Java.util.Scanner.nextLong() Java.util.Scanner.nextLine() Java.util.Scanner.nextLong() 描述 所述java.util.Scanner.nextLong()方法扫描输入作为表单nextLong(此方法的一个long.An调用的下一个标记)的行为以完全相同的方式调用nextLong(),其中基数是此扫描仪的默认基数。 声明 以下是java.util.Scanner.nextLong()方法的声明 public long nextLong() 参数 NA 返回值 此方法返回从输入扫描的长扫描 异常 InputMismatchException - 如果下一个标记与Integer正则表达式不匹配,或者超出范围 NoSuchElementException - 如果输入用尽 IllegalStateException - 如果此扫描程序已关闭 实例 以下示例显示了java.util.Scanner.nextLong()方法的用法。 package com.tutorialspoint; import java.util.*; public class ScannerDemo { public static void main(String[] args) { String s = "Hello World! 3 + 3.0 = 6.0 true "; Long l = 13964599874l; s = s + l; // create a new scanner with the specified String Object Scanner scanner = new Scanner(s); // find the next long token and print it // loop for the whole scanner while (scanner.hasNext()) { // if no long is found, print "Not Found:" and the token System.out.println("Not Found :" + scanner.next()); // if the next is a long, print found and the long if (scanner.hasNextLong()) { System.out.println("Found :" + scanner.nextLong()); } } // close the scanner scanner.close(); } } 让我们编译并运行上面的程序,这将产生以下结果 Not Found :Hello Not Found :World! Found :3 Not Found :+ Not Found :3.0 Not Found := Not Found :6.0 Not Found :true Found :13964599874 Java.util.Scanner.nextLine() Java.util.Scanner.nextLong()