Java：如何读取输入 int
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Java: how to read an input int
提问by Marco
So, I was looking for an efficient way, using Java's standard packages, to read an input integer... For example, I came across the class "Scanner", but I found two main difficulties:
所以，我正在寻找一种有效的方法，使用 Java 的标准包来读取输入整数......例如，我遇到了“扫描仪”类，但我发现了两个主要困难：
- if I don't insert an int, I'm not actually able to solve the exception;
- this class works with tokens, but my aim is to load the string in its full length.
- 如果我不插入 int，我实际上无法解决异常；
- 这个类使用令牌，但我的目标是加载完整长度的字符串。
This is an example of execution I would like to realize:
这是我想实现的执行示例：
Integer: eight
Input error - Invalid value for an int.
Reinsert: 8 secondtoken
Input error - Invalid value for an int.
Reinsert: 8
8 + 7 = 15
And this is the (incorrect) code I tried to implement:
这是我试图实现的（不正确的）代码：
import java.util.Scanner;
import java.util.InputMismatchException;
class ReadInt{
public static void main(String[] args){
Scanner in = new Scanner(System.in);
boolean check;
int i = 0;
System.out.print("Integer: ");
do{
check = true;
try{
i = in.nextInt();
} catch (InputMismatchException e){
System.err.println("Input error - Invalid value for an int.");
System.out.print("Reinsert: ");
check = false;
}
} while (!check);
System.out.print(i + " + 7 = " + (i+7));
}
}
采纳答案by xagyg
Use a BufferedReader. Check NumberFormatException. Otherwise very similar to what you have. Like so ...
使用 BufferedReader。检查 NumberFormatException。否则与您拥有的非常相似。像这样...
import java.io.*;
public class ReadInt{
public static void main(String[] args) throws Exception {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
boolean check;
int i = 0;
System.out.print("Integer: ");
do{
check = true;
try{
i = Integer.parseInt(in.readLine());
} catch (NumberFormatException e){
System.err.println("Input error - Invalid value for an int.");
System.out.print("Reinsert: ");
check = false;
}
} while (!check);
System.out.print(i + " + 7 = " + (i+7));
}
}
回答by Doorknob
To use with tokens:
与令牌一起使用：
int i = Integer.parseInt(in.next());
Then you could do:
那么你可以这样做：
int i;
while (true) {
System.out.print("Enter a number: ");
try {
i = Integer.parseInt(in.next());
break;
} catch (NumberFormatException e) {
System.out.println("Not a valid number");
}
}
//do stuff with i
That above code works with tokens.
上面的代码适用于令牌。
