将元组列表从一个附加到另一个

appending tuple list from one to another

我正在创建一个 UI,我从 SQL 数据库文件中获取一组数据并将其插入列表框。我正在使用 Tkinter 框架。我遇到一个问题,我想将元组附加到另​​一个空数组,这样我就可以附加有限数量的数据并设置延迟。我显示的代码是我到目前为止创建的代码,但我收到以下错误并且我不知道如何修复它。我也不太熟悉使用元组。

    from tkinter import *
    import sqlite3 as sq
    import time

    def Record():
        conn = sq.connect('brian_UI_test.db')
        c = conn.cursor()
        c.execute("SELECT*FROM TEST")

       rows = c.fetchall() # Gets the data from the table

    counter = 0
    list1 = []
    for row in rows:
        counter = counter + 1
        Lb.insert(END,row)# Inserts record row by row in list box
        list1.append(row)
        print(row)
        time.sleep(1)
        if counter%4==0:
           for counter in list1:
            print(counter)
            Lb.insert(counter)



    c.close()
    conn.close()

    window = Tk()
    frame = Frame(window)

    Lb = Listbox(frame, height = 8, width = 25,font=("arial", 12)) 

    scroll = Scrollbar(frame, orient = VERTICAL) # set scrollbar to listbox for when entries exceed size of list box
    scroll.config(command = Lb.yview)

    scroll1 = Scrollbar(frame, orient = HORIZONTAL)
    scroll1.config(command = Lb.xview)

    Lb.config(yscrollcommand = scroll.set)
    Lb.config(xscrollcommand = scroll1.set) 

    scroll1.pack(side = BOTTOM, fill = X)
    Lb.pack(side = LEFT, fill = Y)
    scroll.pack(side = RIGHT, fill = Y)

    Lb.insert(0, 'Time,   Message') #first row in listbox

    b13 = Button(window, text = "OPEN DB File", width= 14, command=lambda:Record())
    b13.grid(row = 0, column = 7, padx = 10)

错误:

Traceback (most recent call last):
File "C:\Program Files (x86)\Microsoft Visual 
Studio\Shared\Python37_64\lib\tkinter\__init__.py",line 1705, in __call__return self.func(*args)
File "Interface.py", line 122, in <lambda>
b13 = Button(window, text = "OPEN DB File", width= 14, command=lambda:Record())
File "Interface.py", line 25, in Record
counter = counter + 1
TypeError: can only concatenate tuple (not "int") to tuple

您的问题可以简化如下:

counter = 0
list1 = []
rows = [(1,2),(3,4,),(5,6,),(7,8),(9,10)]

for row in rows:
    counter = counter + 1 #after the first loop, counter is a tuple and cannot concatenate with an int
    list1.append(row)
    if counter % 4 == 0:
        for counter in list1: #you redefined counter here which is a tuple
            print(counter)

简单的解决方案是不在列表 1 的 for 循环中使用名称 counter

for item in list1:
    print(item)