python3 判断url链接是否为404

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()