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python-3.x - 如何修复python sqlite数据库不存储数据

问题描述

我的代码将只存储数据“长号”和按钮单击的第一个“x”值，但是一旦再次单击按钮，它将不会用正确的值替换 x 的当前值，该值作为一个输出成为 2

import sqlite3
import tkinter as tk
from tkinter import *

connecter = sqlite3.connect('IAworker.db')
c = connecter.cursor()

def create_security():
    c.execute('CREATE TABLE IF NOT EXISTS security(username TEXT, password INT, teacherID INT, studentID INT, instrumentID INT)')
def create_Teacher_Info():
    c.execute('CREATE TABLE IF NOT EXISTS teacher_info(teacherName TEXT, teacherID INT PRIMARY KEY)')

def create_Student_Info():
    c.execute(
        'CREATE TABLE IF NOT EXISTS student_info(Form TEXT, studentID INT PRIMARY KEY, studentName TEXT)')

def create_Instrument_Info():
    c.execute('CREATE TABLE IF NOT EXISTS instrument_info( instrumentName TEXT, instrumentID INT PRIMARY KEY)')
    c.close
create_security()
create_Teacher_Info()
create_Student_Info()
create_Instrument_Info()

def inserterInstTest(name, instID):
     c = connecter.cursor()
     c.execute('SELECT * FROM instrument_info')
     [print(row) for row in c.fetchall()]
     c.execute('INSERT OR REPLACE INTO instrument_info (instrumentName, instrumentID) VALUES (?, ?)', (name, instID))
     # c.execute('UPDATE instrument_info SET instrumentName = (?) WHERE instrumentID = (?)', (name, instID))

     connecter.commit()
     c.close()
     connecter.close


butt4 = tk.Button(root3_2, text="trombone+1",)#instrument buttons
                butt4.grid(row=6, column=1, columnspan=3)#placement on window for buttons



def buttonClick3(x): #commands for buttons
                    print('button was clicked')
                    tromName = 'trombone'
                    x = x + 1
                    x = str(x)
                    print('days read =',x)
                    SqliteIA.inserterInstTest(tromName,x)
                butt4.config(command=buttonClick3(trombones2))#configuring button commands


SQLiteIA.InserterInstTest(tromName, x) was used to call the function as its in another py file. but the code should replace the current

标签： python-3.xtkintersqlite

解决方案

您在函数内部增加 x ，它是不可变的。

考虑下面的代码

def buttonClick(x):
    x = x + 1
    print "X = " + str(x)
    return

x = 0

buttonClick(x)
buttonClick(x)
buttonClick(x)

结果是

X = 1
X = 1
X = 1

有关不可变与可变的解释，请参见此答案

这是一种解决方法

def buttonClick(x):
    x[0] = x[0] + 1
    print "X = " + str(x[0])
    return

x = [0]

buttonClick(x)
buttonClick(x)
buttonClick(x)

结果是

X = 1
X = 2
X = 3

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