bukket插件例子（4）代码分析

试着分析一下/get命令，命令类似/give，/get x y代表给用户y个x物件

原代码如下

初始化函数

super(plugin, "get", 1, 2);其中最少包含1个参数，最多2个

doPlayerCommand来自BaseCommand的实现

逻辑如下，如果参数长度是2，将倍数times存起来

如果x输入的是名字，那么就把有这个名字的全都加给用户

if (c.toString().toLowerCase().contains(cur.toLowerCase()))例如/get diamond 20就会把所有是砖石的都给用户，包括diamond_boots, diamond_pickaxe
如果x输入的是数字，那么就把符合这个id要求的给用户

例如/get 262 2就会把所有id是262的物件给用户。

代码如下，

public class CmdGetblock extends BaseCommand {

Map<String, ConfigurationSection> allKits;

public CmdGetblock(MainPlugin plugin) {
super(plugin, "get", 1, 2);
}

@Override
@SuppressWarnings("deprecation")
protected boolean doPlayerCommand(Player player, Command cmd, String commandLabel, String[] split) {

Integer times = Integer.valueOf(1);
if (split.length == 2)
{
try
{
times = Integer.valueOf(Integer.parseInt(split[1]));
}
catch (Exception localException) {}
if (times.intValue() < 0) {
times = Integer.valueOf(1);
}
}
if ((split.length == 1) || (split.length == 2))
{
String cur = split[0];
boolean found = false;
Material[] arrayOfMaterial;
int j = (arrayOfMaterial = Material.values()).length;

for (int ai = 0; ai < j; ai++)
{
Material c = arrayOfMaterial[ai];
if (c.toString().toLowerCase().contains(cur.toLowerCase()))
{
found = true;
for (int ki = 0; ki < times.intValue(); ki++) {
player.getInventory().addItem(new ItemStack[] { new ItemStack(c.getId(), 1) });
}
player.sendMessage("Material added maybe: " + c.toString() + " with id " + String.valueOf(c.getId()));
}
}
if (!found) {
try
{
Integer x = Integer.valueOf(Integer.parseInt(split[0]));
Material[] all = Material.values();
int k = all.length;
for (j = 0; j < k; j++)
{
Material c = all[j];
if (c.getId() == x.intValue())
{
found = true;
for (int i = 0; i < times.intValue(); i++) {
player.getInventory().addItem(new ItemStack[] { new ItemStack(c.getId(), 1) });
}
player.sendMessage("Material added maybe: " + c.toString() + " with id " + String.valueOf(c.getId()));
}
}
}
catch (Exception localException1) {}
}
return true;
}
return false;
}
}