Question

我目前正在进行一个项目，我本来应该用eclipse在java中创建一个电话簿，我差不多完成了但是我的代码仍然是我的代码的一部分。＃39;理解。

以下代码示例是给我一些麻烦的部分，我的条件是（... contains（...））总是回答为true，即使这些联系人都没有在其名称中包含这部分字符串。此外，我还会显示手机的整个联系人，而我只想要匹配的那些。

我已尝试使用if（... contains（..）== true），但它也无效。

另一方面，寻找与ID匹配的第二部分完美地工作并且只回馈有关的＆＃34; i＆＃34;。使用相同的代码但适应我的字符串也不会产生任何好的结果。

public static void display(){
    System.out.println("You selected display a contact, choose an option :");
    System.out.println("1. by the first name");
    System.out.println("2. by the last name");
    System.out.println("3. by the ID");
    System.out.println("4. by the phone number");

    Scanner sc = new Scanner(System.in);

    int str3 = sc.nextInt();

    if(str3 == 1){
        System.out.println("Please enter the first name of the contact");
        String fname = sc.nextLine();
        sc.nextLine();
        for(int i = 0; i <= (phoneBook.size()-1) ; i++){
            if(((phoneBook.get(i)).getFirstName()).toLowerCase().contains(fname.toLowerCase()) ){
                displayContact(phoneBook.get(i));
            }
        }}

    if(str3 == 2){
        System.out.println("Please enter the last name of the contact");
        String lname = sc.nextLine();
        sc.nextLine();
        for(int i = 0; i <= (phoneBook.size()-1) ; i++){
            if(((phoneBook.get(i)).getLastName()).toLowerCase().contains(lname.toLowerCase())){
                displayContact(phoneBook.get(i));
            }
        }
    }
    if(str3 == 3){
        System.out.println("Please enter the ID of the contact");
        int id = sc.nextInt();
        for(int i = 0; i <= (phoneBook.size()-1) ; i++){
            if(id == ((phoneBook.get(i)).getID())){
                displayContact(phoneBook.get(i));
            }
        }
    }

    if(str3 == 4){
        System.out.println("Please enter the phone number of the contact");
        int phnb = sc.nextInt();
        for(int i = 0; i <= (phoneBook.size()-1) ; i++){
            if(phnb == phoneBook.get(i).getPhoneNb()){
                displayContact(phoneBook.get(i));
            }

        }
    }
    System.out.println(" ");
    System.out.println("If you want to go back at main menu press 1, else press 2 to quit.");
    int str8 = sc.nextInt();
    if(str8 == 1){
        menu();
    }
    if(str8 == 2){
        quit();
    }

}

Answer 1

我认为使用Scanner，首先读取nextInt，然后两次使用nextLine可能会导致空字符串，并且始终包含该字符串。您可以使用第二个nextLine，或者使用BufferedReader，它不易出错。

public static void display(){
    System.out.println("You selected display a contact, choose an option :");
    System.out.println("1. by the first name");
    System.out.println("2. by the last name");
    System.out.println("3. by the ID");
    System.out.println("4. by the phone number");

    BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));

    int str3 = Integer.parseInt(sc.readLine());

    switch (str3) {
    case 1:
        System.out.println("Please enter the first name of the contact");
        String fname = sc.readLine();
        for (int i = 0; i < phoneBook.size(); i++) {
            if (phoneBook.get(i).getFirstName().toLowerCase().contains(fname.toLowerCase())) {
                displayContact(phoneBook.get(i));
            }
        }
        break;

    case 2:
        System.out.println("Please enter the last name of the contact");
        String lname = sc.readLine();
        for (int i = 0; i < phoneBook.size(); i++){
            if (phoneBook.get(i).getLastName().toLowerCase().contains(lname.toLowerCase())) {
                displayContact(phoneBook.get(i));
            }
        }
        break;

    case 3:
        System.out.println("Please enter the ID of the contact");
        int id = Integer.parseInt(sc.readLine());
        for (int i = 0; i < phoneBook.size(); i++) {
            if (id == phoneBook.get(i)).getID()) {
                displayContact(phoneBook.get(i));
            }
        }
        break;

    case 4:
        System.out.println("Please enter the phone number of the contact");
        int phnb = Integer.parseInt(sc.readLine());
        for(int i = 0; i < phoneBook.size(); i++){
            if(phnb == phoneBook.get(i).getPhoneNb()){
                displayContact(phoneBook.get(i));
            }
        }
        break;
    }
    System.out.println(" ");
    System.out.println("If you want to go back at main menu press 1, else press 2 to quit.");
    int str8 = Integer.parseInt(sc.readLine());
    if(str8 == 1){
        menu();
    }
    if(str8 == 2){
        quit();
    }
}

另外开关声明。

使用int而不是String（电话簿）

1 个答案: