mysql和sqlite3多次查询问题，搞了几次没搞定

2020-03-17 10:03:13 360 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

SQLite 和 MySQL语法区别 SQLite 和 MySQL 在 SQL 语法上有一些差异，这些差异主要体现在数据类型、函数、表和索引的管理等方面。阅读详情

两个mysql表：table1，table2，两个表的id值相同，目的是从mysql表table1中查出数据，然后在sqlite3的表table3中查出和mysql的table1具有相同biaoti 、company 、kanwu值的字段filename，再用filename的值去更新mysql表table2中downpath值，现在第一步已经完成，最后一步更新table2的downpath值提示错误，很奇怪，TypeError: not all arguments converted during string formatting



import time

import urllib.request

import urllib.parse

import sqlite3

import pymysql



if __name__ == "__main__":

    connmysql = pymysql.connect(user='root', password='123456', database='empiredd', charset='utf8')

    cursormysql = connmysql.cursor()

    query = 'select id, title , company, qikanname from table1'

    cursormysql.execute(query)

    for (id, title, company, qikanname) in cursormysql:

        print(id, title, company, qikanname)       

        conn = sqlite3.connect(r"E:\网站\SpiderResult.db3")

        cursor = conn.cursor()

        sql = "select filename from table3 where biaoti = ? and company = ? and kanwu = ?"

        cursor.execute(sql, (title, company, qikanname))

        values = cursor.fetchone()

        print(values[0])

        print(id)

       cursormysql.execute('update table2  set downpath = ? where id = ?', (values[0], id))

        connmysql.commit()

    cursor.close()

    conn.close()

    cursormysql.close()

    connmysql.close()