Question

我试图弄清楚如何处理过滤项目。我能够一次按一个字段过滤我的收藏集（由于使用了番石榴Immutable.of），但是我不得不创建一种方法来一次又一次地过滤我的收藏集，直到用户告知为止（例如，切换），他已经完成并想返回主菜单。

private String filterOption() {
    System.out.println("What fields should I filter by? :\n" +
            "title\n" +
            "author\n" +
            "content\n" +
            "date\n");
    String filteringOption = scanner.nextLine();
    return filteringOption;
}

private String filterValue() {
    System.out.println("What value you want to use for filtering? :\n");
    String userInput = scanner.nextLine();
    return userInput;
}

public List<Message> filterMessages() {

    String filteringOption = filterOption();
    String userInput = filterValue();

    Map<String, BiPredicate<String, Message>> criteria = ImmutableMap.of(
            "title", (userTitle, message) -> userInput.equals(message.getTitle()),
            "author", (userAuthor, message) -> userInput.equals(message.getAuthor()),
            "content", (userContent, message) -> userInput.equals(message.getContent()),
            "date", (userDate, message) -> userInput.equals(message.getCreationDate())
    );

    BiPredicate<String, Message> predicate = criteria.get(filteringOption);

    filteredMessages = messageStorage.getAll().stream() //getAll() is to get entire collection
            .filter(m -> predicate.test(userInput, m))
            .collect(Collectors.toList());

    return filteredMessages;
}

有人可以告诉我一种方法，允许用户使用从filterMessages（）获得的值再次使用相同的方法对其进行过滤吗？简而言之-过滤一下曾经过滤的内容，直到感到满意或返回列表中没有任何内容为止。

Answer 1

可能类似：

List<Message> messageStorage = new ArrayList<>(); // messageStorage.getAll()
String userChoice = "";
while (!userChoice.equalsIgnoreCase("No")) {
    messageStorage = filterMessages(messageStorage); // modifying the existing list
    System.out.println("Want to continue filtering? Yes/No");
    userChoice = scanner.next();
}

我刚刚将您的filterMessages方法调整为：

List<Message> filterMessages(List<Message> messageStorage) {
    String filteringOption = filterOption();
    String userInput = filterValue();
    Map<String, BiPredicate<String, Message>> criteria = ImmutableMap.of(
            "title", (userTitle, message) -> userInput.equals(message.getTitle()),
            "author", (userAuthor, message) -> userInput.equals(message.getAuthor()),
            "content", (userContent, message) -> userInput.equals(message.getContent()),
            "date", (userDate, message) -> userInput.equals(message.getCreationDate())
    );

    return messageStorage.stream()
            .filter(m -> criteria.get(filteringOption).test(userInput, m))
            .collect(Collectors.toList());
}

Java-为过滤函数创建循环

1 个答案: