由于某种原因,Java循环运行了4次
所以我有一个任务,我需要循环用户的输入6次。循环完成后,再次循环3次。我没有在它之前添加lop,所以我不知道如何处理它 下面是该方法的代码:由于某种原因,Java循环运行了4次,java,arrays,for-loop,Java,Arrays,For Loop,所以我有一个任务,我需要循环用户的输入6次。循环完成后,再次循环3次。我没有在它之前添加lop,所以我不知道如何处理它 下面是该方法的代码: public static int[] getPlayerNumbers(int[] playNums) { Scanner input = new Scanner(System.in); for (int i = 0; i < playNums.length; i++) { System.out.println("
public static int[] getPlayerNumbers(int[] playNums) {
Scanner input = new Scanner(System.in);
for (int i = 0; i < playNums.length; i++) {
System.out.println("Please enter numbers from 1-9: " + i);
playNums[i] = input.nextInt();
while (playNums[i] < 1 || playNums[i] > 9) {
System.out.println("Invlaid input. Please only enter 1-9. ");
playNums[i] = input.nextInt();
}
}
return playNums;
}
公共静态int[]getPlayerNumber(int[]playNums){
扫描仪输入=新扫描仪(System.in);
for(int i=0;i9){
System.out.println(“输入输入,请只输入1-9”);
playNums[i]=input.nextInt();
}
}
返回玩偶;
}
我放置了
I
来查看索引,它转到0
到5
,然后返回到0
。我没有主意了,请帮帮我。看来你的玩偶不止6个。试一试
public static int[] getPlayerNumbers(int[] playNums) {
Scanner input = new Scanner(System.in);
for (int i = 0; i < 6; i++) {
System.out.println("Please enter numbers from 1-9: " + i);
playNums[i] = input.nextInt();
while (playNums[i] < 1 || playNums[i] > 9) {
System.out.println("Invlaid input. Please only enter 1-9. ");
playNums[i] = input.nextInt();
}
}
return playNums;}
公共静态int[]getPlayerNumber(int[]playNums){
扫描仪输入=新扫描仪(System.in);
对于(int i=0;i<6;i++){
System.out.println(“请输入1-9的数字:+i”);
playNums[i]=input.nextInt();
while(playNums[i]<1 | | playNums[i]>9){
System.out.println(“输入输入,请只输入1-9”);
playNums[i]=input.nextInt();
}
}
返回playNums;}
我已经测试了你的代码,它似乎适合我。PlaySums肯定包含6个以上的值
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
int[] playNums ={1, 2, 3, 4, 5, 6};
getPlayerNumbers(playNums);
for (int playNum: playNums) {
System.out.println(playNum);
}
}
public static int[] getPlayerNumbers(int[] playNums)
{
Scanner input = new Scanner(System.in);
for (int i = 0; i < playNums.length; i++) {
System.out.println("Please enter numbers from 1-9: " + i);
playNums[i] = input.nextInt();
while (playNums[i] < 1 || playNums[i] > 9) {
System.out.println("Invalid input. Please only enter 1-9. ");
playNums[i] = input.nextInt();
}
}
return playNums;
}
}
import java.util.Scanner;
公共班机{
公共静态void main(字符串[]args){
int[]playNums={1,2,3,4,5,6};
GetPlayerNumber(playNums);
for(int playNum:playNums){
System.out.println(playNum);
}
}
公共静态int[]getPlayerNumber(int[]playNums)
{
扫描仪输入=新扫描仪(System.in);
for(int i=0;i9){
System.out.println(“输入无效,请只输入1-9”);
playNums[i]=input.nextInt();
}
}
返回玩偶;
}
}
试试这个
public static int[] getPlayerNumbers(int[] playNums)
{
Scanner input = new Scanner(System.in);
for (int i = 0; i < 6; i++)
{
playNums[i] = input.nextInt();
if(playNums[i] < 1 || playNums[i] > 9){
System.out.println("Invlaid input. Please only enter 1-9. ");
getPlayerNumbers(playNums);
} else{
System.out.println("Please enter numbers from 1-9: " + i);
playNums[i] = input.nextInt();
}
}
return playNums;
公共静态int[]getPlayerNumber(int[]playNums)
{
扫描仪输入=新扫描仪(System.in);
对于(int i=0;i<6;i++)
{
playNums[i]=input.nextInt();
if(playNums[i]<1 | | playNums[i]>9){
System.out.println(“输入输入,请只输入1-9”);
GetPlayerNumber(playNums);
}否则{
System.out.println(“请输入1-9的数字:+i”);
playNums[i]=input.nextInt();
}
}
返回玩偶;
请告诉我们您输入了什么,当应用程序要求您输入一个数字(可能是应用程序执行的屏幕截图?)时,我觉得代码很好。我唯一的猜测是您在两个地方调用了该方法。