ValueError: invalid literal for int() with base 10: '' (Tkinter)

Question

当我尝试运行这段代码时，出现 ValueError 暗指函数 numRandom。我虽然 Python 可以将 int 的字符串表示形式传递给 int.

import tkinter
import random

window = tkinter.Tk()
window.geometry('600x500')

x = random.randint(1,300)
remainingTime = True
Attempts = 4

def numRamdom():
    global Attempts
    while Attempts > 0:
        numWritten = int(entryWriteNumber.get())
        if numWritten > x:
            lblClue.configure(text = 'Its a bigger number')
            Attempts = Attempts -1
        if numWritten < x:
            lblClue.configure(text = 'Its a smaller number')
            Attempts = Attempts -1
        if numWritten == x:
            lblClue.configure(text = 'Congratulations ;)')
            remainingTime = False
            return remainingTime, countdown(0)
        if Attempts == 0:
            remainingTime = False
            return remainingTime, countdown(0), Attempts, gameOver()

entryWriteNumber = tkinter.Entry(window)
entryWriteNumber.grid(column = 0, row = 1, padx = 10, pady = 10)

numRamdom()

window.mainloop()

Answer 1

问题是因为当代码是运行时，它直接调用numRamdom()，也就是说，最初入口小部件是空的，它们运行它与那些空入口小部件，因此错误。所以只需分配一个按钮和一个命令，例如：

b = tkinter.Button(root,text='Click me',command=numRamdom)
b.grid(row=1,column=0)

确保在 def numRamdom(): 之后的 mainloop() 之前说这句话。该按钮只是运行只有在单击该按钮时才起作用。

或者如果你想要 button-less 那么试试：

方法一：

root.after(5000,numRamdom) #after 5 sec it will execute function

但请记住，如果用户没有在 5 秒内正确输入，则会弹出一些错误消息。

方法二：

def numRamdom(event):
......

entryWriteNumber.bind('<Return>',numRamdom)

这样一来，如果您在输入小部件中按回车键（在输入数据后），它将运行该函数。

希望对您有所帮助，如有错误请告诉我。

干杯

Answer 2

这是一个基于您的代码的完整示例。您的问题是试图在 Entry 的内容包含任何内容之前转换它的内容。要解决此问题，您可以添加一个调用命令 numRamdom()

的按钮

import tkinter
import random

window = tkinter.Tk()
window.geometry('600x500')

x = random.randint(1,300)
remainingTime = True
Attempts = 4

def numRamdom():
    global Attempts, lblClue, x
    if Attempts > 0:
            numWritten = int(entryWriteNumber.get())
            if numWritten < x:
                lblClue.configure(text = 'Its a bigger number')
                Attempts = Attempts -1
            elif numWritten > x:
                lblClue.configure(text = 'Its a smaller number')
                Attempts = Attempts -1
            else:
                lblClue.configure(text = 'Congratulations ;)')
                remainingTime = False
                #return remainingTime, countdown(0)
            if Attempts == 0:
                remainingTime = False
                #return remainingTime, countdown(0), Attempts, gameOver()
    else:
        lblClue.configure(text = "You ran out of attempts!")

entryWriteNumber = tkinter.Entry(window)
entryWriteNumber.grid(column = 0, row = 1, padx = 10, pady = 10)

entryWriteButton = tkinter.Button(window, text = "Push me!", command = numRamdom)
entryWriteButton.grid(column = 1, row = 1)

lblClue = tkinter.Label(window)
lblClue.grid(row = 2, column = 1)


window.mainloop()

如果传递的值无法转换为整数，您仍然会收到错误消息，但这很容易通过 if 语句修复。

ValueError: invalid literal for int() with base 10: '' (Tkinter)

ValueError: invalid literal for int() with base 10: '' (Tkinter)

python

tkinter

valueerror