注意:我在 Netbeans 8.2 和 Windows 7 上运行此程序
该程序要求用户输入,他们可以输入一个字符,按空格键,或输入一个句点来停止程序。
1) 当我输入字符时,收到以下错误消息:“您输入了 java.util.Scanner[delimiters=\p{javaWhitespace}+][position=1][match valid=true] [需要输入=假][源关闭=假][跳过=假][组分隔符=\,][小数分隔符=\.][正前缀=][负前缀=\Q-\E][正后缀= ][负数后缀=][NaN字符串=\Q?\E][无穷大字符串=\Q8\E]"
2)当我点击空格键时,我没有得到任何反馈,直到我输入一个句点,然后我收到类似于上面的错误消息,但程序确实停止了。
3)如果我输入句点,我也会收到类似的错误消息,但程序确实会停止。
我期待的是以下内容:
a)如果我按空格键,它会返回一条消息,说明我按空格键并增加两个计数器
b) 如果我输入一个字符,它会返回一条消息,说明输入的字符并增加 ctr
计数器
c) 如果输入句点,则返回一条消息,说明该消息加上停止程序的次数
我猜问题出在 keytrip = userInput.next().charAt(0);
语句上。我认为使用 userInput.next().charAt(0)
会起作用,因为它们都是单个击键和字符。空间是一个字符,对吗?错误的?因此,如果有人能给我指出正确的方向来解决这个问题,我将不胜感激。
/* reads a char, a space, or a period from keyboard, returns user input,
counts number of spaces and total number of entries */
package ch03_36_exercise_01;
import java.util.Scanner;
public class Ch03_36_Exercise_01 {
public static void main(String args[]) throws java.io.IOException {
Scanner userInput = new Scanner(System.in);
char keystroke; // character that user enters
int ctr = 0, spaces = 0; // num of tries to stop run, num of spaces entered
do {
// ask for user input
System.out.print("Enter a character, or hit the space bar," +
" or enter a period to stop: ");
keystroke = userInput.next().charAt(0);
if (keystroke == ' ') {
System.out.println("You entered a space");
spaces++; // increment space bar count
}
else
System.out.println("You entered a " + keystroke);
// increment keystroke count
ctr++;
}
while (keystroke != '.');
System.out.print("It took " + ctr + " tries to stop");
if (spaces > 0)
System.out.println(" and you hit the space bar " + spaces + " times\n");
else
System.out.println();
}
}
最佳答案
您必须使用 nextLine()
而不是 next()
来读取空格。在这里查看更多详细信息:Scanner doesn't see after space 。使用 isSpaceChar 检查变量的空间。在这里查看更多详细信息:Checking Character Properties 。更正后的代码是......
/* reads a char, a space, or a period from keyboard, returns user input,
counts number of spaces and total number of entries */
package ch03_36_exercise_01;
import java.util.Scanner;
public class Ch03_36_Exercise_01 {
public static void main(String args[]) throws java.io.IOException {
Scanner userInput = new Scanner(System.in);
char keystroke; // character that user enters
int ctr = 0, spaces = 0; // num of tries to stop run, num of spaces entered
do {
// ask for user input
System.out.print("Enter a character, or hit the space bar,"
+ " or enter a period to stop: ");
keystroke = userInput.nextLine().charAt(0);
if (Character.isSpaceChar(keystroke)) {
System.out.println("You entered a space");
spaces++; // increment space bar count
} else {
System.out.println("You entered a " + keystroke);
}
// increment keystroke count
ctr++;
} while (keystroke != '.');
System.out.print("It took " + ctr + " tries to stop");
if (spaces > 0) {
System.out.println(" and you hit the space bar " + spaces + " times\n");
}
}
}
关于java - 无法使用 'Scanner' 和 '.next().charAt(0)' 正确读取字符,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45263248/