我使用扫描器方法nextInt()和nextLine()读取输入。
它是这样的:
System.out.println("Enter numerical value");
int option;
option = input.nextInt(); // Read numerical value from input
System.out.println("Enter 1st string");
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)
问题是在输入数值之后，第一个input.nextLine()被跳过，第二个input.nextLine()被执行，因此我的输出看起来像这样:
Enter numerical value
3 // This is my input
Enter 1st string // The program is supposed to stop here and wait for my input, but is skipped
Enter 2nd string // ...and this line is executed and waits for my input
我测试了我的应用程序，看起来问题在于使用input.nextInt()。如果我删除它，那么string1 = input.nextLine()和string2 = input.nextLine()都按照我想要的方式执行。
如果我期望一个非空的输入
避免:
-如果以下输入被未检查的scan.nextLine()作为解决方法吃掉，则数据丢失
因为scan.nextLine()被scan.next()取代而导致的数据丢失(输入:"yippie ya yeah")
-使用Scanner方法解析输入时抛出的异常(先读取，然后解析)
public static Function<Scanner,String> scanLine = (scan -> {
String s = scan.nextLine();
return( s.length() == 0 ? scan.nextLine() : s );
});
用于上面的例子:
System.out.println("Enter numerical value");
int option = input.nextInt(); // read numerical value from input
System.out.println("Enter 1st string");
String string1 = scanLine.apply( input ); // read 1st string
System.out.println("Enter 2nd string");
String string2 = scanLine.apply( input ); // read 2nd string
那是因为扫描仪。nextInt方法不读取通过按“Enter”创建的输入中的换行符，因此调用Scanner。nextLine在读取换行符后返回。
当您使用Scanner时，您将遇到类似的行为。在Scanner.next()或任何Scanner之后的nextLine。nextFoo方法(除了nextLine本身)。
处理:
Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline
int option = input.nextInt();
input.nextLine(); // Consume newline left-over
String str1 = input.nextLine();
Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.
int option = 0;
try {
option = Integer.parseInt(input.nextLine());
} catch (NumberFormatException e) {
e.printStackTrace();
}
String str1 = input.nextLine();
