我试图将键盘输入转换为整数,但我的程序总是崩溃。当输入诸如"k“之类的字符时,它可以工作,但当我输入"5”时,它就崩溃了。知道我做错了什么吗?
// Getting an integer value.
public static int getInt() {
int numberEntered = 0;
String entry = "";
Scanner keyboard = new Scanner(System.in);
while (!keyboard.hasNextInt()) {
entry = keyboard.next();
System.out.println("That is not an integer. " + "Please try again.");
}
numberEntered = Integer.parseInt(entry);
System.out.print(numberEntered);
return numberEntered;
}
输出:
Error given: k That is not an integer.
Please try again.
8
Exception in thread "main" java.lang.NumberFormatException: For input string: "k" at
java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.base/java.lang.Integer.parseInt(Integer.java:652)
at java.base/java.lang.Integer.parseInt(Integer.java:770)
at Program2.getInt(Program2.java:56)
at Program2.problemSelectionMenu(Program2.java:40)
at Program2.main(Program2.java:14)
发布于 2018-10-28 02:09:15
您检查以确保输入有下一个int,但是一旦Scanner
有了下一个int,您就永远不会将int解析为entry
,所以它仍然是错误的输入。您需要将用户输入的int
分配给entry
。只需调用nextInt()
即可轻松完成此操作:
while (!keyboard.hasNextInt()) {
entry = keyboard.next();
System.out.println("That is not an integer. " + "Please try again.");
}
numberEntered = keyboard.nextInt();
System.out.print(numberEntered);
return numberEntered;
发布于 2018-10-28 02:33:27
当您提供字符作为输入时,while循环条件为真,因此它进入while循环并扫描字符和打印值,但当您提供整数时,while循环条件变为false,它不会进入while循环。在while循环之外,您正在解析整数,您不需要这样做,因为您将integer作为输入。您需要做的是,在integer.parseInt( input.all )的位置,您必须扫描整数,即int I =keyboard.nextInt();在while循环条件中,您只需检查给定的输入是否为整数。但是在整数的情况下,您还没有扫描输入。
试试这个!
int numberEntered = 0;
String entry = "";
Scanner keyboard = new Scanner(System.in);
while (!keyboard.hasNextInt()) {
entry = keyboard.next();
System.out.println("That is not an integer. " +"Please try again.");
}
numberEntered = keyboard.nextInt();
System.out.print(numberEntered);
https://stackoverflow.com/questions/53024808
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