Question

这是我的主文件：

ExcelAppl = win32com.client.Dispatch('Excel.Application')
    Workbook = ExcelAppl.Workbooks.Open('excel file path')
    Sheet = Workbook.Worksheets.Item(1)

这是 Python GUI ：

root = Tk()
def callback():
    file_path = tkFileDialog.askopenfilename(filetypes=[("Excel files","*.xls")])
    print "the file path  %s",file_path
    Text_button.insert(INSERT,file_path)
def execute():
    execfile("MainLibrary.py")
Browse_button = Button(text='Browse', command=callback).pack(side=LEFT, padx=10, pady=20, ipadx=20, ipady=10)
Execution_button = Button(text='Convert', command=execute).pack(side=BOTTOM, padx=10, pady=20, ipadx=20, ipady=10)

root.mainloop()

这是我的查询。我需要从GUI获取file_path并将其分配给Workbook = ExcelAppl.Workbooks.Open('excel file path')。我需要从GUI获取路径并将路径分配给我的主代码。有人可以帮忙吗？

Answer 1

此处您不需要execfile()。

您可以在main_library.py中定义一个函数：

import win32com.client

def do_something_with_excel_file(filename):
    app = win32com.client.Dispatch('Excel.Application')
    workbook = app.Workbooks.Open(filename)
    worksheet = workbook.Worksheets.Item(1)
    # use `worksheet` ...

if __name__=="__main__":
   import sys
   # get filename from command-line if called as a script
   do_something_with_excel_file(sys.argv[1])

然后您可以在GUI代码中导入它：

import tkFileDialog
from Tkinter import as tk

import main_library

def browse():
    path = tkFileDialog.askopenfilename(filetypes=[("Excel files","*.xls")])
    file_path.set(path)
    print "the file path  %s", path

def convert():
    root.destroy() # quit gui
    main_library.do_something_with_excel_file(file_path.get())

root = tk.Tk()
file_path = tk.StringVar()
tk.Label(root, textvariable=file_path).pack()
tk.Button(text='Browse', command=browse).pack()
tk.Button(text='Convert', command=convert).pack()
root.mainloop()

如何从Python GUI获取文件路径并在另一个python程序中使用文件路径

1 个答案: