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研究我的双向链表的搜索方法。我遇到了异常，但我似乎无法弄清楚如何在没有得到它们的情况下遍历列表..

    public void searchEntryNode() {
    System.out.println("I'll search through each entry to pull up what you're looking for ");
    System.out.println("Type in what you want ");
    String searchEntry = keyboard.next();

    EntryNode n = head;
    while (head != null) {
        if (head.getFirstName().contains(searchEntry) || head.getLastName().contains(searchEntry) || head.getPhoneNum().contains(searchEntry) || head.getEmail().contains(searchEntry)) { 
            System.out.println("Found a matching entry");
            System.out.println(n.getFirstName() + " " + n.getLastName() + " " + n.getEmail() + " " + n.getPhoneNum());
        }

        if (head.getNext() != null) {
            head = head.getNext();
        }   
        else {
                System.out.println("That's all we found ");
                System.out.println();
                menu();
        }
    }
}
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1 回答 1

2

我没有你的行数，所以我在这里盲目猜测，但我会猜测这条线是你的问题：

if (head.getFirstName().contains(searchEntry) || head.getLastName().contains(searchEntry) || head.getPhoneNum().contains(searchEntry) || head.getEmail().contains(searchEntry))

您的其中一个条目可能会返回null、getFirstName、getLastName或。getPhoneNumgetEmail

您必须在取消引用之前检查每个不是null（您不能这样做null.someMethod()）

这样做的一种方法：

string firstName = head.getFirstName();
string lastName = head.getLastName();
string phoneNum = head.getPhoneNum();
string email = head.getEmail();

if ((firstName != null && firstName.contains(searchEntry)) 
     || (lastName != null && lastName.contains(searchEntry)) 
     || (phoneNum != null && phoneNum.contains(searchEntry)) 
     || (email != null && email.contains(searchEntry)))
于 2012-03-04T23:37:04.060 回答