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
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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()
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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服务器之家。
原文链接:https://blog.csdn.net/u013091109/article/details/80737422