Question

我会喜欢一些帮助。这里的问题是当请求的狗碰巧索引为0时，循环工作正常并且只通过第一个if语句。如果请求的狗碰巧有另一个索引，那么两个if语句都被读取......任何可以看到解决方案的人？：）

protected void removeDog() {
    String k = readString("Name of dog you want to delete from list: ");
    boolean found = false;

    for (int x = 0; x < dogRegister.size(); x++) {
        if (dogRegister.get(x).getName().equalsIgnoreCase(k)) {
            dogRegister.remove(x);

            System.out.println(k + " has been deleted from list.");

            found = true;
            break;
        }
        else if (!found) {
            System.out.println(k + " could not be found in list.");
        }
    }
}

Answer 1

每次迭代if循环时，您都会执行else if或for。如果它不是第一个，那么你将无法找到＆＃34;＆＃34;为每次迭代发送一次消息，直到找到并删除它或者您已经用完该列表。

无法找到＆＃34;＆＃34;只有在您根本找不到该消息时才打印消息，将else if移到for循环之外，删除其他内容：

for (int x = 0; x < dogRegister.size(); x++) {
    if (dogRegister.get(x).getName().equalsIgnoreCase(k)) {
        dogRegister.remove(x);
        System.out.println(k + ” has been deleted from list.”);
        found = true;
        break;
     }
}
if (!found) {
    System.out.println(k + ” could not be found in list.");
}

Answer 2

我会避免使用.get(i)读取Java集合，即使它是ArrayList。您可以使用迭代器，它们可以简化任务。

protected void removeDog() {

    String dogName = readString("Name of dog you want to delete from list:");

    boolean found = false;

    // I assume there is a class "Dog"
    for (Iterator<Dog> it = dogRegister.iterator(); it.hasNext(); ) {
        Dog dog = it.next();
        if (dog.getName().equalsIgnoreCase(dogName)) {
            it.remove();
            found = true;
            break;
        }
    }

    if (found) //print something...
    else //print something else...

}

Answer 3

protected void removeDog() {

    String k = readString(”Name of dog you want to delete from list: ");

    boolean found = false;

    for (int x = dogRegister.size(); x > 0; x--) {

        if (dogRegister.get(x).getName().equalsIgnoreCase(k)) {


            dogRegister.remove(x);
            System.out.println(k + ” has been deleted from list.”);
            found = true;
            break;

        }
        else if ( x == 0){
            System.out.println(k + ” could not be found in list.");
                }


        }

}

如果到达列表的末尾，将会出现“在列表中找不到K”。之前发生的事情是，每次运行循环时，除非在第一个位置找到您的狗，否则它将始终显示NOT FOUND你的阵列。它将一直持续到最后找到或退出。

麻烦制作循环只运行一个if语句而不是几个

3 个答案: