Python 在 tkinter 中打开多个文件名并将文件名添加到列表中
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open multiple filenames in tkinter and add the filesnames to a list
提问by faraz
what I want to do is to select multiple files using the tkinter filedialog and then add those items to a list. After that I want to use the list to process each file one by one.
我想要做的是使用 tkinter 文件对话框选择多个文件，然后将这些项目添加到列表中。之后我想使用列表来一个一个地处理每个文件。
#replace.py
import string
def main():
#import tkFileDialog
#import re
#ff = tkFileDialog.askopenfilenames()
#filez = re.findall('{(.*?)}', ff)
import Tkinter,tkFileDialog
root = Tkinter.Tk()
filez = tkFileDialog.askopenfilenames(parent=root,title='Choose a file')
Now, I am able to select multiple files, but I dont know how to add those filenames to the list. any ideas?
现在，我可以选择多个文件，但我不知道如何将这些文件名添加到列表中。有任何想法吗？
采纳答案by A. Rodas
askopenfilenamesreturns a string instead of a list, that problem is still open in the issue tracker, and the best solution so far is to use splitlist:
askopenfilenames返回一个字符串而不是一个列表，该问题在问题跟踪器中仍然存在，目前最好的解决方案是使用splitlist：
import Tkinter,tkFileDialog
root = Tkinter.Tk()
filez = tkFileDialog.askopenfilenames(parent=root,title='Choose a file')
print root.tk.splitlist(filez)
回答by Belial
askopenfilenames
returns a tuple of strings, not a string. Simply store the the output of askopenfilenamesinto filez (as you've done) and pass it to the python's listmethod to get a list.
返回一个字符串元组，而不是一个字符串。只需将askopenfilenames的输出存储到 filez 中（如您所做的那样）并将其传递给 python 的list方法以获取列表。
filez = tkFileDialog.askopenfilenames(parent=root,title='Choose a file')
lst = list(filez)
>>> type(lst)
<type 'list'>
回答by mvx
Putting together parts from above solution along with few lines to error proof the code for tkinter file selection dialog box (as I also described here).
将上述解决方案中的部分与几行放在一起，以防止 tkinter 文件选择对话框的代码出错（正如我在此处所述）。
import tkinter as tk
from tkinter import filedialog
root = tk.Tk()
root.withdraw()
root.call('wm', 'attributes', '.', '-topmost', True)
files = filedialog.askopenfilename(multiple=True)
%gui tk
var = root.tk.splitlist(files)
filePaths = []
for f in var:
filePaths.append(f)
filePaths
Returns a list of the paths of the files. Can be strippedto show only the actual file name for further use by using the following code:
返回文件路径的列表。可以stripped使用以下代码仅显示实际文件名以供进一步使用：
fileNames = []
for path in filePaths:
name = path[46:].strip()
name2 = name[:-5].strip()
fileNames.append(name2)
fileNames
where the integers (46) and (-5) can be altered depending on the file path.
其中整数 (46) 和 (-5) 可以根据文件路径进行更改。
