本文实例为大家分享了python3判断url链接是否为404的具体代码，供大家参考，具体内容如下
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()
以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持找一找教程网。
