python中如何清空列表数据
Python批量删除只保留最近几天table的代码实例Python批量删除table,只保留最近几天的table
代码如下:
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#!/usr/bin/python3 """ 批量删除table,只保留最近几天的table """ import pymysql import re def conn_(host = ' ',usr=' ',passwd=' ',db=' ',port = 3306 ,): conn = pymysql.connect(host, usr, passwd, db, port,charset = 'utf8' ) return conn def del_table(conn_,table_pre = ' ',table_suff=' % Y % m % d',keep_count = 3 ): date_form = None if table_suff = = "%Y%m%d" : date_form = "_(\d{4}\d{1,2}\d{1,2})$" date_len = 8 elif table_suff = = "%Y-%m-%d" : date_form = "_(\d{4}-\d{1,2}-\d{1,2})$" date_len = 10 elif table_suff = = "%Y%m" : date_form = "_(\d{4}\d{1,2})$" date_len = 6 elif table_suff = = "%Y-%m" : date_form = "_(\d{4}-\d{1,2})$" date_len = 7 else : raise Exception( "暂时不支持其他类型的时间后缀" ) curs = conn_.cursor() curs.execute( 'SHOW TABLES' ) data = curs.fetchall() table_ = r '%s' % table_pre + date_form list_table = [] i = 0 for table in data: mt = re.search(table_, table[ 0 ]) if mt: if len (mt.groups()[ 0 ]) = = date_len: list_table.append((table[ 0 ], mt.groups()[ 0 ])) i + = 1 sorted (list_table, key = lambda date: date[ 1 ]) #按照表结构后缀时间升序排序 for j in range (i - keep_count): sql = 'DROP TABLE if exists %s' % list_table[j][ 0 ] curs.execute(sql) curs.close() conn_.close() if __name__ = = '__main__' : table_pre = "tree_product" table_suff = "%Y%m%d" # table_suff = "%Y-%m-%d" # table_suff = "%Y%m" # table_suff = "%Y-%m" conn = conn_( '10.0.0.11' , 'root' , 'sctele@root' , 'sxf' ,port = 3306 ) del_table(conn,table_pre = table_pre,table_suff = table_suff,keep_count = 1 ) |
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原文链接:https://blog.csdn.net/sxf_123456/article/details/79224404