英文:
Weird interaction of cget() in tkinter.ttk
问题
在使用tkinter玩耍时，我偶然遇到了这个问题。显然，ttk.Entry在打印状态之前似乎无法识别已配置的状态。我的代码如下：
import tkinter as tkfrom tkinter import ttkdef on_focus_in(entry):if entry['state'] == 'disabled': #请注意这个代码块entry.configure(state='normal')entry.delete(0, 'end')print('1')print(entry_x.cget("state"))if entry.cget('state') == 'disabled':entry.configure(state='normal')entry.delete(0, 'end')print('2')def on_focus_out(entry, placeholder):if entry.get() == "":entry.insert(0, placeholder)entry.configure(state='disabled')root = tk.Tk()entry_x = ttk.Entry(root, font=('Arial', 13), width=50)entry_x.pack(pady=10, ipady=7)entry_x.insert(0, "占位符X")entry_x.configure(state='disabled')entry_y = tk.Entry(root, width=50, font=('Arial', 13))entry_y.pack(pady=10, ipady=7)entry_y.insert(0, "占位符Y")entry_y.configure(state='disabled')x_focus_in = entry_x.bind('<Button-1>', lambda x: on_focus_in(entry_x))x_focus_out = entry_x.bind('<FocusOut>', lambda x: on_focus_out(entry_x, '占位符X'))y_focus_in = entry_y.bind('<Button-1>', lambda x: on_focus_in(entry_y))y_focus_out = entry_y.bind('<FocusOut>', lambda x: on_focus_out(entry_y, '占位符Y'))root.mainloop()
有人可以解释为什么会出现这种情况吗？
显然，当我点击X输入框时，日志返回"2"，这意味着应该打印"1"的代码块未运行。Y输入框正常工作。
英文:
Been playing out with tkinter until I stumble upon this. Apparently, it seems that the ttk.Entry does not recognize the configured state until the state is printed.
My code is as follow:
import tkinter as tkfrom tkinter import ttkdef on_focus_in(entry):if entry['state'] == 'disabled': #pay attention to this code blockentry.configure(state='normal')entry.delete(0, 'end')print('1')print(entry_x.cget("state"))if entry.cget('state') == 'disabled':entry.configure(state='normal')entry.delete(0, 'end')print('2')def on_focus_out(entry, placeholder):if entry.get() == "":entry.insert(0, placeholder)entry.configure(state='disabled')root = tk.Tk()entry_x = ttk.Entry(root, font=('Arial', 13), width=50)entry_x.pack(pady=10, ipady=7)entry_x.insert(0, "Place Holder X")entry_x.configure(state='disabled')entry_y = tk.Entry(root, width=50, font=('Arial', 13))entry_y.pack(pady=10, ipady = 7)entry_y.insert(0, "Place Holder Y")entry_y.configure(state='disabled')x_focus_in = entry_x.bind('<Button-1>', lambda x: on_focus_in(entry_x))x_focus_out = entry_x.bind('<FocusOut>', lambda x: on_focus_out(entry_x, 'Place Holder X'))y_focus_in = entry_y.bind('<Button-1>', lambda x: on_focus_in(entry_y))y_focus_out = entry_y.bind('<FocusOut>', lambda x: on_focus_out(entry_y, 'Place Holder Y'))root.mainloop()
Can anyone explain to me why is this the case?
Apparently, when I click the X entry, the log returns "2", which means that the code block that should print "1" does not run. The Y entry works fine.
答案1
得分: 0
问题的根本在于 entry['state'] 不返回一个字符串，而是返回一个类型为 <class '_tkinter.Tcl_Obj'> 的对象。您正在将其与一个字符串进行比较，因此条件将始终为假。您可以在您的 if 语句中将其强制转换为字符串，或使用 string 属性：
if str(entry['state']) == 'disabled':
或
if entry['state'].string == 'disabled':
英文:
The root of the problem is that entry['state'] does not return a string. Instead, it returns an object of type <class '_tkinter.Tcl_Obj'>. You're comparing it to a string, so the condition will always be false. You can cast it to the string in your if statement or use the string attribute:
if str(entry['state']) == 'disabled':
or
if entry['state'].string == 'disabled':
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