.split() Method is NOT Allowed
我从一个乐于助人的人那里得到了一些帮助！我只是想知道是否有人可以帮我修改一下这个，
代码用于分配，它基于扫描仪的输入。它有另外两个类，但这是感兴趣的类。
此代码正在运行，但是必须输入的内容是** U5, D10**等等。这样可行。但是我需要代码能够从一行中读取多个String，同时将它们分开，就像它们现在一样。比如说**D10 U5 L4**，例如，只有一个玩家中的两个玩家。目前的代码并没有将其识别为一行，而是将第二种类型的东西分配给第二个玩家。
任何提示？
由于
import java.util.Scanner;
class Asgn2
{
public static void main(String[] args)
{
Scanner scan = new Scanner (System.in);
Player me = new Player("Player1");
Player opponent = new Player("player2");
int startingLoop = 0;
String strA;
int turn =1;
System.out.print("How many turns will the game have: ");
int turnsInGame = scan.nextInt();
System.out.print("How many moves does each player have each turn: ");
int numberOfTurns = scan.nextInt();
for(int i = turnsInGame; startingLoop < i; startingLoop++)
{
System.out.print("Turn " + turn++ + "\n");
System.out.print("Player 1 what are your " + numberOfTurns + " move(s): ");
String userInput = scan.next();
System.out.print("Player 2 what are your " + numberOfTurns + " move(s): ");
String userInputOne = scan.next();
for (int j = 0; j < userInput.length() - 1; j++)
{
char letter = userInput.charAt(j);
String num = "";
for(int k= j + 1; k < userInput.length(); k++)
{
j++;
if(userInput.charAt(k)!=' ')
{
num+=userInput.charAt(k);
}
else
{
break;
}
}
int integer = Integer.parseInt(num + "");
strA = Character.toString(letter);
switch(strA) //For player oneChooses which value to add or subtract from based on what is input.
{
case "U":
me.move(moveSteps.UP , integer);
break;
case "D":
me.move(moveSteps.DOWN, integer);
break;
case "L":
me.move(moveSteps.LEFT, integer);
break;
case "R":
me.move(moveSteps.RIGHT, integer);
break;
}
//Player 2
for (int playerTwo = 0; playerTwo < userInputOne.length() - 1; playerTwo++)
{
char letterTwo = userInputOne.charAt(0);
String numTwo = "";
String strB = Character.toString(letterTwo);
for(int m= playerTwo + 1; m<userInput.length(); m++)
{
playerTwo++;
if(userInputOne.charAt(playerTwo)!=' ')
{
numTwo+=userInputOne.charAt(playerTwo);
}
else
{
break;
}
}
int stepsMoved = Integer.parseInt(numTwo + "");
switch(strB) //For player two
{
case "U":
opponent.move(moveSteps.UP , stepsMoved);
break;
case "D":
opponent.move(moveSteps.DOWN, stepsMoved);
break;
case "L":
opponent.move(moveSteps.LEFT, stepsMoved);
break;
case "R":
opponent.move(moveSteps.RIGHT, stepsMoved);
break;
}
}
}
System.out.print(me);
System.out.print(opponent);
}
}
}
答案 0 :(得分：0)
将输入分配给String后，使用.split()方法将字符串拆分为数组。要使用.split()，请输入要将其拆分的字符。在这种情况下一个空间。例如，将此用于当前项目：.split(" ")。拆分后，您可以像任何阵列一样访问它。
更新
首先使用.nextLine()并将其分配给临时字符串变量。然后
你可以创建另一个扫描仪并输入一个字符串。例如：
Scanner sc = new Scanner(YOUR TEMPORARY VARIABLE);
您现在可以使用.next()获取单个字符串。
答案 1 :(得分：0)
这是Asgn2类
import java.util.Scanner;
public class Asgn2 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print("What is your name player 1: ");
String p1name = scan.nextLine();
System.out.print("What is your name player 2: ");
String p2name = scan.nextLine();
Player p1 = new Player(p1name);
Player p2 = new Player(p2name);
System.out.print("How many turns will the game have: ");
int numTurns = scan.nextInt();
System.out.print("How many moves does each player have each turn: ");
int numMoves = scan.nextInt();
for (int turn = 1; turn <= numTurns; turn++) {
System.out.println("----------------");
System.out.println("Turn number " + turn);
System.out.println("----------------");
for (int player = 1; player <= 2; player++) {
System.out.print("Player " + player + " what are your " + numMoves + " move(s): ");
for(int move=1;move<=numMoves;move++){
String currMove = scan.next();//splits at space;
char dir = currMove.charAt(0);//gets dir
String temp="";
for(int index=1;index<currMove.length();index++){
temp+=currMove.charAt(index);
}
int dist = Integer.parseInt(temp);
if(player==1){
p1.move(dir, dist);
}else if(player==2){
p2.move(dir, dist);
}
}
System.out.println("Player 1 is at " + p1.getPos() + " and Player 2 is at " + p2.getPos());
System.out.println();
}
}
}
}
这是Player类
public class Player {
private String name;
private int locX = 0;
private int locY = 0;
public Player(String name) {
this.name = name;
}
public void move(char dir, int numSteps) {
switch (dir) {
case 'U':
locY += numSteps;
break;
case 'D':
locY -= numSteps;
break;
case 'L':
locX -= numSteps;
break;
case 'R':
locX += numSteps;
break;
}
}
public String getPos() {
return "(" + locX + ", " + locY + ")";
}
public String getName() {
return name;
}
}
在任何人去之前说发布代码块对OP没有帮助，我和他在聊天室里解释这个东西，所以不要讨厌：）
答案 2 :(得分：-1)
调用方法.nextLine()而不是.next()。我认为这应该可以解决你的问题。