CODEWITHSUNDEEP
pythontkinter
uncleshelby
uncleshelby

Reputation: 553

How to make a button that returns the input filename of tkFileDialog.askopenfilename() as a string?

How can I create a snippet of code that will create a Tkinter Button widget that opens a tkFileDialog.askopenfilename() window, and when you click the "Open" button in the window, make it get the filename as a string, and insert it into an entry.

Here is what I have.

iconEntry = Entry(iconRow)
iconEntry.pack()

def getFileName()
    fileName = tkFileDialog.askopenfilename()
    iconEntry.insert(0, fileName)

iconButton = Button(iconRow, text="Browse", command=getFileName)
iconButton.pack(side=RIGHT)

Upvotes: 2

Views: 2553

Answers (1)

uncleshelby
uncleshelby

Reputation: 553

I've got it myself.

root = Tk()

def getFIleName(varName, entryName):
    varName = tkFileDialog.askopenfilename()
    entryName.insert(0, varName)

iconButton = Button(root, text="Browse", command=(lambda: getImageName(campIcon, iconEntry)))
iconButton.pack()

Upvotes: 4

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