測試環境：

Win10

Python 3.5.4

實現思路

利用62個可列印字元，通過隨機生成32位UUID，由於UUID都為十六進位制，所以將UUID分成8組，每4個為一組，然後通過模62（字元0-9，a-z，A-Z總數量62個字元）操作，結果作為索引取出字元，這樣重複率大大降低，實踐測試，執行20000000次，僅出現2個重複id(僅測試過一次)。

當然，這樣還達不到唯一id，因為還是有重複的。解決方法呢，可以考慮結合資料庫、或者其它儲存來實現，以結合資料庫為例，我們可以新建一張資料庫表，並給表設定一個id欄位，並且設定為主鍵、或者增加唯一約束，每次獲取8 id後，往表裡插入一條資料，如果可以成功插入，說明不重複，否則說明是重複id，再次嘗試獲取。

核心程式碼

#!/usr/bin/env python
# -*- coding:utf-8 -*- 
'''
@CreateTime: 2020/07/14 11:04
@Author : shouke
''' 
import uuid
array = [ "0","1","2","3","4","5","6","7","8","9","a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"
     ]
 
def get_short_id():
  id = str(uuid.uuid4()).replace("-",'') # 注意這裡需要用uuid4
  buffer = []
  for i in range(0,8):
    start = i * 4
    end = i * 4 + 4
    val = int(id[start:end],16)
    buffer.append(array[val % 62])
  return "".join(buffer)

測試驗證

id_set = set() # 用於存放生成的唯一id
count = 0 # 用於統計出現重複的次數
index = [] # 記錄第幾次呼叫生成8位id出現重複
for i in range(0,20000000):
  id = get_short_id()
  if id in id_set:
    count += 1
    index.append(str(i+1))
  else:
    id_set.add(id)
  print('id：%s,執行第 %s 次,重複數:%s,重複率:%s,出現重複次序 %s' % (id,i+1,count,count/(i+1)*100,','.join(index)))
 

補充：用python隨機生成以2019開頭的10個8位數的學號

import random意思是引入內建模組random,j代表著

行數，range()是一個隨機生成數字的函式，i控制著

每行的個數，str()表示是轉化為字串的型別

以上為個人經驗，希望能給大家一個參考，也希望大家多多支援我們。如有錯誤或未考慮完全的地方，望不吝賜教。