Python tkinter从ttk.combobox检索SQLite ID

Question

我有一个ttk.combobox，其中填充了来自数据库的数据。当用户选择数据时，我需要检索所选数据的ID

我有一个解决方案，但我相信有一个更优雅和简单的解决方案，但我找不到。我用“.”拆分SQL行ID，然后从组合框中将字符串转换为按“.”拆分的列表，并检索ID的列表[0]

代码示例：

from tkinter import *
from tkinter import ttk
import sqlite3
import DataBasePM

DBProjectManager2=DataBasePM.DBProjectManager2

def DropDownProjectView():
    con=sqlite3.connect(DBProjectManager2)
    con.row_factory = lambda curs, row:str(row[0])+". "+ row[1] #split ID with '.' 
    curs= con.cursor()
    curs.execute( """SELECT idProject, ProjectName 
                FROM Project WHERE idStatus=1""")
    rows=curs.fetchall()
    con.close()
    return rows

def GetIDFromDropDown(pickedString):
                 GetID=pickedString
                 GetID = list(GetID.split(".")) #id is before '.'
                 GetID=(int(GetID[0])) 
                 print(GetID)


root = Tk()
root.title("Tkinter ex")
root.geometry("400x400")

project_name_drop =  ttk.Combobox (root, value=DropDownProjectView() )
project_name_drop.pack()

buttonA=Button(root, text="get ID",command=lambda: GetIDFromDropDown(project_name_drop.get()))
buttonA.pack()

root.mainloop()

Answer 1

您可以从DropDownProjectView()返回字典而不是列表：

def DropDownProjectView():
    con=sqlite3.connect(DBProjectManager2)
    # return two items in each record: dropdown-item, id
    con.row_factory = lambda curs, row: (str(row[0])+". "+row[1], row[0])
    curs= con.cursor()
    curs.execute("SELECT idProject, ProjectName FROM Project WHERE idStatus=1")
    # build a dictionary with dropdown-item as key and id as value
    rows = {r[0]:r[1] for r in curs}
    con.close()
    return rows

然后使用list(rows.keys())作为下拉项：

rows = DropDownProjectView()
project_name_drop =  ttk.Combobox (root, value=list(rows.keys()))
project_name_drop.pack()

最后，使用rows[pickedString]获取GetIDFromDropDown()中的ID：

def GetIDFromDropDown(pickedString):
    # cater exceptional case
    if pickedString in rows:
        id = rows[pickedString]
        print(id)
    else:
        print("invalid option: '%s'" % pickedString)