我目前正在尝试弄清楚如何搜索我的数组以查找数组中的特定元素。用户将输入他们要查找的名称,我的程序应该将他们所坐的座位返回给用户。添加乘客是通过另一种方法完成的。所以,如果我有一个叫“Tim Jones”的人坐在 5 号座位上,如果我使用这种方法,我会输入“Tim Jones”,程序应该告诉我 Tim Jones 坐在 5 号座位上。我得到的当前输出只是 else 语句,无论我尝试什么都找不到乘客。有小费吗?提前致谢。private static void findPassengerSeat(String airplaneRef[]) { String passengerName; Scanner input = new Scanner(System.in); System.out.println("Enter passenger's name."); // asks user for passenger name. passengerName = input.next(); // user input stored under variable passengerName. for (int i = 0; i < airplaneRef.length; i++) { // for i, if i is less than the length of the array airplaneRef, increment i. if (airplaneRef[i].equalsIgnoreCase(passengerName)) { // iterates through all elements in the array, ignoring case sensitivity and looks for the passenger. int seat = i; System.out.println(passengerName + " is sitting in seat s" + seat); // prints name of passenger and what seat they are sitting in. } else { System.out.println("Passenger not found."); break; } }}
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慕桂英3389331
TA贡献2036条经验 获得超8个赞
你应该摆脱,else否则你只评估你的数组的第一个索引并传递所有其余的。
以下应该发挥它的魅力:您基本上浏览了所有乘客的列表,只有当没有一个与输入相对应时,您才声明未找到该乘客。
for (int i = 0; i < airplaneRef.length; i++) { // for i, if i is less than the length of the array airplaneRef, increment i.
if (airplaneRef[i].equalsIgnoreCase(passengerName)) { // iterates through all elements in the array, ignoring case sensitivity and looks for the passenger.
int seat = i;
System.out.println(passengerName + " is sitting in seat s" + seat); // prints name of passenger and what seat they are sitting in.
return;
}
}
System.out.println("Passenger not found.");
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