Question

所以，我正在使用Bukkit API。基本上，我正在寻找要调用的PlayerInteractEvent，之后我会做一堆东西。但是，当我收到通知我实际踢了块时，我没有得到任何消息，即使它在我的代码中编译没有错误。我也没有从控制台得到任何例外。这是我的代码：

@EventHandler(priority=EventPriority.HIGH)
public void onPlayerInteract(PlayerInteractEvent event, final Player who, final Action action, 
        final ItemStack item, final Block clickedBlock, final BlockFace clickedFace) {

    if (who != null) {

        if (clickedBlock != null) {

            if ((action == Action.LEFT_CLICK_BLOCK) && (clickedBlock.getType() == Material.ANVIL)) {

                if (item == null) {

                    who.sendMessage(ChatColor.YELLOW + "To repair an item, hold it in your inventory and " + 
                            ChatColor.UNDERLINE + "RIGHT CLICK" + ChatColor.RESET + "" + ChatColor.YELLOW + 
                            " the anvil with the item.");

                    event.setCancelled(true);

                }
                else {

                    Material type = item.getType();
                    short durability = item.getDurability();
                    short maximum = type.getMaxDurability();

                    if (maximum == 0) {

                        who.sendMessage(ChatColor.RED + "You can " + ChatColor.UNDERLINE + "NOT" + 
                                ChatColor.RESET + "" + ChatColor.RED + " repair that item.");

                    }
                    else {

                        short add = (short) Math.round(maximum * 0.03);

                        int result = (maximum - durability) / add;

                        int gems = getAmount(who, 388);

                        if (gems < result) {

                            who.sendMessage(ChatColor.RED + "You do " + ChatColor.UNDERLINE + "NOT" + 
                                    ChatColor.RESET + "" + ChatColor.RED + " have enough Gems to repair "
                                    + "that item.");
                            who.sendMessage(ChatColor.RED + "" + ChatColor.BOLD + "Gems Needed: " + result);

                        }
                        else {

                            who.sendMessage(ChatColor.YELLOW + "It will cost " + ChatColor.WHITE + 
                                    result + "g " + ChatColor.GREEN + "to repair this item.");
                            who.sendMessage(ChatColor.YELLOW + "Continue repairing? " + ChatColor.GREEN + 
                                    "" + ChatColor.BOLD + "Y" + ChatColor.RESET + "" + ChatColor.WHITE + 
                                    " / " + ChatColor.RESET + "" + ChatColor.RED + "" + ChatColor.BOLD + 
                                    "N");

                            map.put(who, item);

                        }

                    }

                }

            }

        }

    }

}

Answer 1

因此，执行此操作的标准方法是创建一个implements Listener

的类

然后在你内部以

的形式创建方法

@EventHandler
public void nameDontMatter(PlayerInteractEvent event) {
 // do stuff, you can get all that information you passed in from the event

}

此外，您需要确保告诉插件在哪里找到您的PlayerListener，所以通常您要做的是onEnable()方法：

PlayerListenerClass PL = new PlayerListenerClass();
//instantiate an instance of you player listener 

public void onEnable() {
    PluginManager pm = this.getServer().getPluginManager();
    pm.registerEvents(InspiredNationsPL, this);
    //Tell the plugin manager where it can find the player listener
}

这应该让它发挥作用。

Answer 2

该方法唯一的参数是PlayerIneractEvent。您无法在此处设置更多参数。

您设置的参数多于方法本身的参数。

@EventHandler
public void onPlayerInteract(PlayerInteractEvent event, Player who, ...) {
    ...
}

改为使用：

@EventHandler
public void onPlayerInteract(PlayerInteractEvent event) {
    // do stuff here
}

在方法中，您可以获得您设置为参数的所有其他内容。例如：

event.getPlayer(); // gets the player
event.getAction(); // gets the action

另请注意，您已在插件主类中注册了听众。

registerEventHandler(yourEventHandler, this);

Answer 3

你只能拥有PlayerInteractEvent，否则它不会工作。此外，您不需要将播放器作为参数提供，您可以使用getPlayer（）方法查看触发事件的人。

@EventHandler
public void onPlayerInteract(PlayerInteractEvent e) {
Player player = e.getPlayer();
}

为什么PlayerInteractEvent不适用于Bukkit API？

3 个答案: