pymysql

cursor = conn.cursor(cursor=pymysql.cursors.DictCursor) 返回字典格式

 

lastrowid 获取自增id

import pymysql

conn = pymysql.connect(host="127.0.0.1", port=3306, user="root", passwd="root", db="pymysql")
cursor = conn.cursor()
cursor.execute("insert into user(name,age ) values('vincen',22)")
print(cursor.lastrowid)

执行结果:
2

 

查询操作:

import pymysql

#数据库连接 conn = pymysql.connect(host="127.0.0.1", port=3306, user="root", passwd="root", db="mydb")
#使用cursor()方法创建一个游标对象,并配置获取数据返回的类型 cursor = conn.cursor(cursor=pymysql.cursors.DictCursor)
#执行sql语句 cursor.execute("select * from department")
#获取查询结果 print(cursor.fetchall()) cursor.close() conn.close()

 

更新操作:

import pymysql

# 打开数据库连接
db = pymysql.connect("localhost","testuser","test123","TESTDB" )

# 使用cursor()方法获取操作游标 
cursor = db.cursor()

# SQL 更新语句
sql = "UPDATE EMPLOYEE SET AGE = AGE + 1 WHERE SEX = '%c'" % ('M')   不能使用SQL字符串拼接,否则会造成SQL注入
try:
   # 执行SQL语句
   cursor.execute(sql)
   # 提交到数据库执行
   db.commit()
except:
   # 发生错误时回滚
   db.rollback()

# 关闭数据库连接
db.close()

 

删除操作

import pymysql

# 打开数据库连接
db = pymysql.connect("localhost","testuser","test123","TESTDB" )

# 使用cursor()方法获取操作游标 
cursor = db.cursor()

# SQL 删除语句
sql = "DELETE FROM EMPLOYEE WHERE AGE > '%d'" % (20)
try:
   # 执行SQL语句
   cursor.execute(sql)
   # 提交修改
   db.commit()
except:
   # 发生错误时回滚
   db.rollback()

# 关闭连接
db.close()

 

posted @ 2017-03-30 14:44  Vincen_shen  阅读(193)  评论(0)    收藏  举报