如何从命令 tkinter 中获取价值

Question

这可能是一个新手问题，但我在弄清楚如何从函数中获取值以用作另一个函数的输入时遇到问题。我还想了解一些有关代码组织的技巧。在此示例中，我试图获取要在函数 processFile 中使用的 filePath 和 outPut Folder 路径。

谢谢

代码：

    from tkinter import *

from tkinter import ttk

from tkinter.filedialog import askopenfilename

from tkinter.filedialog import askdirectory

import PyPDF2

from PyPDF2 import PdfFileWriter, PdfFileReader

version = "v0.0.01"

root = Tk()

def window (main):

    main.title("File Opener " + version)

    width = main.winfo_width()

    main.geometry('500x200')

numero = str(1)

def OpenFile():

    fileName = askopenfilename(initialdir="C:/",

                            filetypes =(("PDF File", "*.pdf"),("All Files","*.*")),

                            title = "Select PDF File",

                           )

    labelName = Label(text="File Path: " + fileName)

    labelName.pack()

    print(fileName)

    return fileName

def outputFolder():                                         #Bug, label keeps adding paths

    outPath = askdirectory()

    labelName2 = Label(text="Output Folder: " + outPath)

    labelName2.pack()

    print(outPath)

    return outPath

def processFile(inFile,outFolder):

   ''' print(fileName) get input name and output variables

    print(outPath)'''

label = ttk.Label(root, text ="",foreground="black",font=("Helvetica", 16))

label.pack()

#Button Open-----------

button1 = Button (text = "Open File", command = OpenFile)

button1.pack()

#Button Start---------

buttonStart = Button (text = "Start Process", command = processFile)#give as parameter inputFile and link

buttonStart.place(height=50, width=100)

#Button Open-----------

button3 = Button (text = "Output Folder", command = outputFolder)

button3.pack()

#Menu Bar ----------------

menu = Menu(root)

root.config(menu=menu)

file = Menu(menu)

file.add_command(label = 'Open', command = OpenFile)

file.add_command(label = 'Exit', command = lambda:exit())

menu.add_cascade(label = 'File', menu = file)

window(root)

root.mainloop()

Answer 1

如果您在变量中使用该函数，则该函数的返回结果将存储在该变量中。def Function1(x):    return x*2var1 = Function1(5)然后 var1 = 10。但是关于您的按钮，您是想从用户那里获取在按下按钮时保存的输入还是希望按钮发送一个设定值？

Answer 2

from tkinter import *from tkinter import ttkfrom tkinter.filedialog import askopenfilenamefrom tkinter.filedialog import askdirectoryimport PyPDF2from PyPDF2 import PdfFileWriter, PdfFileReaderversion = "v0.0.01"root = Tk()def window (main):    main.title("File Opener " + version)    width = main.winfo_width()    main.geometry('500x200')    main.configure(background = 'black')numero = str(1)#openFiledef OpenFile():    global fileName    fileName = askopenfilename(initialdir="C:/",                            filetypes =(("PDF File", "*.pdf"),("All Files","*.*")),                            title = "Select PDF File",                           )    labelName = Label(text="File Path: " + fileName)    labelName.pack()    print(fileName)    return str(fileName)#getOutputPathdef outputFolder():                                         #Bug, label keeps adding paths    outPath = askdirectory()    labelName2 = Label(text="Output Folder: " + outPath)    labelName2.pack()    print(outPath)    return outPath#testFunctdef printFilename(inArg):    print(inArg)    x = OpenFile()    print (OpenFile())lambda: printFilename(fileName)#processRealDealdef processFile(): #THIS IS THE MAIN FUNC   ''' print(fileName) get input name and output variables    print(outPath)'''print (OpenFile)label = ttk.Label(root, text ="",foreground="black",font=("Helvetica", 16))label.pack()#Button Open-----------button1 = Button (text = "Open File", command = OpenFile)button1.pack()#Button Start---------buttonStart = Button (text = "Start Process", command = lambda: printFilename("Hello World!!!"))#give as parameter inputFile and linkbuttonStart.place(height=50, width=100)#Button Open-----------button3 = Button (text = "Output Folder", command = outputFolder)button3.pack()#Menu Bar ----------------menu = Menu(root)root.config(menu=menu)file = Menu(menu)file.add_command(label = 'Open', command = OpenFile)file.add_command(label = 'Exit', command = lambda:exit())menu.add_cascade(label = 'File', menu = file)window(root)root.mainloop()