to identify it) and the name of the default menu that you want to use (DEFAULT_MENU).
Create keyboard shortcuts to this shell script passing as a parameter the name of the custom menu that you want to use. For example:
Super+1 = /path/to/script work.menu
Super+2 = /path/to/script play.menu
...etc
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