2020-03-23 B站推出了全新的稿件视频idbvid来接替之前的avid，其意义与之相同
详见：
“bvid”恒为长度为 12 的字符串，前两个字母为大写“BV”，后 10 个为 base58 计算结果
“bvid"为“avid”的base58编码，可通过算法进行相互转化
从 2009-09-09 09:09:09 av2 的发布到 2020-03-28 19:45:02 av99999999 的发布B站结束了以投稿时间为顺序的avid发放，改为随机发放avid
暗示B站东方要完？泪目
算法以及程序主要参考知乎@mcfx的回答
实际上该算法并不完整,新的算法参考自【揭秘】av号转bv号的过程
注：本算法及示例程序仅能编解码avid < 29460791296，且暂无法验证avid >= 29460791296的正确性
再注：本人不清楚新算法能否编解码avid >= 29460791296
- a = (avid ⊕ 177451812) + 100618342136696320
- 以 i 为循环变量循环 6 次 b[i] = (a / 58 ^ i) % 58
- 将 b[i] 中各个数字转换为以下码表中的字符
码表：
fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF
-
初始化字符串 b[i]=
-
按照以下字符顺序编码表编码并填充至 b[i]
字符顺序编码表：
0 -> 9
1 -> 8
2 -> 1
3 -> 6
4 -> 2
5 -> 4
6 -> 0
7 -> 7
8 -> 3
9 -> 5
为以上算法的逆运算
使用 Python C TypeScript Java Kotlin Golang Rust 等语言作为示例，欢迎社区提交更多例程
XOR = 177451812
ADD = 100618342136696320
TABLE = "fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF"
MAP = {
0:9,
1:8,
2:1,
3:6,
4:2,
5:4,
6:0,
7:7,
8:3,
9:5
}
def av2bv(av):
av = (av ^ XOR) + ADD
bv = [""]*10
for i in range(10):
bv[MAP[i]] = TABLE[(av//58**i)%58]
return "".join(bv)
def bv2av(bv):
av = [""]*10
s = 0
for i in range(10):
s += TABLE.find(bv[MAP[i]])*58**i
av=(s-ADD)^XOR
return(av)
def main():
while True:
mod = input("1.AV2BV\n2.BV2AV\n3.Exit\n你的选择:")
if mod == "1":
print("BV号是: BV"+av2bv(int(input("AV号是:"))))
elif mod == "2":
print("AV号是 AV"+str(bv2av(input("BV号是"))))
elif mod == "3":
break
else:
print("输入错误请重新输入")
main()#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
const char table[] = "fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF"; // 码表
char tr[124]; // 反查码表
const unsigned long long XOR = 177451812; // 固定异或值
const unsigned long long ADD = 8728348608; // 固定加法值
const int s[] = {11, 10, 3, 8, 4, 6}; // 位置编码表
// 初始化反查码表
void tr_init() {
for (int i = 0; i < 58; i++)
tr[table[i]] = i;
}
unsigned long long bv2av(char bv[]) {
unsigned long long r = 0;
unsigned long long av;
for (int i = 0; i < 6; i++)
r += tr[bv[s[i]]] * (unsigned long long)pow(58, i);
av = (r - ADD) ^ XOR;
return av;
}
char *av2bv(unsigned long long av) {
char *result = (char*)malloc(13);
strcpy(result,"BV1 4 1 7 ");
av = (av ^ XOR) + ADD;
for (int i = 0; i < 6; i++)
result[s[i]] = table[(unsigned long long)(av / (unsigned long long)pow(58, i)) % 58];
return result;
}
int main() {
tr_init();
printf("%s\n", av2bv(170001));
printf("%u\n", bv2av("BV17x411w7KC"));
return 0;
}输出为：
BV17x411w7KC
170001
感谢#417提供
export default class BvCode {
private TABEL = 'fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF'; // 码表
private TR: Record<string, number> = {}; // 反查码表
private S = [11, 10, 3, 8, 4, 6]; // 位置编码表
private XOR = 177451812; // 固定异或值
private ADD = 8728348608; // 固定加法值
constructor() {
// 初始化反查码表
const len = this.TABEL.length;
for (let i = 0; i < len; i++) {
this.TR[this.TABEL[i]] = i;
}
}
av2bv(av: number): string {
const x_ = (av ^ this.XOR) + this.ADD;
const r = ['B', 'V', '1', , , '4', , '1', , '7'];
for (let i = 0; i < 6; i++) {
r[this.S[i]] = this.TABEL[Math.floor(x_ / 58 ** i) % 58];
}
return r.join('');
}
bv2av(bv: string): number {
let r = 0;
for (let i = 0; i < 6; i++) {
r += this.TR[bv[this.S[i]]] * 58 ** i;
}
return (r - this.ADD) ^ this.XOR;
}
}
const bvcode = new BvCode();
console.log(bvcode.av2bv(170001));
console.log(bvcode.bv2av('BV17x411w7KC'));输出为：
BV17x411w7KC
170001
/**
* 算法来自：https://www.zhihu.com/question/381784377/answer/1099438784
*/
public class Util {
private static final String TABLE = "fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF";
private static final int[] S = new int[]{11, 10, 3, 8, 4, 6};
private static final int XOR = 177451812;
private static final long ADD = 8728348608L;
private static final Map<Character, Integer> MAP = new HashMap<>();
static {
for (int i = 0; i < 58; i++) {
MAP.put(TABLE.charAt(i), i);
}
}
public static String aidToBvid(int aid) {
long x = (aid ^ XOR) + ADD;
char[] chars = new char[]{'B', 'V', '1', ' ', ' ', '4', ' ', '1', ' ', '7', ' ', ' '};
for (int i = 0; i < 6; i++) {
int pow = (int) Math.pow(58, i);
long i1 = x / pow;
int index = (int) (i1 % 58);
chars[S[i]] = TABLE.charAt(index);
}
return String.valueOf(chars);
}
public static int bvidToAid(String bvid) {
long r = 0;
for (int i = 0; i < 6; i++) {
r += MAP.get(bvid.charAt(S[i])) * Math.pow(58, i);
}
return (int) ((r - ADD) ^ XOR);
}
}/**
* 此程序非完全原创，改编自GH站内某大佬的Java程序，修改了部分代码，且转换为Kotlin
* 算法来源同上
*/
object VideoUtils {
//这里是由知乎大佬不知道用什么方法得出的转换用数字
var ss = intArrayOf(11, 10, 3, 8, 4, 6, 2, 9, 5, 7)
var xor: Long = 177451812 //二进制时加减数1
var add = 8728348608L //十进制时加减数2
//变量初始化工作，加载哈希表
private const val table = "fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF"
private val mp = HashMap<String, Int>()
private val mp2 = HashMap<Int, String>()
//现在，定义av号和bv号互转的方法
//定义一个power乘方方法，这是转换进制必要的
fun power(a: Int, b: Int): Long {
var power: Long = 1
for (c in 0 until b) power *= a.toLong()
return power
}
//bv转av方法
fun bv2av(s: String): String {
var r: Long = 0
//58进制转换
for (i in 0..57) {
val s1 = table.substring(i, i + 1)
mp[s1] = i
}
for (i in 0..5) {
r += mp[s.substring(ss[i], ss[i] + 1)]!! * power(58, i)
}
//转换完成后，需要处理，带上两个随机数
return (r - add xor xor).toString()
}
//av转bv方法
fun av2bv(st: String): String {
try {
var s = java.lang.Long.valueOf(st.split("av".toRegex()).dropLastWhile { it.isEmpty() }
.toTypedArray()[1])
val sb = StringBuffer("BV1 4 1 7 ")
//逆向思路，先将随机数还原
s = (s xor xor) + add
//58进制转回
for (i in 0..57) {
val s1 = table.substring(i, i + 1)
mp2[i] = s1
}
for (i in 0..5) {
val r = mp2[(s / power(58, i) % 58).toInt()]
sb.replace(ss[i], ss[i] + 1, r!!)
}
return sb.toString()
} catch (e: ArrayIndexOutOfBoundsException) {
return ""
}
}
}package main
import "math"
const TABLE = "fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF"
var S = [11]uint{11, 10, 3, 8, 4, 6}
const XOR = 177451812
const ADD = 8728348608
var TR = map[string]int64{}
// 初始化 TR
func init() {
for i := 0; i < 58; i++ {
TR[TABLE[i:i+1]] = int64(i)
}
}
func BV2AV(bv string) int64 {
r := int64(0)
for i := 0; i < 6; i++ {
r += TR[bv[S[i]:S[i]+1]] * int64(math.Pow(58, float64(i)))
}
return (r - ADD) ^ XOR
}
func AV2BV(av int64) string {
x := (av ^ XOR) + ADD
r := []rune("BV1 4 1 7 ")
for i := 0; i < 6; i++ {
r[S[i]] = rune(TABLE[x/int64(math.Pow(58, float64(i)))%58])
}
return string(r)
}
func main() {
println(AV2BV(170001))
println(BV2AV("BV17x411w7KC"))
}输出为：
BV17x411w7KC
170001
crate: https://github.com/stackinspector/bvid
// Copyright (c) 2023 stackinspector. MIT license.
const XORN: u64 = 177451812;
const ADDN: u64 = 100618342136696320;
const TABLE: [u8; 58] = *b"fZodR9XQDSUm21yCkr6zBqiveYah8bt4xsWpHnJE7jL5VG3guMTKNPAwcF";
const MAP: [usize; 10] = [9, 8, 1, 6, 2, 4, 0, 7, 3, 5];
const REV_TABLE: [u8; 74] = [
13, 12, 46, 31, 43, 18, 40, 28, 5, 0, 0, 0, 0, 0, 0, 0, 54, 20, 15, 8,
39, 57, 45, 36, 0, 38, 51, 42, 49, 52, 0, 53, 7, 4, 9, 50, 10, 44, 34, 6,
25, 1, 0, 0, 0, 0, 0, 0, 26, 29, 56, 3, 24, 0, 47, 27, 22, 41, 16, 0,
11, 37, 2, 35, 21, 17, 33, 30, 48, 23, 55, 32, 14, 19,
];
const POW58: [u64; 10] = [
1, 58, 3364, 195112, 11316496, 656356768, 38068692544,
2207984167552, 128063081718016, 7427658739644928,
];
fn av2bv(avid: u64) -> [u8; 10] {
let a = (avid ^ XORN) + ADDN;
let mut bvid = [0; 10];
for i in 0..10 {
bvid[MAP[i]] = TABLE[(a / POW58[i]) as usize % 58];
}
bvid
}
fn bv2av(bvid: [u8; 10]) -> u64 {
let mut a = 0;
for i in 0..10 {
a += REV_TABLE[bvid[MAP[i]] as usize - 49] as u64 * POW58[i];
}
(a - ADDN) ^ XORN
}
// assert_eq!(*b"17x411w7KC", av2bv(170001));
// assert_eq!(170001, bv2av(*b"17x411w7KC"));