-1 
    String name;
String count;
Scanner keyboardn = new Scanner(System.in);
Scanner keyboard = new Scanner(System.in);
System.out.println("Enter Full Product Name");
name = keyboard.nextLine();
count = keyboard.nextLine();
String [] p = name.split(" ");
for(String s : p) { 
    if (count == "1"){  System.out.println(p[0]);}
    if (count == "2"){  System.out.println(p[1]);}
    if (count == "3"){System.out.println(p[1]);}
    if (count == "4"){System.out.println(p[1]);}
    if (count == "5"){System.out.println(p[1]);}
    if (count == "6"){System.out.println(p[1]);}
    if (count == "7"){System.out.println(p[1]);}
    if (count == "8"){System.out.println(p[1]);}
    if (count == "9"){System.out.println(p[1]);}
    if (count == "10"){System.out.println(p[0]+", "+p[1]+", "+ p[2]+", "+p[3]+", "+p[4]+", "+p[5]+", "+p[6]+", "+p[7]+", "+p[8]+", "
                + ""+p[9]+", "+p[1]+p[2]+", "+p[1]+p[2]+p[4]+", "+p[1]+p[2]+p[4]+p[5]+", "  );
    }
}

该程序没有使用如果我认为最有效的语句做出任何回应，如果有任何其他建议，请告诉我

4

1 回答 1

1

您已声明count为String. 您需要将其声明为 anint或 use count.equals("1")。您无法Strings在 Java 中使用==.

于 2013-09-26T03:23:46.623 回答