如果函数不起作用的 Java 扫描器

Question

现在选择菜单上的每个选项都可以正常工作(感谢 Tarek Salah 和 dasblinkenlight 的帮助)。现在的问题是，当我选择一个需要我输入新单词的选项(例如选项 3，用户必须输入歌曲名称)时，它会跳过该选项并返回到菜单。有谁知道如何阻止这种情况发生，以便用户可以实际输入某些内容？

import java.util.Scanner;
public class JukeboxApp { 
    public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    Jukebox jb = new Jukebox();
    boolean check = false;
    System.out.println("Please enter the corresponding number to perform said action.");
    while ( check == false ) {
        System.out.println("1: Add a song to the JukeBox\n" +
                            "2: Remove a song from the JukeBox\n" +
                            "3: Search for a specific song\n" +
                            "4: Display total price to play all songs\n" +
                            "5: Display the most expensive song\n" +
                            "6: Display the shortest song\n" +
                            "7: Display the most played song\n" +
                            "8: Display all songs in the JukeBox\n" +
                            "9: Display all songs by a specific artist\n" +
                            "10:\n" +
                            "11: Exit the JukeBox");
        int num = sc.nextInt();
        if ( num == 1 ) {
            System.out.println("Please enter the name of the artist");
            String artist = sc.next();
            sc.nextLine();
            System.out.println("Please enter the title of the song");
            String title = sc.nextLine();
            System.out.println("Please enter the price of the song");
            double price = sc.nextDouble();
            System.out.println("Please enter the length of the song");
            double length = sc.nextDouble();
            Song s1 = new Song(artist, title, price, length);
            jb.addSong(s1);
        } else if ( num == 2 ) {
            System.out.println("Please enter the title of the song you would like to remove");
            sc.nextLine();
            jb.removeSong(sc.nextLine());
        } else if ( num == 3 ) {
            System.out.println("Enter the title of the song you are searching for");
            jb.searchSong(sc.nextLine());
        } else if ( num == 4 ) {
            System.out.println(jb.calcTotal());
        } else if ( num == 5 ) {
            System.out.println(jb.showMostExpensive());
        } else if ( num == 6 ) {
            System.out.println(jb.showShortest());
        } else if ( num == 7 ) {
            System.out.println(jb.mostPlayed());
        } else if ( num == 8 ) {
            jb.displaySongs();
        } else if ( num == 9 ) {
            System.out.println("Please enter the artist you are searching for");
            System.out.println(jb.searchArtist(sc.nextLine()));
        } else if ( num == 10 ) {
            System.out.println("");
        } else if ( num == 11 ) {
            System.out.println("Thank you for using my JukeBox.");
            check = true;
        }
    }
}

}

Answer 1

这不起作用的原因是，当您希望用户仅输入一个值时，您可能会多次调用 sc.nextInt()。

您应该将结果存储在变量中，并在所有 if 语句中使用该变量。或者，您可以使用 switch 语句。

var userEntry = sc.nextInt();
sc.nextLine(); // skip to the end of the line
if ( userEntry  == 1 ) {
    ...
} else if ( userEntry  == 2 ) {
    ...
} else if ( userEntry  == 3 ) {
    ...
} else ...

或

var userEntry = sc.nextInt();
sc.nextLine(); // skip to the end of the line
switch ( userEntry ) {
    case 1:
        ...
    break;
    case 2:
        ...
    break;
    case 3:
        ...
    break;
    default:
        ...
    break;
}