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Python批量删除只保留最近几天table的代码实例

yipeiwu_com5年前 (2020-03-06)Python基础

Python批量删除table,只保留最近几天的table

代码如下：

#!/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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