Question

我该怎么做？我想测试标志线2（1）是否等于“已关闭”或“打开”，如果不是我想说请指明它是打开还是关闭它说如果我打开或关闭在哪里||是行不通的......

package me.mcmatt.shops;

import org.bukkit.ChatColor;
import org.bukkit.Material;
import org.bukkit.Sound;
import org.bukkit.block.Block;
import org.bukkit.block.Sign;
import org.bukkit.entity.Player;
import org.bukkit.event.EventHandler;
import org.bukkit.event.Listener;
import org.bukkit.event.block.Action;
import org.bukkit.event.block.SignChangeEvent;
import org.bukkit.event.player.PlayerInteractEvent;

public class Signs implements Listener {


    @EventHandler
    public void onSignChange(SignChangeEvent e) {
        if (e.getLine(0).equalsIgnoreCase("[shop]")) {
            Block attached = e.getBlock().getRelative(0, -1, 0);
            String name = e.getPlayer().getDisplayName();
            if (!(attached.getType() == Material.CHEST))
                e.getPlayer().sendMessage(ChatColor.RED + "Please place the shop on a chest!");
            else {
                if (!e.getPlayer().hasPermission("shops.create"))
                    e.getPlayer().sendMessage(ChatColor.RED + "You don't have permission to create a shop! (shops.create)");
                else {
                    if (!e.getLine(1).equalsIgnoreCase("open") || (!e.getLine(1).equalsIgnoreCase("closed"))) {
                        e.getPlayer().sendMessage(ChatColor.RED + "You must specify if the shop is open or closed on the second line!");
                    } else {
                        Sign o = (Sign) e.getBlock().getState();
                        String p = o.getLine(1);
                        e.setLine(0, "§9[Shop]");
                        e.setLine(1, "§4" + name + "'s");
                        e.setLine(2, "§4Shop");
                        e.setLine(3, p);
                        e.getPlayer().sendMessage(ChatColor.GREEN + "Shop Created!");
                        e.getPlayer().playSound(e.getPlayer().getLocation(), Sound.LEVEL_UP, 10, 10);
                    }
                }
            }
        }
    }

    @EventHandler
    public void onPlayerInteract(PlayerInteractEvent e) {
        if (e.getAction().equals(Action.RIGHT_CLICK_BLOCK)) {
            Player p = e.getPlayer();
            Block b = e.getClickedBlock();
            Material m = b.getType();
            if (!(m == Material.SIGN_POST)) {
                return;
            } else {
                Sign sign = (Sign) e.getClickedBlock().getState();
                if ((sign.getLine(0).equalsIgnoreCase("§9[Shop]"))) {
                    p.sendMessage("I right clicked the sign!");
                }
            }
        }
    }
}

Answer 1

您可以创建要与之比较的字符串列表，然后使用List.contains方法：

if (!Arrays.asList("open", "closed").contains(e.getLine(1).toLowerCase()) {...

Answer 2

您的代码无效的原因是因为De Morgan的法律

给出两个陈述，A和B.如果你不想要A或B为真

(!A) or (!B)不正确

(!A) and (!B)是正确的，或!(A or B)在逻辑上是等效的。

我需要知道如何测试一个值是否不等于多个字符串java

2 个答案: