Java中的 Scanner hasNextLine() 方法及示例

如果此扫描仪的输入中有另一行， Java.util.Scanner类的hasNextLine()方法将返回 true。此方法可能会在等待输入时阻塞。扫描仪不会超过任何输入。

句法：

public boolean hasNextLine()

参数：该函数不接受任何参数。

返回值：当且仅当此扫描仪有另一行输入时，此函数才返回 true

异常：如果此扫描仪关闭，该函数将抛出IllegalStateException 。

下面的程序说明了上述函数：

方案一：

// Java program to illustrate the
// hasNextLine() method of Scanner class in Java
// without parameter
  
import java.util.*;
  
public class GFG1 {
    public static void main(String[] argv)
        throws Exception
    {
  
        String s = "gfg 2 geeks!";
  
        // new scanner with the
        // specified String Object
        Scanner scanner = new Scanner(s);
  
        // use US locale to interpret Lines in a string
        scanner.useLocale(Locale.US);
  
        // iterate till end
        while (scanner.hasNextLine()) {
  
            // print what is scanned
            System.out.println(scanner.nextLine());
        }
  
        // close the scanner
        scanner.close();
    }
}

输出：

gfg 2 geeks!

程序 2：演示异常的程序

// Java program to illustrate the
// hasNextLine() method of Scanner class in Java
// without parameter
  
import java.util.*;
  
public class GFG1 {
    public static void main(String[] argv)
        throws Exception
    {
        try {
  
            String s = "gfg 2 geeks!";
  
            // new scanner with the
            // specified String Object
            Scanner scanner = new Scanner(s);
  
            // use US locale to interpret Lines in a string
            scanner.useLocale(Locale.US);
  
            scanner.close();
  
            // iterate till end
            while (scanner.hasNextLine()) {
  
                // print what is scanned
                System.out.println(scanner.nextLine());
            }
  
            // close the scanner
            scanner.close();
        }
        catch (IllegalStateException e) {
            System.out.println("Exception is: " + e);
        }
    }
}

输出：

Exception is: java.lang.IllegalStateException: Scanner closed

参考： https: Java/util/Scanner.html#hasNextLine()