Android java自定义排序使用Comparable

Question

  @Override
      public int compareTo(final myRow another) {

        final int BEFORE    =-1;
        final int EQUAL     = 0;
        final int AFTER     = 1;

        if (this==another) return EQUAL;

        if (sorttype==sort_type.SORT_ABC) {

            int rv      =0;
            int sorted  =row.toLowerCase().compareTo(another.getRow().toLowerCase());

            if (this.getUserType()==user_type.USER_TYPE_BANNED) rv=AFTER;
            if (this.getUserType()==user_type.USER_TYPE_NORMAL) rv=sorted;
            if (this.getUserType()==user_type.USER_TYPE_FRIEND) rv=BEFORE;

            Log.e("sorted", row+" "+this.getUserType()+" - "+rv+" ");

            return rv;

        } else if (sorttype==sort_type.SORT_LOGINTIME) {

            if (this.getUserType()==user_type.USER_TYPE_BANNED) {
                return AFTER;
            }
            if (this.getUserType()==user_type.USER_TYPE_NORMAL) return EQUAL;
            if (this.getUserType()==user_type.USER_TYPE_FRIEND) {
                //int sorted    =row.toLowerCase().compareTo(another.getRow().toLowerCase());
                return BEFORE;
            }

            //return 0;
        }

        return 0;

        //return row.toLowerCase().compareTo(another.getRow().toLowerCase());
      }

我有一个带有昵称的字符串用户列表。我已经禁止了用户、普通用户和朋友用户。

我想将我的好友用户排序到列表顶部，将被禁止的用户排序到列表底部，我想显示它们是 ASC 样式。

我该怎么做？

所以结构是：

Abc  (friend)
abrt (friend)
dfgh (friend)

abdfg (normal user)
bnmm  (normal user)
wert  (normal user)

Andgh  (banned user)
Dfghhj (banned user)
Qwer   (banned user)

我明白了：

06-19 14:43:46.586: ERROR/AndroidRuntime(23434): java.lang.IllegalArgumentException: Comparison method violates its general contract!

Answer 1

您的代码看起来有点复杂。最简单的方法是从优先级最高的字段开始，然后向下移动到其他字段。示例代码如下：

public class Member implements Comparable<Member> {

    static enum Status {
        NORMAL(1), FRIEND(2), BANNED(3);

        private final int order;

        Status(int order) {
            this.order = order;
        }

        public int getOrder() {
            return this.order;
        }

    };

    private final String name;

    private final Status status;

    public Member(final String name, final Status status) {
        this.name = name;
        this.status = status;
    }

    @Override
    public int compareTo(Member o) {
        if (this.status.equals(o.status)) {
            return this.name.compareTo(o.name);
        } else {
            return this.status.compareTo(o.status);
        }
    }

    @Override
    public String toString() {
        return "Member [name=" + name + ", status=" + status + "]";
    }

    public static void main(String[] args) throws Throwable {
        Member[] members = {
                        new Member("abrt", Status.FRIEND),
                        new Member("dfgh", Status.FRIEND),
                        new Member("abdf", Status.NORMAL),
                        new Member("wert", Status.NORMAL),
                        new Member("andgh", Status.BANNED),
                        new Member("qwer", Status.BANNED)
        };
        List<Member> lst = Arrays.asList(members);
        Collections.sort(lst);
        System.out.println(lst);
    }


}

Answer 2

您的 ABC 排序逻辑不起作用。例如，如果您传入两个 USER_TYPE_FRIEND 对象，无论它们各自的顺序如何，compareTo 将始终返回 BEFORE。

您需要通过首先比较用户类型来实现这一点。

如果它们相等，您可以返回您的 row.compareTo(...) 表达式。

如果不是，您需要返回之前/之后仅取决于这些类型在您的逻辑中“比较”的方式（即朋友

Answer 3

我现在的代码：

  public enum user_type {USER_TYPE_FRIEND(1), USER_TYPE_NORMAL(2), USER_TYPE_BANNED(3);
      private final int order;
      user_type(int order) {
          this.order = order;
      }
      public int getOrder() {
          return this.order;
      }
  }

...

  @Override
  public int compareTo(final myRow another) {

    if (sorttype==sort_type.SORT_ABC) {
        if (this.getUserType().equals(another.getUserType())) {
            return this.getRow().toLowerCase().compareTo(another.getRow().toLowerCase());
        } else {
            return this.getUserType().compareTo(another.getUserType());
        }
    } 
    else {
        //LOGINTIME
        return this.getUserType().compareTo(another.getUserType());
    }
  }

谢谢，桑杰！