用“FILEPICKER”python打开一个数据文件（csv，xls，xlsx，ods等）？

当我们知道文件的名称和types时，我知道如何打开一个数据文件，但是我们如何使用文件select器来select文件？

import pyexcel as pe records = pe.get_records(file_name="your_file.xls") for record in records: print("%s is aged at %d" % (record['Name'], record['Age'])) 

Answer 1

您可以定义一个函数来返回您可以用作pe.get_records（）的input的函数的名称。 from tkinter import * root = Tk() def get_file_name(): global root root.filename = filedialog.askopenfilename(initialdir = "D:/",title = "choose your excel file",filetypes = (("excel files","*.xls"),("all files","*.*"))) print (root.filename) root.withdraw() return root.filename

Answer 2

我得到了问题，现在这个代码运行！

 filters = ( 'Data Files (*.csv *.txt *.xls *.xml *.xlsx *.xlsm)', ) path, filter = QtGui.QFileDialog.getOpenFileNameAndFilter( self, 'Open File', '', ';;'.join(filters))