python3判断url链接是否为404的方法

import pymysql
import threading
import time
import urllib
import requests
 
# 打开数据库连接
db = pymysql.connect("192.168.0.*", "username", "password", "databasename")
# 使用 cursor() 方法创建一个游标对象 cursor
cursor = db.cursor()
# SQL 查询语句
sql = "SELECT sku,url_6 FROM url_new where flag_6 is null and url_6<>'' "
  # 执行SQL语句
cursor.execute(sql)
  # 获取所有记录列表
results = cursor.fetchall()
num = 0
for row in results:
  sku = row[0]
  url = row[1]
  html = requests.head(url) # 用head方法去请求资源头
  re=html.status_code
  num = num + 1
  print(num,re)
  if re == 200:
    sql_2 = "UPDATE url_new SET flag_6 = 0 WHERE sku = '%s'" % sku
    try:
      # 执行SQL语句
      cursor.execute(sql_2)
      #print(cursor.rowcount)
    except Exception as e:
      print(e)
      conn.rollback()
  if re == 404:
    sql_3 = "UPDATE url_new SET flag_6 = 1 WHERE sku = '%s'" % sku
    try:
      # 执行SQL语句
      cursor.execute(sql_3)
      print(cursor.rowcount)
    except Exception as e:
      print(e)
      conn.rollback()
  db.commit()
db.close()