我需要一个程序,要求用户引入最多10个名字(最后,用户可以键入"fim"[葡萄牙语结尾](。
我目前的问题是如果用户达到 10 个名称,如何终止程序。
这是我的主要功能:
public static void main(String[] args) {
Scanner keyboard = new Scanner (System.in);
System.out.println("Introduza até 10 nomes completos com até 120 caracteres e pelo menos dois nomes com pelo menos 4 caracteres: ");
String nome = keyboard.next();
for(int i = 0; i < 10; i++) {
while(!nome.equalsIgnoreCase("fim") && i<10) {
nome = keyboard.next();
}
}
keyboard.close();
}
您正在按原样运行while
的无限循环。您想将其更改为if
语句,并仅要求fim
,如果发生这种情况,请致电break;
。
所以它应该以:
for(int i = 0; i < 10; i++) { //This will run 10 times
nome = keyboard.next();
if(nome.equalsIgnoreCase("fim")) { //This will verify if last input was "fim"
break; //This breaks the for-loop
}
}
或者,如果您真的想在for
循环中使用while
循环(不推荐(,则需要在其中增加i
:
for(int i = 0; i < 10; i++) {
while(!nome.equalsIgnoreCase("fim") && i<10) {
nome = keyboard.next();
i++;
}
}
我不是break
的忠实粉丝,所以除了 Frakcool 的出色答案之外,您还可以使用 do-while
循环:
String nome;
int i = 0;
do {
nome = keyboard.next();
i++;
}
while(!nome.equalsIgnoreCase("fim") && i<10);
同样现在,您正在覆盖所有以前输入的名称。因此,您要么必须直接在循环中处理它们,要么将它们收集在某种容器中,例如列表。我会这样重写循环:
String nome;
int i = 0;
while(i<10 && !(nome = keyboard.next()).equalsIgnoreCase("fim")) {
i++;
// Either handle nome here directly, or add it to a list for later handling.
}
您可能想尝试一下这段代码(我在注释中对正确的行进行了一些解释(:
public static void main(String[] args) {
Scanner keyboard = new Scanner (System.in);
System.out.println("Introduza até 10 nomes completos com até 120 caracteres e pelo menos dois nomes com pelo menos 4 caracteres: ");
String nome = keyboard.next();
int i = 0; // Here I instroduce a counter, to increment it after each input given
while(!nome.equalsIgnoreCase("fim") && i!=10) { // stop performing while-loop when nome is equal to "fim"
// or if i==10 (if any of these conditions is false, entire condition is false)
nome = keyboard.nextLine();
i++; // increment counter after input
}
keyboard.close();
System.out.println("End of input"); // Just to confirm that you exited while-loop
}