tkinter callback through an error: "Index 0 out of range"

Question

我搜索了几个小时，到底是什么原因导致出现此错误消息：我有一个搜索条目，它根据我的搜索使用回调函数更新列表框：

列表框：

self.name_search=tk.StringVar()
self.name_search.trace_add('write', self.my_callback)
self.e_name_search_text = tk.Label(search_f, text="Name: ").grid(row=0, column=0, padx=10, pady=5, sticky='E') 
self.e_name_search = ttk.Entry(search_f, width = 35, textvariable=self.name_search)
self.e_name_search.grid(row=0, column=1, padx=5, pady=5, sticky='W')

self.lbox = tk.Listbox(search_f, width=35, height=8)
self.lbox.bind("<Double-Button-1>", self.show_name_search) 
self.lbox.bind('<Return>', self.show_name_search)          
self.scrollbar = tk.Scrollbar(search_f)
self.lbox.grid(row=1, column=1, rowspan=3, padx=10, pady=1)
self.lbox.config(yscrollcommand = self.scrollbar.set)
self.scrollbar.grid(row=1, column=2, rowspan=3, padx=1, pady=1, sticky='ns')
self.scrollbar.config(command=self.lbox.yview)

因此，如果我键入搜索，列表框会显示我的 sqlite 数据库中的简化值列表，我很感兴趣。如果我 select 双击一个。另一个 sqlite 查询更新我的组合框。

如果我 select 一个我得到这个错误：

Exception in Tkinter callback
Traceback (most recent call last):
  File "C:\Python38-32\lib\tkinter\__init__.py", line 1883, in __call__
    return self.func(*args)
  File "D:\... name.py", line 337, in show_name_search
    self.e_fax.current(0)
  File "C:\Python38-32\lib\tkinter\ttk.py", line 717, in current
    return self.tk.call(self._w, "current", newindex)
_tkinter.TclError: Index 0 out of range

第 337 行来自另一个函数：

def show_name_search(self, event):
    self.clear_field()
    widget = event.widget
    selection = widget.curselection()
    
    indName = widget.get(selection[0])
    print(indName)
    print("selktierter Wert: {}".format(indName))
    
    self.realName.set(indName)
    
    
    connection = sqlite3.connect(select_connect_db)
    print('Database connected.')
    with connection: 
        cursor = connection.cursor()
        cursor.execute("SELECT number, type, prio, id, uniqueid FROM numbers WHERE realName=?;",(indName,))
        data = cursor.fetchall()
        print(data)
        
        for row in data:
            if row[1] == 'home':
                self.phone_home.append(row[0])
                print('HOME:',self.phone_home)
            
            if row[1] == 'mobile':
                self.mobile.append(row[0])
                print('Mobile:',self.mobile)
            
            if row[1] == 'work':
                self.business.append(row[0])
                print(row[0])
                print('WORK:',self.business)
                
            if row[1] == 'fax_work':
                self.fax.append(row[0])
                print(row[0])
                print('FAX_WORK:',self.fax)
                
            self.uid_name.set(row[4])
                
        if len(self.phone_home) != 0: 
            self.e_phone['values'] = self.phone_home
            self.e_phone.current(0)
        if len(self.mobile) != 0:
            self.e_mobile['values'] = self.mobile
            self.e_mobile.current(0)
        if len(self.business) != 0:
            self.e_business['values'] = self.business # Set the value to the new list
            self.e_business.current(0) # Set the first item of the list as current item
        if len(self.business) != 0:
            self.e_fax['values'] = self.fax
            self.e_fax.current(0)      ### Line 337 - No entry for this value in my sqlite database

知道吗，我可以搜索什么？

Answer 1

所以 self.e_fax 对我来说就像 ttk.Combobox。考虑这里的代码：

import tkinter as tk
from tkinter import ttk
   
root = tk.Tk()
values = []
lb = ttk.Combobox(root,values=values)
lb.current(0)
lb.pack()
root.mainloop()

它通过同样的错误：

_tkinter.TclError: Index 0 out of range

原因是列表 values 是空的，在其中插入任何常规字符串即可。确保如果你想设置默认值，有一个值。

import tkinter as tk
from tkinter import ttk
   
root = tk.Tk()
values = ['see']
lb = ttk.Combobox(root,values=values)
lb.current(0)
lb.pack()
root.mainloop()

tkinter callback through an error: "Index 0 out of range"

tkinter callback through an error: "Index 0 out of range"

sqlite

tkinter

python-3.8