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 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()

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。