我不確定如何驗證用戶是否只輸入一個字符進行輸入。我知道我對長度檢查的看法根本不正確。我只是用它來填充物。請幫忙。我嘗試了許多不同的方法，并搜索了這個網(wǎng)站和其他網(wǎng)站好幾天，以找到答案。final char SIZE = 10;char [] letter = new char [SIZE];// initiallizing input deviceScanner scan = new Scanner(System.in);for (char index = 0; index < SIZE;){ System.out.println ("Please enter Letter #" + (index + 1));// gets letter from user while ((!scan.hasNext("[A-Za-z]+")) || (!scan.hasNext(length(1)))){ if(!scan.hasNext(length (1))){ System.out.println ("Please only enter one Letter at a time: "); letter [index] = scan.next().charAt(0); // accepts first character entered by user } if(!scan.hasNext("[A-Za-z]+")){ System.out.println ("Please enter a valid Letter: "); letter [index] = scan.next().charAt(0); // accepts first character entered by user } else if((scan.hasNext("[A-Za-z]+")) && (scan.next(length(1)))){// makes sure letter entered is a letter letter [index] = scan.next().charAt(0); // accepts first character entered by user index++;// increases index if proper letter entered } }}for (char index = 0; index < SIZE; index++){ System.out.println ("Letter #" + (index + 1) + ": " + letter [index]);// prints characters entered by user in order}
3 回答
開滿天機
TA貢獻1786條經(jīng)驗 獲得超13個贊
有很多方法可以做到這一點。一種方式是A.K.發(fā)布的內(nèi)容。一種方式如下：
for (char index = 0; index < SIZE;)
{
String temp = scan.nextLine();
if (temp.length() != 1)
{
// This means their input was too big
}
else
{
// This means their input was one character
}
}
您還可以使用此行代碼僅從控制臺中選取一個字符作為輸入。scan.next().charAt(0);
如果您有任何疑問或此答案不是您要找的，請在下面發(fā)表評論，我將很樂意為您提供幫助。
ABOUTYOU
TA貢獻1812條經(jīng)驗 獲得超5個贊
輸入僅在用戶按下后放在緩沖區(qū)上，因此在現(xiàn)實中它始終是一次行。您應(yīng)該始終將代碼與現(xiàn)實保持一致，因此請始終按行讀取用戶輸入。Enter
此實用程序方法可能使用有用的：
private static String read(Scanner scanner, String message, String regex) {
System.out.println(message);
String result = "";
while (true) {
result = scanner.nextLine();
if (result.matches(regex)) {
break;
}
System.out.println("Invalid input. " + message);
}
}
你可以這樣稱呼它：
String letter = read(scan, "Please enter Letter #" + (index + 1)", "[A-Za-z]");
該方法在僅輸入一個字母之前不會返回，并且它將返回該字母。read()
慕尼黑的夜晚無繁華
TA貢獻1864條經(jīng)驗 獲得超6個贊
一種解決方案是這樣的：
Scanner sc = new Scanner(System.in);
System.out.println("Enter char");
String in = sc.next();
while(in.length()!=1){
System.out.println("Enter only a single char");
in = sc.next();
}
// Print the read value
System.out.println("Your char is = " + in);
如果您還想檢查它是否是字母而不是數(shù)字，只需在 while 循環(huán)中添加一個檢查即可。
添加回答
舉報
0/150
提交
取消