python3判断url链接是否为404的方法

import pymysqlimport threadingimport timeimport urllibimport 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 cursorcursor = 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 = 0for 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()