Java Scanner需要多个条目
下面是我用Java编写的一些代码,用于在游戏中设置玩家。我为atk、def和hp设置的扫描仪需要2-3个条目。很抱歉代码太乱,我才刚刚开始学习JavaJava Scanner需要多个条目,java,java.util.scanner,Java,Java.util.scanner,下面是我用Java编写的一些代码,用于在游戏中设置玩家。我为atk、def和hp设置的扫描仪需要2-3个条目。很抱歉代码太乱,我才刚刚开始学习Java String NAME; int ATK; int DEF; int HP; int remainingPoints = 10; // pick name System.out.println("Enter name:"); Scanner p2Name = new Scanner(System.in); NAME = p2Name.nextLi
String NAME;
int ATK;
int DEF;
int HP;
int remainingPoints = 10;
// pick name
System.out.println("Enter name:");
Scanner p2Name = new Scanner(System.in);
NAME = p2Name.nextLine();
// pick type
System.out.println("Choose element theme: (water,fire,ice)");
Scanner p2Type = new Scanner(System.in);
String TYPE = p2Type.nextLine();
// atk
System.out.println("You have ten points to disperse " + "among your atk, def and hp");
System.out.println("Enter atk:");
Scanner p2Atk = new Scanner(System.in);
if (p2Atk.nextInt() <= 10) {
ATK = p2Atk.nextInt();
remainingPoints = remainingPoints - ATK;
} else {
System.out.println("Your entry was invalid, your " + "atk is now five by default.");
ATK = 5;
remainingPoints = remainingPoints - ATK;
}
// def
System.out.println("You now have " + remainingPoints + ". Please enter def:");
Scanner p2Def = new Scanner(System.in);
if (p2Def.nextInt() <= remainingPoints) {
remainingPoints = remainingPoints - p2Def.nextInt();
DEF = p2Def.nextInt();
} else {
System.out.println("Your entry was invalid, your " + "def is now one by default.");
DEF = 1;
remainingPoints = remainingPoints - DEF;
}
// hp
System.out.println("You now have " + remainingPoints + ". Please enter hp:");
Scanner p2Hp = new Scanner(System.in);
if (p2Hp.nextInt() <= remainingPoints) {
remainingPoints = remainingPoints - p2Hp.nextInt();
HP = p2Hp.nextInt();
} else {
System.out.println("Your entry was invalid, your " + "hp is now one by default.");
HP = 1;
remainingPoints = remainingPoints - HP;
}
Player p2 = new Player(NAME, TYPE, ATK, DEF, HP);
System.out.println(p2.n);
System.out.println(p2.type);
System.out.println(p2.atk);
System.out.println(p2.def);
System.out.println(p2.hp);
字符串名;
int ATK;
int-DEF;
int-HP;
整数剩余点=10;
//挑选名字
System.out.println(“输入名称:”);
扫描仪p2Name=新扫描仪(System.in);
NAME=p2Name.nextLine();
//拾取类型
println(“选择元素主题:(水、火、冰)”;
扫描仪p2Type=新扫描仪(System.in);
字符串类型=p2Type.nextLine();
//atk
System.out.println(“在atk、def和hp中,您有十个点要分散“+”);
System.out.println(“输入atk:”);
扫描仪p2Atk=新扫描仪(System.in);
如果(p2Atk.nextInt()创建了太多不正确的扫描仪对象
关于需要2-3个条目-您应该在temp变量中输入一次值,然后多次使用它
PFB校正代码:
String NAME = null;
String TYPE = null;
int ATK = 0;
int DEF = 0;
int HP = 0;
int remainingPoints = 10;
try {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter name:");
NAME = scanner.nextLine();
System.out.println("Choose element theme: (water,fire,ice)");
TYPE = scanner.nextLine();
System.out.println("You have ten points to disperse "
+ "among your atk, def and hp");
System.out.println("Enter atk:");
int temp = scanner.nextInt();
if (temp <= 10)
ATK = temp;
else {
System.out.println("Your entry was invalid, your "
+ "atk is now five by default.");
ATK = 5;
}
remainingPoints = remainingPoints - ATK;
System.out.println("You now have " + remainingPoints
+ ". Please enter def:");
temp = scanner.nextInt();
if (temp <= remainingPoints)
DEF = temp;
else {
System.out.println("Your entry was invalid, your "
+ "def is now one by default.");
DEF = 1;
}
remainingPoints = remainingPoints - DEF;
System.out.println("You now have " + remainingPoints
+ ". Please enter hp:");
temp = scanner.nextInt();
if (temp <= remainingPoints)
HP = temp;
else {
System.out.println("Your entry was invalid, your "
+ "hp is now one by default.");
HP = 1;
}
remainingPoints = remainingPoints - HP;
scanner.close();
} catch (Exception e) {
e.printStackTrace();
}
Player p2 = new Player(NAME, TYPE, ATK, DEF, HP);
System.out.println(p2.n);
System.out.println(p2.type);
System.out.println(p2.atk);
System.out.println(p2.def);
System.out.println(p2.hp);
String NAME=null;
字符串类型=null;
int-ATK=0;
int DEF=0;
int HP=0;
整数剩余点=10;
试一试{
扫描仪=新的扫描仪(System.in);
System.out.println(“输入名称:”);
NAME=scanner.nextLine();
println(“选择元素主题:(水、火、冰)”;
类型=scanner.nextLine();
System.out.println(“你有10点要分散”
+“在您的atk、def和hp中”);
System.out.println(“输入atk:”);
int temp=scanner.nextInt();
如果(temp,则无需在每次提示输入时创建新的扫描仪对象
提示您输入的数据类型,无论是int
、double
还是String
,您只需创建扫描对象一次——即:对于system.in
但是,Java初学者在使用scn.nextInt()
时可能会遇到问题。
您可以在提示解决该问题后放置scn.nextLine()
但就我个人而言,我更喜欢使用scn.nextLine()
以String
的形式接收每个输入。如果是int
输入,只需执行以下操作:
int num = Integer.parseInt(scn.nextLine());
示例:
Scanner scn = new Scanner(System.in);
String name = scn.nextLine();
String type = scn.nextLine();
String spell = scn.nextLine();
int health = Integer.parseInt(scn.nextLine());
问题是什么?为什么要创建多个Scanner
对象?神圣的代码格式化蝙蝠侠!这是一堵需要编辑的文本墙!当你“刚刚学习Java”的时候,从一开始就没有养成良好的代码格式化/编写习惯,这对你自己来说是一种伤害。我强烈建议你修改代码的格式,不管你的技能水平如何。有没有理由调用scanner.close()
在您的代码中?当您使用完扫描仪时,最好总是将其关闭@Tom@EvanBechtol你认为OP通过了系统。在中,你可以关闭该流?我不这么认为,因为该代码仅用于设置游戏玩家,所以OP很可能希望在游戏中获得更多用户输入正在关闭此游戏,因此关闭扫描仪是非常错误的。说得好!如果他计划获得更多输入,他不应该关闭。但是,在游戏结束时,他应该关闭它。@EvanBechtol如果他绝对肯定不再需要系统。
,那么是的,他应该清理使用过的资源。“但是,Java初学者在使用scn.nextInt()
时可能会遇到问题。在提示解决该问题后,您可以放置一个scn.next()
。”调用next()
不会解决该问题,因为如果“新行”分隔符后面没有其他标记,“新行”分隔符不会被占用。