如何通过python分隔sqlite3中的分组项目

Question

iv 开始为我的程序使用数据库而不是 .txt 文件，我想做一个搜索功能

iv 了解了搜索功能的基础知识（包含所有可能选择的下拉菜单），但大多数项目在 1 列/行中有多个项目，例如

列“食物名称”“成分”“食谱”

           potato casserole    potato,cheese,ect

但在我的搜索功能土豆中，奶酪是 1 项，我该如何分离它？？

我的第二个问题显然会有成分相似的食物，我怎样才能阻止它显示相同成分的倍数，并且仍然显示所有具有相同成分的食物的结果？？

option 1 has cheese selected option 2 has potato and cheese selected ( but i want it to be seprated) option 3 at the top it shows cheese and at the buttom it shows cheese (from 2 differnt foods) and i want to only have 1 instance of that ingredient but return all the foods with that ingredient

from tkinter import *
from tkinter.ttk import Combobox
import sqlite3

con = sqlite3.connect("Food.db")
c = con.cursor()

c.execute("SELECT Ingredients FROM Dinner_Order")
order = c.fetchall()
c.execute("SELECT Ingredients FROM Dinner_Stove")
stove = c.fetchall()
c.execute("SELECT Ingredients FROM Dinner_Oven")
oven = c.fetchall()
c.execute("SELECT Ingredients FROM Dinner_Cold")
cold = c.fetchall()
c.execute("SELECT Ingredients FROM Dinner_Simple")
simple = c.fetchall()
con.commit()

dinner_order = order
dinner_stove = stove
dinner_oven = oven
dinner_cold = cold
dinner_simple = simple

ws = Tk()
ws.title("Python Guides")
ws.geometry("200x200")



item_names = dinner_order + dinner_stove + dinner_oven + dinner_simple + dinner_cold

combo = Combobox(ws, state='readonly')
combo['values'] = item_names
combo.pack()

combo = Combobox(ws, state='readonly')
combo['values'] = item_names
combo.pack()

combo = Combobox(ws, state='readonly')
combo['values'] = item_names
combo.pack()

combo = Combobox(ws, state='readonly')
combo['values'] = item_names
combo.pack()

button = Button(ws, text="search", )#command=search_items)
button.pack()

ws.mainloop()

Answer 1

我建议split, 上的结果，例如，如果查询返回"potato,cheese,ect"，你会这样做：

"potato,cheese,ect".split(",")

类似这样的：

>>> c.execute("SELECT Ingredients FROM Dinner_Simple WHERE food_name ='potato casserole'")
>>> ingredients = c.fetchall()
>>> print(ingredients)
"potato,cheese,ect"
>>> ingredients = ingredients.split(",")
>>> print(ingredients)
["potato", "cheese", "ect"]