在 Tkinter 中制作更新加载栏（标签）时遇到问题
Trouble with making a updating loading bar (Label) in Tkinter
我正在对加载栏（标签）进行百分比更新，它应该会在我的程序启动时自动启动。当 'Loading bar' 达到 100% 时，'Loading bar' 应该被删除，程序的其余部分应该继续，尽管当我启动程序时，window 直到大约 5 点才响应秒。
number = 0
number1 = str(number)
Labe = tk.Label(root, width=10,bd=1, bg="#ADA5A5", font=("Arial Bold", 36), fg="White")
Labe.grid(row=0, column=0)
def work():
global number1
Labe['text'] = "Loading" + " " + number1 + "%"
global number
number = number + 1
number1 = str(number)
if number == 102:
root.after(1000)
print("end")
Labe.grid_remove()
num = 101
for x in range(num):
print("He")
root.after(50, work)
root.after(50)
#Adding stuff to root
rootHelloLabel = tk.Label(root, text="Ordering system, please sign in", bg="#7271AD",
fg="White", font=("Arial Bold", 40), width=48, height=2)
placeholder = tk.Label(root, text="", pady=20, bg="#ACAD71")
placeholder.grid(row=0, column=0)
rootHelloLabel.grid(row=1, column=0)
placeholder1 = tk.Label(root, text="", pady=70, bg="#ACAD71")
placeholder1.grid(row=2, column=0)
loginButton = tk.Button(root, pady=30, padx=80, text="Login", font=("Arial Bold", 27),
bg="#71AD90", fg="#D0CFD1", command=login)
loginButton.grid(row=3, column=0)
adminButton = tk.Button(root, pady = 10, padx = 40, text="Admin",
font=("Arial Bold", 27), bg="#71AD90", fg="#D0CFD1")
adminButton.grid(row=5, column=0)
我做了三个小改动，我认为它有效。
number = 0
number1 = str(number)
Labe = tk.Label(root, width=10,bd=1, bg="#ADA5A5", font=("Arial Bold", 36), fg="White")
Labe.grid(row=0, column=0)
def work():
global number1
Labe['text'] = "Loading" + " " + number1 + "%"
global number
number = number + 1
number1 = str(number)
if number == 102:
root.after(1000)
print("end")
Labe.grid_remove()
num = 101
for x in range(num):
print("He")
root.after(50, work)
root.after(50)
root.update() # added this so the window displays for one frame
# it's similar to mainloop() but only runs once
#Adding stuff to root
rootHelloLabel = tk.Label(root, text="Ordering system, please sign in", bg="#7271AD",
fg="White", font=("Arial Bold", 40), width=48, height=2)
placeholder = tk.Label(root, text="", pady=20, bg="#ACAD71")
placeholder.grid(row=0, column=0)
rootHelloLabel.grid(row=1, column=0)
placeholder1 = tk.Label(root, text="", pady=70, bg="#ACAD71")
placeholder1.grid(row=2, column=0)
loginButton = tk.Button(root, pady=30, padx=80, text="Login", font=("Arial Bold", 27),
bg="#71AD90", fg="#D0CFD1", command=lambda: login())
loginButton.grid(row=3, column=0) # added lambda function so this only runs when clicked
adminButton = tk.Button(root, pady = 10, padx = 40, text="Admin",
font=("Arial Bold", 27), bg="#71AD90", fg="#D0CFD1")
adminButton.grid(row=5, column=0)
root.mainloop() # added this so that the program runs when done (can remove aswell)
我希望这对你有用。我不太明白这个问题，但是一旦我添加了 update() 函数，它就可以正常工作，我意识到你所说的加载栏是什么意思。希望对您有所帮助！
你在 for 循环中 运行s 100 次 after(50) 所以它可以停止程序 5 秒。而且我不知道你是否真的需要它。
我会 运行 after(50, work) 直接在 work
def work():
global number # PEP8: all `global` at the beginning of function
labe['text'] = f"Loading {number}%"
number += 1
if number > 100:
root.update() # to update last text in label because `after(1000)` will block it
root.after(1000)
print("end")
labe.grid_remove()
else:
root.after(50, work)
完整的工作代码（有其他更改）：
PEP 8 -- Style Guide for Python Code
import tkinter as tk
# --- functions --- # PEP8: all functions directly after imports (and after all classes)
def work():
global number # PEP8: all `global` at the beginning of function
labe['text'] = f"Loading {number}%"
number += 1
if number > 100:
root.update() # to update last text in label because `after(1000)` will block it
root.after(1000)
print("end")
labe.grid_remove()
else:
# run again
root.after(50, work)
def login():
pass
# --- main ---
number = 0
root = tk.Tk()
labe = tk.Label(root, width=10, bd=1, bg="#ADA5A5", font=("Arial Bold", 36), fg="White")
labe.grid(row=0, column=0)
root_hello_label = tk.Label(root, text="Ordering system, please sign in", bg="#7271AD",
fg="White", font=("Arial Bold", 40), width=48, height=2)
root_hello_label.grid(row=1, column=0)
placeholder1 = tk.Label(root, text="", pady=20, bg="#ACAD71")
placeholder1.grid(row=0, column=0)
placeholder2 = tk.Label(root, text="", pady=70, bg="#ACAD71")
placeholder2.grid(row=2, column=0)
login_button = tk.Button(root, pady=30, padx=80, text="Login", font=("Arial Bold", 27),
bg="#71AD90", fg="#D0CFD1", command=login)
login_button.grid(row=3, column=0)
admin_button = tk.Button(root, pady = 10, padx = 40, text="Admin",
font=("Arial Bold", 27), bg="#71AD90", fg="#D0CFD1")
admin_button.grid(row=5, column=0)
# run first time
root.after(50, work)
root.mainloop()
编辑：
如果您需要 运行 和 for 循环，那么您应该在 after(..., work)
中设置不同的时间
for i in range(1, 102):
root.after(50*i, work)
完整的工作代码：
import tkinter as tk
# --- functions --- # PEP8: all functions directly after imports (and after all classes)
def work():
global number # PEP8: all `global` at the beginning of function
labe['text'] = f"Loading {number}%"
number += 1
if number > 100:
root.update() # to update last text in label because `after(1000)` will block it
root.after(1000)
print("end")
labe.grid_remove()
def login():
pass
# --- main ---
number = 0
root = tk.Tk()
labe = tk.Label(root, width=10, bd=1, bg="#ADA5A5", font=("Arial Bold", 36), fg="White")
labe.grid(row=0, column=0)
root_hello_label = tk.Label(root, text="Ordering system, please sign in", bg="#7271AD",
fg="White", font=("Arial Bold", 40), width=48, height=2)
root_hello_label.grid(row=1, column=0)
placeholder1 = tk.Label(root, text="", pady=20, bg="#ACAD71")
placeholder1.grid(row=0, column=0)
placeholder2 = tk.Label(root, text="", pady=70, bg="#ACAD71")
placeholder2.grid(row=2, column=0)
login_button = tk.Button(root, pady=30, padx=80, text="Login", font=("Arial Bold", 27),
bg="#71AD90", fg="#D0CFD1", command=login)
login_button.grid(row=3, column=0)
admin_button = tk.Button(root, pady = 10, padx = 40, text="Admin",
font=("Arial Bold", 27), bg="#71AD90", fg="#D0CFD1")
admin_button.grid(row=5, column=0)
# 102-1 = 101 loops, because it needs to display values 0...100 which gives 101 values
for i in range(1, 102):
root.after(50*i, work)
root.mainloop()
编辑：
因为 after(1000) 冻结所有 GUI 1 秒，所以最好使用 after(1000, remove_label) 在 1 秒后执行其他功能，这不会冻结 GUI。
def work():
global number
labe['text'] = f"Loading {number}%"
number += 1
if number > 100:
root.after(1000, remove_label)
else:
root.after(50, work)
def remove_label():
print("end")
labe.grid_remove()
我正在对加载栏（标签）进行百分比更新，它应该会在我的程序启动时自动启动。当 'Loading bar' 达到 100% 时，'Loading bar' 应该被删除，程序的其余部分应该继续，尽管当我启动程序时，window 直到大约 5 点才响应秒。
number = 0
number1 = str(number)
Labe = tk.Label(root, width=10,bd=1, bg="#ADA5A5", font=("Arial Bold", 36), fg="White")
Labe.grid(row=0, column=0)
def work():
global number1
Labe['text'] = "Loading" + " " + number1 + "%"
global number
number = number + 1
number1 = str(number)
if number == 102:
root.after(1000)
print("end")
Labe.grid_remove()
num = 101
for x in range(num):
print("He")
root.after(50, work)
root.after(50)
#Adding stuff to root
rootHelloLabel = tk.Label(root, text="Ordering system, please sign in", bg="#7271AD",
fg="White", font=("Arial Bold", 40), width=48, height=2)
placeholder = tk.Label(root, text="", pady=20, bg="#ACAD71")
placeholder.grid(row=0, column=0)
rootHelloLabel.grid(row=1, column=0)
placeholder1 = tk.Label(root, text="", pady=70, bg="#ACAD71")
placeholder1.grid(row=2, column=0)
loginButton = tk.Button(root, pady=30, padx=80, text="Login", font=("Arial Bold", 27),
bg="#71AD90", fg="#D0CFD1", command=login)
loginButton.grid(row=3, column=0)
adminButton = tk.Button(root, pady = 10, padx = 40, text="Admin",
font=("Arial Bold", 27), bg="#71AD90", fg="#D0CFD1")
adminButton.grid(row=5, column=0)
我做了三个小改动，我认为它有效。
number = 0
number1 = str(number)
Labe = tk.Label(root, width=10,bd=1, bg="#ADA5A5", font=("Arial Bold", 36), fg="White")
Labe.grid(row=0, column=0)
def work():
global number1
Labe['text'] = "Loading" + " " + number1 + "%"
global number
number = number + 1
number1 = str(number)
if number == 102:
root.after(1000)
print("end")
Labe.grid_remove()
num = 101
for x in range(num):
print("He")
root.after(50, work)
root.after(50)
root.update() # added this so the window displays for one frame
# it's similar to mainloop() but only runs once
#Adding stuff to root
rootHelloLabel = tk.Label(root, text="Ordering system, please sign in", bg="#7271AD",
fg="White", font=("Arial Bold", 40), width=48, height=2)
placeholder = tk.Label(root, text="", pady=20, bg="#ACAD71")
placeholder.grid(row=0, column=0)
rootHelloLabel.grid(row=1, column=0)
placeholder1 = tk.Label(root, text="", pady=70, bg="#ACAD71")
placeholder1.grid(row=2, column=0)
loginButton = tk.Button(root, pady=30, padx=80, text="Login", font=("Arial Bold", 27),
bg="#71AD90", fg="#D0CFD1", command=lambda: login())
loginButton.grid(row=3, column=0) # added lambda function so this only runs when clicked
adminButton = tk.Button(root, pady = 10, padx = 40, text="Admin",
font=("Arial Bold", 27), bg="#71AD90", fg="#D0CFD1")
adminButton.grid(row=5, column=0)
root.mainloop() # added this so that the program runs when done (can remove aswell)
我希望这对你有用。我不太明白这个问题，但是一旦我添加了 update() 函数，它就可以正常工作，我意识到你所说的加载栏是什么意思。希望对您有所帮助！
你在 for 循环中 运行s 100 次 after(50) 所以它可以停止程序 5 秒。而且我不知道你是否真的需要它。
我会 运行 after(50, work) 直接在 work
def work():
global number # PEP8: all `global` at the beginning of function
labe['text'] = f"Loading {number}%"
number += 1
if number > 100:
root.update() # to update last text in label because `after(1000)` will block it
root.after(1000)
print("end")
labe.grid_remove()
else:
root.after(50, work)
完整的工作代码（有其他更改）：
PEP 8 -- Style Guide for Python Code
import tkinter as tk
# --- functions --- # PEP8: all functions directly after imports (and after all classes)
def work():
global number # PEP8: all `global` at the beginning of function
labe['text'] = f"Loading {number}%"
number += 1
if number > 100:
root.update() # to update last text in label because `after(1000)` will block it
root.after(1000)
print("end")
labe.grid_remove()
else:
# run again
root.after(50, work)
def login():
pass
# --- main ---
number = 0
root = tk.Tk()
labe = tk.Label(root, width=10, bd=1, bg="#ADA5A5", font=("Arial Bold", 36), fg="White")
labe.grid(row=0, column=0)
root_hello_label = tk.Label(root, text="Ordering system, please sign in", bg="#7271AD",
fg="White", font=("Arial Bold", 40), width=48, height=2)
root_hello_label.grid(row=1, column=0)
placeholder1 = tk.Label(root, text="", pady=20, bg="#ACAD71")
placeholder1.grid(row=0, column=0)
placeholder2 = tk.Label(root, text="", pady=70, bg="#ACAD71")
placeholder2.grid(row=2, column=0)
login_button = tk.Button(root, pady=30, padx=80, text="Login", font=("Arial Bold", 27),
bg="#71AD90", fg="#D0CFD1", command=login)
login_button.grid(row=3, column=0)
admin_button = tk.Button(root, pady = 10, padx = 40, text="Admin",
font=("Arial Bold", 27), bg="#71AD90", fg="#D0CFD1")
admin_button.grid(row=5, column=0)
# run first time
root.after(50, work)
root.mainloop()
编辑：
如果您需要 运行 和 for 循环，那么您应该在 after(..., work)
for i in range(1, 102):
root.after(50*i, work)
完整的工作代码：
import tkinter as tk
# --- functions --- # PEP8: all functions directly after imports (and after all classes)
def work():
global number # PEP8: all `global` at the beginning of function
labe['text'] = f"Loading {number}%"
number += 1
if number > 100:
root.update() # to update last text in label because `after(1000)` will block it
root.after(1000)
print("end")
labe.grid_remove()
def login():
pass
# --- main ---
number = 0
root = tk.Tk()
labe = tk.Label(root, width=10, bd=1, bg="#ADA5A5", font=("Arial Bold", 36), fg="White")
labe.grid(row=0, column=0)
root_hello_label = tk.Label(root, text="Ordering system, please sign in", bg="#7271AD",
fg="White", font=("Arial Bold", 40), width=48, height=2)
root_hello_label.grid(row=1, column=0)
placeholder1 = tk.Label(root, text="", pady=20, bg="#ACAD71")
placeholder1.grid(row=0, column=0)
placeholder2 = tk.Label(root, text="", pady=70, bg="#ACAD71")
placeholder2.grid(row=2, column=0)
login_button = tk.Button(root, pady=30, padx=80, text="Login", font=("Arial Bold", 27),
bg="#71AD90", fg="#D0CFD1", command=login)
login_button.grid(row=3, column=0)
admin_button = tk.Button(root, pady = 10, padx = 40, text="Admin",
font=("Arial Bold", 27), bg="#71AD90", fg="#D0CFD1")
admin_button.grid(row=5, column=0)
# 102-1 = 101 loops, because it needs to display values 0...100 which gives 101 values
for i in range(1, 102):
root.after(50*i, work)
root.mainloop()
编辑：
因为 after(1000) 冻结所有 GUI 1 秒，所以最好使用 after(1000, remove_label) 在 1 秒后执行其他功能，这不会冻结 GUI。
def work():
global number
labe['text'] = f"Loading {number}%"
number += 1
if number > 100:
root.after(1000, remove_label)
else:
root.after(50, work)
def remove_label():
print("end")
labe.grid_remove()