_tkinter.TclError：预期的整数，但得到“Serif”答案

【问题标题】：_tkinter.TclError: expected integer but got "Serif"_tkinter.TclError：预期的整数，但得到“Serif”
【发布时间】：2018-08-14 02:06:45
【问题描述】：

正在尝试更改 tkinter 中的字体。按钮一直有效，直到我到达数字 7 我的猜测是 MS Serif 是两个词，通常我会用“”包围两个词但我不能像我写程序的方式。

我的问题是如何修复我的函数以接受字体两个单词的名字。

这是输出：

Counter:  1  Fontz:  Terminal 12 bold
Counter:  2  Fontz:  Fixedsys 12 bold
Counter:  3  Fontz:  Modern 12 bold
Counter:  4  Fontz:  Roman 12 bold
Counter:  5  Fontz:  Script 12 bold
Counter:  6  Fontz:  Courier 12 bold
Counter:  7  Fontz:  MS Serif 12 bold
Exception in Tkinter callback
Traceback (most recent call last):
  File "C:\Users\Slim 960\AppData\Local\Programs\Py
    return self.func(*args)
  File "Tkinter001.py", line 83, in button_font
    logo = tk.Label(root, compound = tk.CENTER, tex
.grid(row=0, column=0)
  File "C:\Users\Slim 960\AppData\Local\Programs\Py
    Widget.__init__(self, master, 'label', cnf, kw)
  File "C:\Users\Slim 960\AppData\Local\Programs\Py
    (widgetName, self._w) + extra + self._options(c
_tkinter.TclError: expected integer but got "Serif"

代码：

from tkinter import font

import os

import random
import tkinter as tk

# Give 10 blank lines
print("\n"*10)

# Create path to GURPS gif files
home = os.path.expanduser('~')
dt = desktop = os.path.join(home,'Desktop')
p3p = python_3_projects = os.path.join(dt, "Python 3 Projects")
ilfru = images_labeled_for_reuse = os.path.join(p3p,"Images Labeled for Reuse")
gurps = os.path.join(ilfru,'GURPS')
gg = os.path.join(gurps, 'GIF')

# Count all then number of GIFS in Folder GIFS
fc = 0
for root, dir, files in os.walk(gg):
    for filename in files:
    fc += 1
    print("Filenames: ",filename)
    print("The total number of files is: ",fc)

# Generate Random Number from maximum number of files
rn = random.randint(1,fc)
print("The random number is: ", rn)


# Assign a file path to lip (logo_image_path)
fc = 0
for root, dir, files in os.walk(gg):
    for filename in files:
    fc += 1
    if fc == rn:
        lip = os.path.join(gg,filename)
        print("The Random Number is: ",rn)
        print("The choosen file is: ", filename)
        print("The file counter is: ", fc)
        print("The file path is: ", lip)

f_counter = 0
ff = []

# Cycle through Fonts
def button_font():
    global f_counter
    global ff
    global fontz

    f_counter +=1
    fontz = str(ff[f_counter]) + " 12 bold"
    print("Counter: ",f_counter," Fontz: ",fontz)

    logo = tk.Label(root, compound = tk.CENTER, text = "GURPS\n CHARACTER\n SHEET", font=fontz, fg ="white", image=ggpi).grid(row=0, column=0)

root = tk.Tk()
ggpi =gurps_gif_photo_image = tk.PhotoImage(file=lip)

# List of all tkinter fonts
tk_fonts = font.families()
for font in tk_fonts:
    ff.append(font)
print(ff)
button_font()
logo = tk.Label(root, compound = tk.CENTER, text = "GURPS\n CHARACTER\n SHEET", font=fontz, fg ="white", image=ggpi).grid(row=0, column=0)
Label1 = tk.Label(root, text="Name: ", font ="bold").grid(row=0, column = 1)
button = tk.Button(root, text="Font Changer", command=button_font).grid(row=3,column=3)
root.mainloop()

【问题讨论】：

附带说明（这不是您要问的问题，但这是您需要解决的问题），grid 方法返回None，所以你的变量@ 987654325@、button 等不是Label、Button 等对象，它们都只是None。如果您想将小部件存储在变量中以供以后使用，则必须将小部件分配给变量，然后在变量上调用grid，作为两个单独的语句。
终于遇到问题并找到了您提到的问题的解决方案。感谢您的提醒。

标签： python string tkinter fonts

【解决方案1】：

使用元组而不是字符串 - 当名称有多个空格时，这允许 tkinter 不会将名称与大小混淆：

这行得通：

fontz = (str(ff[f_counter]), 12, "bold")
logo = tk.Label(root, compound=tk.CENTER, 
    text="GURPS\n CHARACTER\n SHEET",
    font=fontz,
    fg ="white", 
    image=ggpi
).grid(row=0, column=0)

【讨论】：