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python - 按下按钮时执行功能 (tkinter)

转载 作者:行者123 更新时间:2023-12-01 03:19:10 26 4
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我正在尝试编写一个小程序,允许用户单击按钮来选择文件,输入库存编号,然后单击按钮来重命名这些文件。

这是我到目前为止所想到的:

import os
from tkinter import *
from tkinter import ttk
from tkinter.filedialog import askopenfilenames

root = Tk()
root.title("Photo Renamer")
mainframe = ttk.Frame(root, padding="3 3 12 12")
mainframe.grid(column=0, row=0, sticky=(N, W, E, S))
mainframe.columnconfigure(0, weight=1)
mainframe.rowconfigure(0, weight=1)

directory = ''
arrayOfFiles = []
stockNum = 0

def getStockNumber():
stockNum = stockNumber.get()

def selectFiles():
selectedFiles = askopenfilenames()
# Switch our files from a tuple to an array.
# arrayOfFiles = []
for i in selectedFiles:
arrayOfFiles.append(i)
return arrayOfFiles

def rename(array): # array will be arrayOfFiles
# Separate file names from the whole file path.
fileNames = []
for i in range(len(array)):
fileNames.append(os.path.basename(i))

count = 1
directory = os.path.dirname(array[1])
for file in directory:
oldFileName = '%s/%s' % (directory, file)
newFileName = '%s/gma%d_%d.jpg' % (directory, stockNum, count)
os.rename(oldFileName, newFileName)
count += 1

# "Stock Number" label
ttk.Label(mainframe, text="Stock Number: ").grid(column=1, row=1, sticky=W)

# Entry box
stockNumber = ttk.Entry(mainframe, width=7)
stockNumber.grid(column=2, row=1, sticky=(W, E))

# "Select Files" button
ttk.Button(mainframe, text="Select Files", command=selectFiles).grid(column=1, row=2, sticky=(W, E))

# "Number of Files" label
ttk.Label(mainframe, text="Number of Files: ").grid(column=1, row=3, sticky=W)

# "Rename" button
ttk.Button(mainframe, text="Rename", command= lambda: rename(arrayOfFiles)).grid(column=3, row=3, sticky=(W, E)) # command=rename

for child in mainframe.winfo_children(): child.grid_configure(padx=5, pady=5)
stockNumber.focus()

root.mainloop()

我可以选择文件,然后输入库存编号,但是当我单击重命名按钮时,它会在控制台中显示此错误:

Exception in Tkinter callback
Traceback (most recent call last):
File "C:\Users\...\AppData\Local\Programs\Python\Python35-32\lib\tkinter\__init__.py", line 1549, in __call__
return self.func(*args)
File "C:/Users/.../Desktop/Python/gui.py", line 56, in <lambda>
ttk.Button(mainframe, text="Rename", command= lambda: rename(arrayOfFiles)).grid(column=3, row=3, sticky=(W, E)) # command=rename
File "C:/Users/.../Desktop/Python/gui.py", line 32, in rename
fileNames.append(os.path.basename(i))
File "C:\Users\...\AppData\Local\Programs\Python\Python35-32\lib\ntpath.py", line 232, in basename
return split(p)[1]
File "C:\Users\...\AppData\Local\Programs\Python\Python35-32\lib\ntpath.py", line 204, in split
d, p = splitdrive(p)
File "C:\Users\...\AppData\Local\Programs\Python\Python35-32\lib\ntpath.py", line 139, in splitdrive
if len(p) >= 2:
TypeError: object of type 'int' has no len()

最佳答案

list.append() 需要一个字符串 - 您给出一个整数。改变

fileNames.append(os.path.basename(int))

fileNames.append(os.path.basename(str))

当 Python 代码(os 模块, see the docs over here ( Python 2 ))尝试获取 len(int) 时,会引发错误,该错误是不可能的(因为 int 没有方法 __len__)

关于python - 按下按钮时执行功能 (tkinter),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42110571/

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