使用 TKinter 创建浏览按钮

Question

I'm writing some code where the user needs to be able to select a file that the program will run on. I have created a browse button that allows the user to select a file but when you hit 'okay' the rest of the program doesn't realize that there has been an input. The file name should also automatically be entered in he browse bar after the file has been selected. Any suggestions?

from Tkinter import *

class Window:       

def __init__(self, master):     

    #Browse Bar
    csvfile=Label(root, text="File").grid(row=1, column=0)
    bar=Entry(master).grid(row=1, column=1) 

    #Buttons  
    y=7
    self.cbutton= Button(root, text="OK", command=master.destroy)       #closes window
    y+=1
    self.cbutton.grid(row=10, column=3, sticky = W + E)
    self.bbutton= Button(root, text="Browse", command=self.browsecsv)
    self.bbutton.grid(row=1, column=3)

#-------------------------------------------------------------------------------------#
def browsecsv(self):
    from tkFileDialog import askopenfilename

    Tk().withdraw() 
    filename = askopenfilename()

#-------------------------------------------------------------------------------------#
import csv

with open('filename', 'rb') as csvfile:
    logreader = csv.reader(csvfile, delimiter=',', quotechar='|')
    rownum=0

    for row in logreader:    
        NumColumns = len(row)        
        rownum += 1

    Matrix = [[0 for x in xrange(NumColumns)] for x in xrange(rownum)] 

csvfile.close()


root = Tk()
window=Window(root)
root.mainloop()  

Answer 1

you can also use tkFileDialog..

import Tkinter,tkFileDialog

root = Tkinter.Tk()
file = tkFileDialog.askopenfile(parent=root,mode='rb',title='Choose a file')
if file:
    data = file.read()
    file.close()
    print "I got %d bytes from this file." % len(data)

Answer 2

filename = askopenfilename() is only known in this scope, you have to return it or use it in any way.

See this site for more examples:

    Tkinter.Button(self, text='Browse', command=self.askopenfile)

...

    def askopenfile(self):
        return tkFileDialog.askopenfile(mode='r', **self.file_opt)

---

EDIT

Bryan Oakley is right of course! That is what I meant when i said "use it in any way" ;) At one point you choose a filename, at anoter you simply use filename .

How about this?

from Tkinter import *
import csv

class Window:       
def __init__(self, master):     
    self.filename=""
    csvfile=Label(root, text="File").grid(row=1, column=0)
    bar=Entry(master).grid(row=1, column=1) 

    #Buttons  
    y=7
    self.cbutton= Button(root, text="OK", command=self.process_csv)
    y+=1
    self.cbutton.grid(row=10, column=3, sticky = W + E)
    self.bbutton= Button(root, text="Browse", command=self.browsecsv)
    self.bbutton.grid(row=1, column=3)

def browsecsv(self):
    from tkFileDialog import askopenfilename

    Tk().withdraw() 
    self.filename = askopenfilename()

def process_csv(self):
    if self.filename:
        with open(self.filename, 'rb') as csvfile:
            logreader = csv.reader(csvfile, delimiter=',', quotechar='|')
            rownum=0

            for row in logreader:    
                NumColumns = len(row)        
                rownum += 1

            Matrix = [[0 for x in xrange(NumColumns)] for x in xrange(rownum)] 

root = Tk()
window=Window(root)
root.mainloop()  

There is still a lot to do with that, but at least you don't try to open a file before having determined its name.

Answer 3

# importing tkinter and tkinter.ttk 
# and all their functions and classes 
from tkinter import * 
from tkinter.ttk import *

# importing askopenfile function 
# from class filedialog 
from tkinter.filedialog import askopenfile 

root = Tk() 
root.geometry('200x100') 

# This function will be used to open 
# file in read mode and only Python files 
# will be opened 
def open_file(): 
    file = askopenfile(mode ='r', filetypes =[('Python Files', '*.docx')]) 
    if file is not None: 
        content = file.read() 
        print(content) 

btn = Button(root, text ='Open', command = lambda:open_file()) 
btn.pack(side = TOP, pady = 10) 

mainloop() 

Answer 4

The root of the problem is that you're trying to process the file before the user has the chance to pick a file.

You need to put the block of code beginning with with open('filename', 'rb') as csvfile: in a function, then call the function as a result of the user pressing the button. For example, you could call it from within the browsecsv function.

Also, you don't need csv.close() , that comes for free when using the with statement.

Answer 5

I have edited above code to use in python 3.6. Only package name changes

    import tkinter
    from tkinter import filedialog
    file = filedialog.askopenfile(parent=root,mode='rb',title='Choose a file')
    if file != None:
        data = file.read()
        file.close()
        print("I got %d bytes from this file." % len(data))