使用tkinter和db连接的Python GUI

Question

我无法从以下program.What获得任何输出。我在这里尝试做的是将员工姓名和薪水存储在数据库中，然后根据姓名删除数据库条目，id.It没有给出任何错误。我认为这是一个数据库错误。它不会在数据库中插入任何值。请帮帮忙。谢谢

import Tkinter
import sqlite3
import ttk


def put(*args):
    try:
        e_name = str(i_name)
        e_pay = int(i_pay)
        conn = sqlite3.connect('medha.db')
        c = conn.cursor()
        c.execute('create table if not exists employee(id integer primary key auto increment, name text, salary integer)')
        c.execute('insert into employee(name, salary) values(?,?)',(e_name,e_pay))
        conn.close()
        notice = 'insertion successful'
    except:
        notice ='insert proper values'

def del_id(*args):
    try:
        e_id = int(d_del)
        conn = sqlite3.connect('medha.db')
        c = conn.cursor()
        c.execute('delete from employee where id =?',(e_id))
        notice = 'deleted successfully'
        conn.close()
    except:
        notice = 'insert proper values'

def del_name(*args):
    try:
        e_name = int(d_del)
        conn = sqlite3.connect('medha.db')
        c = conn.cursor()
        c.execute('delete from employee where name = ?',(e_name))
        notice = 'deleted successfully'
        conn.close()
    except:
        notice = 'insert proper values'

root = Tkinter.Tk()
root.title('Python Projet - gui and db')

mainframe = ttk.Frame(root,padding='3 3 12 12' )
mainframe.grid(column=0, row=0)
mainframe.columnconfigure(0, weight=1)
mainframe.rowconfigure(0, weight =1)

i_name = str()
i_id = str()
i_pay = str()
notice = str()
d_del = str()

i_name_entry = ttk.Entry(mainframe, width =7, textvariable = i_name)
i_name_entry.grid(column = 1, row = 1)

i_id_entry = ttk.Entry(mainframe, width =7, textvariable = i_id)
i_id_entry.grid(column = 2, row = 1)

i_pay_entry = ttk.Entry(mainframe, width =7, textvariable = i_pay)
i_pay_entry.grid(column = 3, row = 1)

i_del_entry = ttk.Entry(mainframe, width =7, textvariable = d_del)
i_del_entry.grid(column = 1, row =4)

ttk.Label(mainframe, text ='Name of Employee').grid(column = 1, row = 0)
ttk.Label(mainframe, text ='Employee ID').grid(column = 2, row =0)
ttk.Label(mainframe, text ='Employee Pay').grid(column = 3, row=0)
ttk.Label(mainframe, text ='Name/ID of Employee to delete').grid(column = 1, row=3)
ttk.Label(mainframe, text = notice).grid(column = 1, row = 6)

ttk.Button(mainframe, text='Insert', command = put).grid(column=3, row=2)
ttk.Button(mainframe, text='Delete By ID', command = del_id).grid(column =1, row = 5)
ttk.Button(mainframe, text='Delete By Name', command = del_name).grid(column = 3, row=5)

for child in mainframe.winfo_children(): child.grid_configure(padx=5,pady=5)
i_name_entry.focus()

root.mainloop()

任何帮助都是非常感谢的。

Answer 1

ttk.Label(mainframe, text = notice).grid(column = 1, row = 6)

更改字符串notice的值不会更新此标签中的文本。为此，您需要使用StringVar。

notice = Tkinter.StringVar()
ttk.Label(mainframe, textvariable = notice).grid(column = 1, row = 6)

然后，在函数中使用notice.set("whatever")，而不是使用notice = "whatever"。

或者，根本没有notice值。只需直接重新配置文本变量即可。

notice_label = ttk.Label(mainframe, text = "")
notice_label.grid(column = 1, row = 6)

(注意:不要试图在同一行中赋值notice_label和执行grid，它不会起作用)

然后用notice_label.config(text = "whatever")代替notice = "whatever"。

Answer 2

首先，尝试查看是否已连接到数据库。使用postgreSQL，您可以通过psycopg2进行连接

import psycopg2

try:    
        conn = psycopg2.connect(database="user", user="postgres", password="silverTip", host="localhost")
        print("connected")
    except:
        print ("I am unable to connect to the database")
    cur =conn.cursor()

因此，如果您已连接到数据库，它将显示connected，否则将显示消息unable to connect to database