[英]How to check if the System.in contains only numbers
有一个任务从控制台获取 N 个数字,找到最长和最短的数字及其长度。 该任务并不困难并且可以正常工作,但是我决定检查一下控制台输入是否符合任务的条件:
我决定编写一个 boolean 方法 isInputCorrect(),它会使用 Scanner 并检查输入是否正确,但它不能正常工作。
public static void main(String[] args) {
int n = 5;
Scanner sc = new Scanner(System.in);
do {
System.out.println("Hello, please enter " + n + " integer numbers:");
while (!isInputCorrect(sc,n)){
System.out.println("Wrong! Try again:");
sc.next();
}
} while (!isInputCorrect(sc, 5));
String scLine = sc.nextLine();
String[] arr = scLine.split("\\s+");
String maxLengthNum = arr[0];
String minLengthNum = arr[0];
for (int i = 1; i < arr.length; i++){
if (maxLengthNum.length() < arr[i].length()){
maxLengthNum = arr[i];
}
if (minLengthNum.length() > arr[i].length()){
minLengthNum = arr[i];
}
}
String equalMaxNum = "";
String equalMinNum = "";
int countMax = 0;
int countMin = 0;
for (String s : arr){
if (maxLengthNum.length() == s.length()){
countMax += 1;
equalMaxNum += s + " ";
}
if (minLengthNum.length() == s.length()){
countMin += 1;
equalMinNum += s + " ";
}
}
if (countMax > 1){
System.out.println("The longest numbers are: " + equalMaxNum + " Their length is: " + maxLengthNum.length());
}
else {
System.out.println("The longest number is: " + maxLengthNum + " Its length is: " + maxLengthNum.length());
}
if (countMin > 1){
System.out.println("The shortest numbers are: " + equalMinNum + " Their length is: " + minLengthNum.length());
}
else {
System.out.println("The shortest number is: " + minLengthNum + " Its length is: " + minLengthNum.length());
}
}
public static boolean isInputCorrect(Scanner sc, int n){
boolean flag = true;
for (int i = 0; i < n; i++){
if (sc.hasNextInt()){
sc.next();
}else {
flag = false;
break;
}
}
return flag;
}
编辑此代码仍然不起作用。 我意识到,问题出在 isDigit() 中。 并且在最后一个 if 语句中完全符合常规条件。 它是这样的:
public static boolean isDigit (String input, int n){
String[] arr = input.split("\\s+");
boolean flag = false;
if (arr.length != n){
flag = false;
}
else {
for (String s : arr) {
if (s.startsWith("-")) {
if (s.substring(1).matches("[0-9]*")) {
flag = true;
}
} else if (s.matches("[0-9]*")) {
flag = true;
} else {
flag = false;
}
}
}
return flag;
}
此方法将控制台输入作为字符串,然后检查它包含多少个数字(字符串),是否有任何负数等等。 但它只能应用于子字符串(没有空格的单词)。 就我而言,它可以应用于 arr[i]。
所以我修改它以将 String 拆分为 array[] 并尝试检查每个元素。 我有:
public static boolean isDigit (String input, int n){
String[] arr = input.split("\\s+");
boolean flag = false;
if (arr.length != n){
flag = false;
}
else {
for (String s : arr) {
if (s.startsWith("-")) {
if (s.substring(1).matches("[0-9]*")) {
flag = true;
}
} else if (s.matches("[0-9]*")) {
flag = true;
} else {
flag = false;
}
}
}
return flag;
}
但即使输入为:1 3213 w 15 3 我也无法理解,它返回 true,有什么问题? 完整的代码是:
public static void main(String[] args) {
int n = 5;
boolean validInput = false;
String input;
do {
System.out.println("Please enter " + n + " integer numbers:");
Scanner sc = new Scanner(System.in);
input = sc.nextLine();
if (isDigit(input, n)) {
validInput = true;
} else {
System.out.println("Wrong input! Try again: ");
}
}
while (!validInput);
String[] arr = input.split("\\s+");
String maxLengthNum = arr[0];
String minLengthNum = arr[0];
for (int i = 1; i < arr.length; i++){
if (maxLengthNum.length() < arr[i].length()){
maxLengthNum = arr[i];
}
if (minLengthNum.length() > arr[i].length()){
minLengthNum = arr[i];
}
}
String equalMaxNum = "";
String equalMinNum = "";
int countMax = 0;
int countMin = 0;
for (String s : arr){
if (maxLengthNum.length() == s.length()){
countMax += 1;
equalMaxNum += s + " ";
}
if (minLengthNum.length() == s.length()){
countMin += 1;
equalMinNum += s + " ";
}
}
if (countMax > 1){
System.out.println("The longest numbers are: " + equalMaxNum + " Their length is: " + maxLengthNum.length());
}
else {
System.out.println("The longest number is: " + maxLengthNum + " Its length is: " + maxLengthNum.length());
}
if (countMin > 1){
System.out.println("The shortest numbers are: " + equalMinNum + " Their length is: " + minLengthNum.length());
}
else {
System.out.println("The shortest number is: " + minLengthNum + " Its length is: " + minLengthNum.length());
}
}
public static boolean isDigit (String input, int n){
String[] arr = input.split("\\s+");
boolean flag = false;
if (arr.length != n){
flag = false;
}
else {
for (String s : arr) {
if (s.startsWith("-")) {
if (s.substring(1).matches("[0-9]*")) {
flag = true;
}
} else if (s.matches("[0-9]*")) {
flag = true;
} else {
flag = false;
}
}
}
return flag;
}
解决了
谢谢大家,你们的帮助真的很有用。 我终于找到了问题它在 isDigit() 方法中。 它正在检查数组的每个元素,并根据最后的结果切换一个标志。 我写了“break”来停止进一步检查循环中是否至少有一个错误标志。
public static boolean isDigit (String input, int n){
String[] arr = input.split("\\s+");
boolean flag = false;
if (arr.length != n){
flag = false;
}
else{
for (String s: arr){
if (s.startsWith("-")) {
if (s.substring(1).matches("[0-9]*")){
flag = true;
}
else {
flag = false;
break;
}
}
else {
if (s.matches("[0-9]*")){
flag = true;
}
else {
flag = false;
break;
}
}
}
}
return flag;
}
您可以使用正则表达式来验证输入是否仅包含 integer 数字:
int n = 5;
// ... your current code
String scLine = sc.nextLine();
if (!scLine.matches("\\d+(?:\\s+\\d+)*")) {
throw new IllegalArgumentException("input contained non-integers");
}
String[] arr = scLine.split("\\s+");
if (arr.length != n) {
throw new IllegalArgumentException("found " + arr.length + " number inputs but expected " + n + ".");
}
您可以检查输入字符串是否只有数字,如下所示
public boolean isDigit(String input) {
if (input == null || input.length() < 0)
return false;
input = input.trim();
if ("".equals(input))
return false;
if (input.startsWith("-")) {
return input.substring(1).matches("[0-9]*");
} else {
return input.matches("[0-9]*");
}
}
编辑:
允许用户重新输入,直到输入有效数字
boolean validInput = false;
do
{
System.out.println("Enter the number ");
// get user input
String input sc.nextLine();
if(isDigit(input))
validInput = true;
else
System.out.println("Enter valid Number");
}
while (!validInput );
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