英文:
Java Scanner Reading Leading Space
问题
我试图创建一个菜单，接受一个字符输入用于开关案例，并循环直到输入为"q"，但在循环运行一次后，我得到了这个错误：
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 0 at java.lang.String.charAt(Unknown Source) at Menu.main(Menu.java:19)
这意味着它正在获取一个空输入，对吗？所以我添加了.trim()但我仍然得到了错误，它甚至不等待输入，我只是得到了错误。
抱歉，如果以前有人回答过这个问题，但我找不到任何答案。我还尝试添加了keyboard.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");，但也不起作用。
程序错误日志的输入：
Enter Option (a,b,c,d,e,f,q):
c
Enter 2 Numbers
First Number:
1
Second Number:
10
1, 2, 3, 4, 5,
6, 7, 8, 9, 10
Enter Option (a,b,c,d,e,f,q):
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 0
at java.lang.String.charAt(Unknown Source)
at Menu.main(Menu.java:19)
代码：
import java.io.*;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner keyboard = new Scanner (System.in);
char ch; int m,n; String y;
do
{
System.out.println("Enter Option (a,b,c,d,e,f,q):");
ch = keyboard.nextLine().toLowerCase().charAt(0);
switch (ch)
{
case 'c':
System.out.println("Enter 2 Numbers");
System.out.println("First Number: ");
m = keyboard.nextInt();
System.out.println("Second Number: ");
n = keyboard.nextInt();
for(int i = 1; i <= n - m + 1; i++)
{
if (i % 5 == 0)
{
System.out.print(i);
if (i != n - m + 1)
{
System.out.println(", ");
}
}else
{
System.out.print(i);
if (i != n - m + 1)
{
System.out.print(", ");
}
}
}
System.out.println("");
break;
}
}while (ch != 'q');
}
}
请注意，我已经移除了HTML编码中的"并将toLowerCase()应用到ch的输入上。
英文:
I am trying to have a menu that takes a character input for a switch case and loops till the input is q but after the loop run once i get this error:
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 0 at java.lang.String.charAt(Unknown Source) at Menu.main(Menu.java:19)
Which means its getting a null input right? So i added the .trim() but i still get the error, it doesn't even wait for an input i just get the error.
Sorry if this has been answered before but i can't find it anywhere. I've also tried adding keyboard.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); which doesn't work as well.
Input given for the program with error log:
Enter Option (a,b,c,d,e,f,q):
c
Enter 2 Numbers
First Number:
1
Second Number:
10
1, 2, 3, 4, 5,
6, 7, 8, 9, 10
Enter Option (a,b,c,d,e,f,q):
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 0
at java.lang.String.charAt(Unknown Source)
at Menu.main(Menu.java:19)
CODE:
import java.io.*;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner keyboard = new Scanner (System.in);
char ch; int m,n; String y;
do
{
System.out.println("Enter Option (a,b,c,d,e,f,q):");
ch = keyboard.nextLine().toLowerCase().charAt(0);
switch (ch)
{
case 'c':
System.out.println("Enter 2 Numbers");
System.out.println("First Number: ");
m = keyboard.nextInt();
System.out.println("Second Number: ");
n = keyboard.nextInt();
for(int i = 1; i <= n - m + 1; i++)
{
if (i % 5 == 0)
{
System.out.print(i);
if (i != n - m + 1)
{
System.out.println(", ");
}
}else
{
System.out.print(i);
if (i != n - m + 1)
{
System.out.print(", ");
}
}
}
System.out.println("");
break;
}
}while (ch != 'q');
}
}
答案1
得分: 2
import java.io.*;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
char ch;
int m, n;
String y;
do {
System.out.println("Enter Option (a,b,c,d,e,f,q):");
ch = keyboard.next().charAt(0);
switch (ch) {
case 'c':
System.out.println("Enter 2 Numbers");
System.out.println("First Number: ");
m = keyboard.nextInt();
System.out.println("Second Number: ");
n = keyboard.nextInt();
for (int i = 1; i <= n - m + 1; i++) {
if (i % 5 == 0) {
System.out.print(i);
if (i != n - m + 1) {
System.out.println(", ");
}
} else {
System.out.print(i);
if (i != n - m + 1) {
System.out.print(", ");
}
}
}
System.out.println("");
break;
}
} while (ch != 'q');
}
}
英文:
Try the line ch = keyboard.next().charAt(0); switch(ch)
import java.io.*;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner keyboard = new Scanner (System.in);
char ch; int m,n; String y;
do
{
System.out.println("Enter Option (a,b,c,d,e,f,q):");
ch = keyboard.next().charAt(0);
switch (ch)
{
case 'c':
System.out.println("Enter 2 Numbers");
System.out.println("First Number: ");
m = keyboard.nextInt();
System.out.println("Second Number: ");
n = keyboard.nextInt();
for(int i = 1; i <= n - m + 1; i++)
{
if (i % 5 == 0)
{
System.out.print(i);
if (i != n - m + 1)
{
System.out.println(", ");
}
}else
{
System.out.print(i);
if (i != n - m + 1)
{
System.out.print(", ");
}
}
}
System.out.println("");
break;
}
}while (ch != 'q');
}
}
Happy coding!
答案2
得分: 1
以下是已更新的代码，请检查：
public static void main(String[] args)
{
Scanner keyboard = new Scanner(System.in);
char ch;
do
{
System.out.println("输入选项 (a,b,c,d,e,f,q)：");
ch = keyboard.nextLine().toLowerCase().charAt(0);
switch (ch)
{
// a、b、c、d、e 和 f 的 case 语句
}
} while (ch != 'q');
}
英文:
Here's the updated code Please check:
public static void main(String[] args)
{
Scanner keyboard = new Scanner (System.in);
char ch;
do
{
System.out.println("Enter Option (a,b,c,d,e,f,q):");
ch = keyboard.nextLine().toLowerCase().charAt(0);
switch (ch)
{
//case statements for a,b,c,d,e and f
}
}while (ch != 'q');
}
答案3
得分: 1
主要问题在于扫描器在你按下"Enter"键后立即与输入一起工作。
因此，如果你的输入是输入'\u0020'（空格），甚至在第19行多次输入，你会对空行进行修剪，根据javadoc中的charAt。
英文:
Main thing here that scanner works with input right just after you hit "enter".
So if your input would be entering '\u0020'(whitespaces) even several times on line 19 you're doing trim to empty line and according to javadoc charAt.
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