Python/MySQL - 通過 Tkinter 將數據插入表中

Question

我試圖通過圖形界面向表中插入數據，但出現以下錯誤：

錯誤

Exception in Tkinter callback
Traceback (most recent call last):
  File "C:\Users\TESTES\AppData\Local\Programs\Python\Python38-32\lib\tkinter\__init__.py", line 1883, in __call__
    return self.func(*args)
TypeError: submit() missing 3 required positional arguments: 'nome', 'idade', and 'cpf'

代碼

from tkinter import *
import pymysql


book = Tk()
book.geometry("400x300")
book.title("test")

##-----------------##
##  CONEXÃO BANCO  ##
##-----------------##

conn = pymysql.connect(host="localhost", port=3306, user="root", password="", database="oneday")
print("connect successfull! goodwork bro screw screww")
cursor = conn.cursor()
cursor.execute("SELECT VERSION()")
versao = cursor.fetchone()
print("Versão do gerenciador Maria DB: %s" % versao)
conn.close()

##-----------------##
##    STRINGVAR    ##
##-----------------##


##-----------------##
##  FUNÇÕES/DEF'S  ##
##-----------------##

def submit(nome, idade, cpf):


    nome = StringVar()
    idade = StringVar()
    cpf = StringVar()

    try:
        conn = pymysql.connect(host="localhost", port=3306, user="root", password="", db="oneday")
        cursor = conn.cursor()
        mysql_submit = ("INSERT INTO novo VALUES (NULL, {}, {}, {})", (nome.get(), idade.get(), cpf.get()))
        cursor.execute(mysql_submit)
        conn.commit()
        print("Dados inseridos com sucesso na tabela.")
        cursor.close()
        conn.close()
    except pymysql.connector.Error as error:
        print("failed".format(error))
    finally:
        if (conn.connect()):
            cursor.close()
            conn.close()
            print("MySQL connection is closed")

        # Clear the text boxes
        e_nome.delete(0, END)
        e_idade.delete(0, END)
        e_cpf.delete(0, END)


def closer():
    try:
        conn = pymysql.connect(host="localhost", port=3306, user="root", password="", db="oneday")
        cursor = conn.cursor()
        mysql_closer = """DELETE FROM novo"""
        cursor.execute(mysql_closer)
        conn.commit()
        print("Dados removidos com sucesso da tabela.")
        cursor.close()
        conn.close()
    except pymysql.connector.Error as error:
        print("failed".format(error))
    finally:
        if (conn.connect()):
            cursor.close()
            conn.close()
            print("MySQL connection is closed")


##-----------------##
##     LABELS      ##
##-----------------##

mylabel = LabelFrame(book, text="Novo")
mylabel.place(x=10, y=10, width=300, height=150)

l_nome = Label(book, text="Nome")
l_nome.place(x=15, y=30)
e_nome = Entry(book)
e_nome.place(x=100, y=30)

l_idade = Label(book, text="Idade")
l_idade.place(x=15, y=60)
e_idade = Entry(book)
e_idade.place(x=100, y=60)

l_cpf = Label(book, text="CPF")
l_cpf.place(x=15, y=90)
e_cpf = Entry(book)
e_cpf.place(x=100, y=90)

##-----------------##
##     BOTOÕES     ##
##-----------------##

bt = Button(book, width=15, text="Gravar", command=submit)
bt.place(x=100, y=175)

bt = Button(book, width=15, text="Sair", command=exit)
bt.place(x=100, y=245)

bt = Button(book, width=15, text="Apagar", command=closer)
bt.place(x=100, y=210)


##-----------------##
##     IMAGEM      ##
##-----------------##

photo = PhotoImage(file="C:/Users/TESTES/PycharmProjects/OMNIA/images/logo.png")
logo = Label(book, image=photo)
logo.place(x=250, y=175, width=100, height=100)
logo.background = "orange"


book.mainloop()

Answer 1

Button運行沒有參數的函數，所以你不能用參數聲明它

def submit():

它應該可以解決錯誤消息的問題，但您仍然會遇到獲取值的問題。

您的StringVar()在submit()中沒用。 您應該直接在 SQL 中使用 Entries

mysql_submit = ("INSERT INTO novo VALUES (NULL, {}, {}, {})", (e_nome.get(), e_idade.get(), e_cpf.get()))

但是您應該將它直接放在execute因為它期望QUERY作為第一個參數，將值作為第二個參數。 您的元組mysql_submit將被視為一個參數，而不是兩個。

cursor.execute("INSERT INTO novo VALUES (NULL, {}, {}, {})", (e_nome.get(), e_idade.get(), e_cpf.get()))