從第一個組合框中選擇后，如何從select查詢中為其他組合框添加值

Question

我試圖在一個combobox1上做出選擇，它用來自sqlite3 db的數據填充combobox2

我做了combobox1，但我不知道為什么它不使用combobox2我試圖使事件=沒有錯誤消失但沒有值的組合框2

import tkinter as tk 
from tkinter import ttk 
import sqlite3

class SchoolProjict(tk.Tk):
    def __init__(self, *args, **kwargs):
        tk.Tk.__init__(self, *args, **kwargs)
        container = tk.Frame(self)
        container.pack(side = "top", fill = "both", expand = True)
        container.grid_rowconfigure(0, weight = 1)
        container.grid_columnconfigure(0, weight = 1)
        self.frames = {}
        for F in (StartPage,):
            frame = F(container, self)
            self.frames[F] = frame
            frame.grid(row = 0, column = 0, sticky = "nsew")
        self.show_frame(StartPage)

    def show_frame(self, cont):
        frame = self.frames[cont]
        frame.tkraise()

    def get_page(self, classname):
        for page in self.frames.values():
            if str(page.__class__.__name__) == classname:
                return page
        return None


class StartPage(tk.Frame):
    def __init__(self, parent, controller):
        self.controller = controller
        tk.Frame.__init__(self, parent)
        lablel = tk.Label(self, text = "Stuident Info")
        lablel.grid(row = 1, columnspan = 3, pady=5, padx=5)
        lable2 = tk.Label(self, text = "gread")
        lable2.grid(row = 2, column = 2, pady=5, padx=5)
        lable3 = tk.Label(self, text = "class")
        lable3.grid(row = 3, column = 2, pady=5, padx=5)
        lable4 = tk.Label(self, text = "Stuident Name")
        lable4.grid(row = 4, column = 2, pady=5, padx=5)
        self.number = tk.StringVar()
        self.combobox1 = ttk.Combobox(self, width = 15)
        self.combobox1.bind("<<ComboboxSelected>>", self.comboclass)
        self.combobox1['value'] = self.combogread()
        self.combobox1.grid(row = 2, column = 1, pady=5, padx=5)
        self.combobox2 = ttk.Combobox(self, width = 15)
        self.combobox2['value'] = self.comboclass()
        self.combobox2.grid(row = 3, column = 1, pady=5, padx=5)


    def combogread(self):
        self.conn = sqlite3.connect("exeldata.db")
        self.cur = self.conn.cursor()
        self.cur = self.conn.execute('SELECT rowid, GradNumber FROM gradelevel')

        result = []

        for row in self.cur.fetchall():
            result.append(row[1])

        return result

    def comboclass(self, event = None):
        greadid = self.combobox1.get() 
        self.conn = sqlite3.connect("exeldata.db")
        self.cur = self.conn.cursor()
        self.cur = self.conn.execute('SELECT rowid, GradNumber FROM gradelevel WHERE GradNumber = (?)', (greadid,))
        result = []
        for row in self.cur.fetchall():
            result.append(row[0])    

        self.cur = self.conn.execute('SELECT rowid , ClassNumb FROM classnumber  WHERE GradID = (?)', (str(result),))
        result = []
        for row in self.cur.fetchall():
            result.append(row[0])

        return result



app = SchoolProjict()
app.mainloop()

我的數據庫是3個表，一對多關系一個用於每個級別的班級的一年級和每個班級的學生信息

Answer 1

問題是comboclass()函數實際上並沒有更新Combobox的值，這是一個簡單的修復。

您只需要創建一個更新值的函數。 這基本上就是這行代碼： self.combobox2['value'] = self.comboclass()

所以這將是您需要更改/添加的代碼：

class StartPage(tk.Frame):
    def __init__(self, parent, controller):
        """All Init Code Here"""

        self.combobox1 = ttk.Combobox(self, width = 15)
        self.combobox1.bind("<<ComboboxSelected>>", self.update_combo)  # Changed binds command to the update the combobox
        self.combobox1['value'] = self.combogread()
        self.combobox1.grid(row = 2, column = 1, pady=5, padx=5)

        self.combobox2 = ttk.Combobox(self, width = 15)
        self.combobox2['value'] = self.comboclass()
        self.combobox2.grid(row = 3, column = 1, pady=5, padx=5)

    def update_combo(self, event=None):  # New function to update the combobox
        self.combobox2['value'] = self.comboclass()

該Combobox ，只要你選擇一個選項2現在應該更新Combobox 1