将元组列表从一个追加到另一个
将元组列表从一个追加到另一个
提问于 2019-09-29 23:23:57
我正在创建一个UI，其中我从SQL数据库文件中获取一组数据并将其插入到列表框中。我使用的是Tkinter框架。我遇到了一个问题，我想把元组追加到另一个空数组上，这样我就可以在设定的延迟内追加有限数量的数据。我已经展示的代码，是我到目前为止创建的代码，但我收到了下面的错误，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回答 1
Stack Overflow用户
回答已采纳
发布于 2019-09-30 02:04:24
您的问题可以简化为：
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list1的for循环中使用名称counter：
for item in list1:
print(item)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/58159967
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