我只是在为 Java 暑期类准备书中的一些练习,因为我有点超前了,这意味着这不是家庭作业。我收到一条错误消息,指出您无法将 String
转换为 int
,但两者都是字符串,一个是变量,一个是数组。
我遇到问题的是这一行... select = items[select];
public class CarpentryChoice {
public static void main(String[] args) {
String items [] = {"Table", "Desk", "Chair", "Couch", "Lazyboy"};
int price [] = {250, 175, 125, 345, 850};
String select;
Scanner scan = new Scanner(System.in);
System.out.println("Please enter an item to view it's price: ");
select = scan.nextLine();
for(int i=0; i < items.length; i++) {
select = items[select];
}
}
}
最佳答案
因为 select
是一个 String
变量,所以它不能用作数组中的索引。
select = items[select];
我相信您打算在 for
循环中使用索引值 i
(其中 i
是 0
到 items.length
)。有点像
select = items[i];
但是,根据您在下面的评论,我相信您确实想要
int select = scan.nextInt();
System.out.println("You selected: " + items[select]);
根据您的编辑,您可以使用两个数组和两个循环来完成。类似的东西,
String items[] = { "Table", "Desk", "Chair", "Couch", "Lazyboy" };
int price[] = { 250, 175, 125, 345, 850 };
Scanner scan = new Scanner(System.in);
int pos = -1;
outer: while (pos == -1) {
System.out.println("Please enter an item to view it's price: ");
String select = scan.nextLine();
for (int i = 0; i < items.length; i++) {
if (items[i].equalsIgnoreCase(select.trim())) {
pos = i;
break outer;
}
}
}
if (pos != -1) {
System.out.printf("The price is %d%n", price[pos]);
}
但是 Map
会(在我看来)是一个更好的解决方案(它肯定更有效)。喜欢,
String[] items = { "Table", "Desk", "Chair", "Couch", "Lazyboy" };
int[] price = { 250, 175, 125, 345, 850 };
Map<String, Integer> map = new HashMap<>();
for (int i = 0; i < items.length; i++) {
map.put(items[i], price[i]);
}
Scanner scanner = new Scanner(System.in);
System.out.println("Please enter an item to view it's price: ");
while (scanner.hasNextLine()) {
String item = scanner.nextLine().trim();
if (map.containsKey(item)) {
System.out.printf("The price is %d%n", map.get(item));
}
System.out.println("Please enter an item to view it's price: ");
}
关于java - 无法将 String 转换为 int,但两者都是字符串?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31107966/