python中sqlite问题和坑

import sqlite3 #导入模块

conn = sqlite3.connect('example.db')

C=conn.cursor()

#创建表

C.execute('''CREATE TABLE stocks(data text,trans text, symple text,qty real,price real)''')

#插入一条数据

C.execute('''INSERT INTO stocks VALUES('2006-10-01,'BUY','RHA',100,35.14)''')

#提交当前事务，保存数据

#关闭数据连接

conn.close()

#查询数据库

for row in c.execute('SELECT * FROM stocks OPRED BY price'):

print(row)#s数据库是列表类型，可进行迭代

import sqlite3

conn = sqlite3.connect(":memory:")

cur=conn.cursor()

cur.execute("CREATE TABLE people(name_last,age)")

who = "li"

age="20"

cur.execute("INSERT INTO people VALUES(?, ?)",(who, age))

cur.execute("SELECT * FROM people WHERE name_last=:who AND age=:age",{"who":who,"age":age})

print(cur.fetchone())

Cursor

#1利用数据库完成小写字母

import sqlite3

class IterChars:

def __init__(self):

self.count=ord('a')

def __iter__(self):

return self

def __next__(self):

if self.count>ord('z'):

raise StopIteration

self.count+=1

return(chr(self.count-1),)

conn = sqlite3.connect(":memory:")

cur=conn.cursor()

cur.execute("CREATE TABLE characters(c)")

theIter=IterChars()

cur.executemany("INSERT INTO characters(c) VALUES(?)",theIter)

cur.execute("SELECT c FROM characters")

print(cur.fetchall())

#2利用yield完成对小写字母的创建

import sqlite3

import string

def char_gernerator():

for c in string.ascii_lowercase:

yield(c,)

conn = sqlite3.connect(":memory:")

cur=conn.cursor()

cur.execute("CREATE TABLE characters(c)")

#theIter=IterChars()

cur.executemany("INSERT INTO characters(c) VALUES(?)",char_gernerator())

cur.execute("SELECT c FROM characters")

print(cur.fetchall())

python的sqlite坑

#bug1表不存在

import sqlite3

conn=sqlite3.connect("D:/addressbook.db")

cur = conn.cursor()

cur.execute('''INSERT INTO addresslist(name,sex,phone,qq,address)VALUES('小王','女','18337257400','18674529','beijing')''')

cur.execute('''INSERT INTO addresslist(name,sex,phone,qq,address)VALUES('小李','女','18337257401','18674528','nanjing')''')

cur.execute('''INSERT INTO addresslist(name,sex,phone,qq,address)VALUES('小l','女','18337257403','18674525','tianjin')''')

conn.commit()

conn.close()

程序执行，我在心里想着这么low的代码应该没问题的，结果竟然报错了，fuck,表不存在什么鬼，于是百度，发现好多都是废话，好吧，不需要了，我在脑中想了会，既然不存在，那我就创建

#修改后

import sqlite3

conn=sqlite3.connect("D:/addressbook.db")

cur = conn.cursor()

cur.execute('''CREATE TABLE addresslist(name text,sex text,phone real,qq real,address text)''')

cur.execute('''INSERT INTO addresslist(name,sex,phone,qq,address)VALUES('小王','女','18337257400','18674529','beijing')''')

cur.execute('''INSERT INTO addresslist(name,sex,phone,qq,address)VALUES('小李','女','18337257401','18674528','nanjing')''')

cur.execute('''INSERT INTO addresslist(name,sex,phone,qq,address)VALUES('小l','女','18337257403','18674525','tianjin')''')

conn.commit()

conn.close()

程序bug完美解决

下面进行遍历查找

import sqlite3

conn=sqlite3.connect("D:/addressbook.db")

cur=conn.cursor()

cur.execute('SELECT * FROM addresslist')

li = cur.fetchall()

for line in li:

for item in line:

print(item,end=' ')

print()

conn.close()

好了,终于完美解决。

#sqllite之row

import sqlite3

conn=sqlite3.connect(":memory:")

c=conn.cursor()

#创建表

c.execute('''CREATE TABLE stocks(data text,trans text,symple text,qty real,price real)''')

c.execute("""INSERT INTO stocks VALUES ('2006-10-01','BUY','RHA',100,35.14)""")

conn.commit()

c.close()

conn.row_factory=sqlite3.Row

c=conn.cursor()

c.execute('SELECT * FROM stocks')

r=c.fetchone()

print(type(r))

print(tuple(r))

print(r[2])

print(r.keys())

print(r['qty'])

for field in r:

print(field)