“打印”；我们没有'；我没有找到这封信；带循环的java

这是我的节目

import java.util.*;
public class LeTter { //open class name
    static Scanner keyboard = new Scanner(System.in); // Scanner
    public static void main(String[] args) { //open main class
        char letter; //decl
        int y, x, z; //decl
        boolean check = false;
        System.out.println("Enter 8 letters: "); // print sentince
        char[] des = new char[8]; //array
        for (y = 0; y < des.length; y++) { // for without semicolon :), open for loop
            des[y] = keyboard.next().charAt(0); //initialize
        } // close for loop
        System.out.println("Enter any letter to check about it: "); //will apper to the user
        letter = keyboard.next().charAt(0); //initi
        for (x = 0; x < des.length; x++) { //for loop opens
            if (des[x] == letter) { //open if loop
                System.out.println("we found it's " + des[x] + " and the index of it is " + x);
                System.out.println("so we will show you this letters " + x + " times");
                for (z = 0; z < x; z++)
                    System.out.println(des[x] + "  ");
                check = false;
            } //close if loop
            if (true)
                System.out.println("are you sure about the letter's you enterd it? we didnt found it ");
        } // close for loop
    } //close main
} //close class

import java.util.*；
公共类字母{//打开类名称
静态扫描仪键盘=新扫描仪（System.in）；//扫描仪
公共静态void main（字符串[]args）{//open main类
字符字母；//decl
int y，x，z；//decl
布尔检查=假；
System.out.println（“输入8个字母：”；//打印感知
char[]des=新字符[8]；//数组
对于（y=0；y

如果用户没有输入看起来不像8个字母的字母，我想打印这句话“我们没有找到它” 我该怎么做

示例： 输入8个字母 a s d f g h j k 输入任何字母 T

---

我们没有找到它我跳过输入，从代码的循环开始，将其更改为：

for (x = 0; x < des.length; x++)
{ // for loop opens
    if (des[x] == letter)
    { // open if loop
        check = true;
        System.out.println("we found it's " + des[x] + " and the index of it is " + x);
        break; // you want to show 1st one only, right?
    } // close if loop
} // close for loop
if (!check)
    System.out.println("are you sure about the letter's you enterd it? we didnt found it ");

for（x=0；x

我缩短了您的代码，因为您只需将用户的输入添加到

字符

数组

，然后删除空格即可

我用了

break

要在找到第一个

char

后退出循环，如果您想知道它的多个实例，可以将其删除

static Scanner keyboard = new Scanner(System.in);

    public static void main(String[] args) {
        System.out.println("Enter 8 letters: ");
        String letters = keyboard.nextLine().replaceAll(" ", "");
        char cArray[] = letters.toCharArray();
        System.out.println("Enter any letter to check about it: ");
        char letter = keyboard.next().charAt(0);
        boolean found = false;
        for (int i = 0; i < cArray.length; i++) {
            if (cArray[i] == letter) {
                System.out.println("We found " + letter + " at position " + i);
                found = true;
                break; //used as you specify only to find the first one
            }
        }
        if (!found) {
            System.out.println("Are you sure about the letter you entered? We didn't find it");
        }
    }

静态扫描仪键盘=新扫描仪（System.in）；
公共静态void main（字符串[]args）{
System.out.println（“输入8个字母：”）；
字符串字母=键盘.nextLine（）.replaceAll（“，”）；
char cArray[]=letters.toCharArray（）；
System.out.println（“输入任何要检查的字母：”）；
字符字母=键盘.next（）.charAt（0）；
布尔值=false；
对于（int i=0；i