微博mid和id转换

mid为62进制编码,id为常见的10进制编码。

id从低位到高位,7个数字为一组,转换为62进制,并顺序合并,即转换为mid。

mid从地位到高位,4个字母为一组,转换为10进制,并右移7位,计算和,得到id。

 

7位的10进制数最大为9999999,转换为62进制后为FXsj;8位的10进制数最大为99999999,转换为62进制后为9FXsj。

7位分隔10进制数,能保证最多获得4位62进制数。

 

 

参考了http://blog.csdn.net/dier4836/article/details/7827908,但是他的mid计算id有问题。

package com.founder.weibocrawler.util;

public class WeiboIDUtils {

    public static void main(String[] args) {
        System.out.println(Mid2Id("D5DVt7A9n"));
        System.out.println(Id2Mid("3913451191807261"));
    }

    public static String Mid2Id(String mid) {
        long id = 0L;
        // 从最后往前以4字节为一组读取字符
        int count = 0;
        for (int i = mid.length() - 4; i > -4; i = i - 4) {
            int offset = i < 0 ? 0 : i;
            int len = i < 0 ? mid.length() % 4 : 4;
            String str = Encode62ToLong(mid.substring(offset, offset + len)).toString();
            // 不足7位补0
            if (offset != 0)
                str = String.format("%07d", Long.valueOf(str));
            id = (long) (Long.valueOf(str) * Math.pow(10, 7 * count) + id);
            count++;
        }
        return id + "";
    }

    public static String Id2Mid(String id) {
        String mid = "";
        for (int i = id.length() - 7; i > -7; i = i - 7) {
            int offset = i < 0 ? 0 : i;
            int len = i < 0 ? id.length() % 7 : 7;
            String str = LongToEnode62(Long.valueOf(id.substring(offset, offset + len)));
            mid = str + mid;
        }
        return mid;
    }

    private static String str62keys = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";

    public static Long Encode62ToLong(String str62) {
        long long10 = 0;
        for (int i = 0; i < str62.length(); i++) {
            double n = str62.length() - i - 1;
            long10 += Long.valueOf(str62keys.indexOf(str62.substring(i, i + 1)) * (long) Math.pow(62, n) + "");
        }
        return long10;
    }

    public static String LongToEnode62(long long10) {
        String str62 = "";
        int offset = 0;
        while (long10 != 0) {
            offset = Integer.valueOf(long10 % 62 + "");
            str62 = str62keys.substring(offset, offset + 1) + str62;
            long10 = (long) Math.floor(long10 / 62.0);
        }
        return str62;
    }
}

 

posted @ 2016-10-11 11:05  mahuan2  阅读(634)  评论(0编辑  收藏  举报