【实现一套爬虫数据抓取平台】[3-5-03] 微博长短地址转换

零、系列目录

写这套文章的时候，不会完全按照目录的顺序一篇一篇写， 大家可以到目录中直接找到对应的章节进行查看。

点我跳转

一、背景

新浪微博有两类地址，分别是：

# pc站地址
https://weibo.com/2214257545/LgUW8DlCG

# mob站地址
https://m.weibo.cn/status/4740321169377914

这两个地址对应的是同一篇微博，实际使用过程中，需要根据拿到的不同地址，灵活判断是否需要转换地址来进行解析。

二、代码

完整代码地址：点击跳转

#!/bin/bash
# -*- coding=utf-8 -*-

ALPHABET = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"


# 10进制转为62进制
def base62_encode(num, alphabet=ALPHABET):
    """Encode a number in Base X
    `num`: The number to encode
    `alphabet`: The alphabet to use for encoding
    """
    if num == 0:
        return alphabet[0]
    arr = []
    base = len(alphabet)
    while num:
        rem = num % base
        num = num // base
        arr.append(alphabet[rem])
    arr.reverse()
    return ''.join(arr)


# 62进制转为10进制
def base62_decode(string, alphabet=ALPHABET):
    """Decode a Base X encoded string into the number
    Arguments:
    - `string`: The encoded string
    - `alphabet`: The alphabet to use for encoding
    """
    base = len(alphabet)
    strlen = len(string)
    num = 0
    idx = 0
    for char in string:
        power = (strlen - (idx + 1))
        num += alphabet.index(char) * (base ** power)
        idx += 1
    return num


# mid转换为id
def mid2id(mid):
    mid = str(mid)[::-1]
    size = int(len(mid) / 7) if len(mid) % 7 == 0 else int(len(mid) / 7 + 1)
    result = []
    for i in range(size):
        s = mid[i * 7: (i + 1) * 7][::-1]
        s = base62_encode(int(s))
        s_len = len(s)
        if i < size - 1 and len(s) < 4:
            s = '0' * (4 - s_len) + s
        result.append(s)
    result.reverse()
    return ''.join(result)


# id转换为mid
def id2mid(id):
    id = str(id)[::-1]
    size = int(len(id) / 4) if len(id) % 4 == 0 else int(len(id) / 4 + 1)
    result = []
    for i in range(size):
        s = id[i * 4: (i + 1) * 4][::-1]
        s = str(base62_decode(str(s)))
        s_len = len(s)
        if i < size - 1 and s_len < 7:
            s = (7 - s_len) * '0' + s
        result.append(s)
    result.reverse()
    return ''.join(result)


if __name__ == '__main__':
    print('mdi2id: ' + mid2id('4740321169377914'))
    print('id2mid: ' + id2mid('LgUW8DlCG'))

三、总结

这个工具本身比较简单，至于具体需要怎么使用，大家根据自身情况灵活选择就好。

祝大家变的更强。