Question

我正在尝试编写一个小程序，允许用户单击按钮选择文件，输入库存号，然后单击按钮重命名这些文件。

这是我到目前为止所提出的：

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，然后输入库存号，但是当我点击重命名按钮时，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()

Answer 1

for i in range(len(array)):
    fileNames.append(os.path.basename(i))

在上面的代码中，i是一个整数，os.path.basename(path)需要一个类似字符串的路径名。

迭代列表时，您可以直接遍历项目。

for file in array:
    fileNames.append(os.path.basename(file))

或者如果你想使用索引，你可以做

for i in range(len(array)):
    fileNames.append(os.path.basename(array[i]))
                                      ^^^^^ notice accessing an item using index

Answer 2

list.append（）需要一个字符串 - 你给出一个整数。变化

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

到

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

在尝试获取os时，Python代码（len(int)模块see the docs over here（Python 2））中出现错误，这是不可能的（因为{ {1}}没有方法int）

Answer 3

试试这个：

for i in range(array):

而不是

for i in range(len(array)):

按下按钮执行功能（tkinter）

3 个答案: