java - 它到达 if 语句并停止工作，我做错了什么(java 新手)

Question

这个问题已经有答案了:

Can't use Scanner.nextInt() and Scanner.nextLine() together [duplicate] (1 个回答)

已关闭 9 年前。

import java.util.Scanner;

public class JavaMovies {
      Scanner det = new Scanner(System.in);
      int age = det.nextInt();

public static void main(String[] args){
    calculator();


}


public static void calculator(){

Scanner det = new Scanner(System.in);

    String rMovie = "Attack of the nerds";
    String mMovie = "Nerd lyfe";
    String pgMovie = "Nerd adventures";
    String gMovie = "Nerd playtime";
    System.out.println("These are the movies that are playing this weekend"
            + rMovie
            + mMovie
            + pgMovie
            + gMovie);
    System.out.println("Please input your age");
   int age = det.nextInt();


System.out.print("Input the movie you wish to see:");
    String x = det.nextLine();                             //Program ends here...

    if(x.equals("Attack of the nerds") & age > 17){
        System.out.println("You can see Attack of the nerds");

    }else if(x.equals("Attack of the nerds") & age < 18){
        System.out.println("You can see attack of the nerds!");

    }if(x.equals("Nerd lyfe")& age > 15){
        System.out.println("You can see Nerd lyfe");

    }else if(x.equals("Nerd lyfe")& age < 16){
        System.out.println("You can not see Nerd lyfe");

    }if (x.equals("Nerd adventures")& age > 10){
        System.out.println("You can see Nerd adventures");

    }else if(x.equals("Nerd adventures")& age < 11){
        System.out.println("You can not see Nerd adventures");

    }if(x.equals("Nerd playtime")& age > 5){
        System.out.println("You can see Nerd playtime");

    }else if(x.equals("Nerd playtime")& age < 6){
        System.out.println("You can not see Nerd playtime");


    }
}
}

在 if 语句开始之前一切都很好，我能做些什么来解决这个问题？

控制台输出(使用netbeans)运行:这些是本周末播放的电影 Nerd 的攻击 Nerd lyfe Nerd 冒险 Nerd 玩耍时间请输入您的年龄14输入您想看的电影:BUILD SUCCESSFUL(总时间:4秒)

Answer 1

我认为问题在于 nextInt 没有用完整行。如果您在提示输入年龄时输入 32，nextInt 将使用 3 和 2 字符，但有输入字符串中仍然剩余一个换行符。然后，当您调用 nextLine 时，效果是返回当前行中剩余的所有内容，直到下一个换行符。由于当前行仍然有换行符，这会导致 String x = det.nextLine() 将 x 设置为空字符串。

在提示播放电影之前尝试det.nextLine();。 (您不需要将其分配给变量；只需丢弃结果即可。)这将导致扫描器消耗换行符，以便下一个 nextLine() 将获得另一行输入.