Gtk event.key.keyval信号未连接

当按下键盘上的“结束”键时，我想更新

菜单按钮的标签

代码编译无误，但信号未连接且
stdout.printf（“Keyval:0x%X\n”，event.key.Keyval）不打印任何内容

标签也没有更新。
我从vala discord服务器的一个了不起的家伙那里得到了这个答案

由于Gtk.LocationEntry是Gtk.Entry的一个子类，因此我们可以使用激活信号

然后我用这个代码来更新标签

public class Epoch.CitiesChooser : Gtk.MenuButton {
    private Epoch.CitiesSearch cities_search;
    
    construct {
        var button_grid = new Gtk.Grid ();
        button_grid.column_spacing = 6;
        var button_label = new Gtk.Label ("Dimapur");
        button_grid.add (button_label);
        button_grid.add (new Gtk.Image.from_icon_name ("pan-down-symbolic", Gtk.IconSize.MENU));
        add (button_grid);
        
        cities_search = new Epoch.CitiesSearch ();
        cities_search.show_all ();
        
        /* Use a event.eventkey.keypress here to add update the label of the menu_button using set_text */
        if (cities_search.location_entry.visible) {
            key_press_event.connect ((event) => {
                stdout.printf ("Keyval: 0x%X\n", event.key.keyval);
                if (event.key.keyval == 0xFF57) {
                    button_label.set_text (cities_search.location_entry.get_text ());
                    cities_search.location_entry.hide ();
                }
                return false;
            });
        }
        
        popover = new Gtk.Popover (this);
        popover.width_request = 310;
        popover.add (cities_search);
    }
}

它满足了我的目的，但不是对原来问题的回答

cities_search.location_entry.activate.connect ((obj) => {
            button_label.set_text (cities_search.location_entry.get_text ());
            if (this.get_active ()) {
                this.set_active (false);
            }
        });