python-3.x - 不同的焦点顺序和堆叠顺序 tkinter

Question

在下面的示例中，我创建了多个文本条目，其中一些文本条目在您聚焦时可以展开/折叠。

但是，当某些扩展(具体来说，名为 entry_2_problem_1 和 entry_5_problem_2 的扩展，我还在“此处”插入了文本)时，它们会在另一个文本“下方”结束条目 - 我的意思是，由于堆叠顺序，它们位于另一个条目下方。

在创建向上堆叠条目(即 entry_3 和entry_6，但这些会改变我的焦点顺序。我想要一个“自然”的焦点顺序，从左到右，从上到下。

下面，您可以看到带有一些注释行的代码:如果您取消注释这些行，您会发现堆叠问题被焦点顺序问题所取代(因为它并不是真正从左到右，正如您在使用时会注意到的那样)选项卡)。

此外，考虑到在我正在处理的实际代码中，由于多种原因，在小部件之间留下更多空白的任何内容都会被丢弃

MRE:

from tkinter import Tk, Text

def focus_next_widget(event):
    event.widget.tk_focusNext().focus()
    return("break")

class iText(Text):
    def __init__(self, stdwidth_mult=2.5, stdheight_mult=3, **kwargs):
        super(iText, self).__init__(**kwargs)

        self.stdwidth = kwargs.get('width')
        self.stdheight = kwargs.get('height')
        self.stdwidth_mult = stdwidth_mult
        self.stdheight_mult = stdheight_mult

def text_resizer(event):
    if event.widget == event.widget.focus_get():
        if not event.widget.stdheight == None:event.widget.configure(height=int(event.widget.stdheight*event.widget.stdheight_mult))
        if not event.widget.stdwidth == None: event.widget.configure(width=int(event.widget.stdwidth*event.widget.stdwidth_mult))
    else:
        if not event.widget.stdheight == None:event.widget.configure(height=event.widget.stdheight)
        if not event.widget.stdwidth == None: event.widget.configure(width=event.widget.stdwidth)

window = Tk()
window.geometry("300x300")

entry1 = iText(width=4, bd=2, font='futura', relief='flat', highlightcolor='#3A3A3A', highlightbackground='#3A3A3A', highlightthickness=2, bg="#D9D9D9", fg="#000716")
entry1.place(x=6.0, y=68.0, height=43.0)
entry1.bind("<Tab>", focus_next_widget)
entry1.bind('<FocusIn>', text_resizer)
entry1.bind('<FocusOut>', text_resizer)

# First problematic entry
entry_2_problem_1 = iText(width=4, bd=2, font='futura', relief='flat', highlightcolor='#3A3A3A', highlightbackground='#3A3A3A', highlightthickness=2, bg="#D9D9D9", fg="#000716")
entry_2_problem_1.place(x=6.0, y=116.0, height=43.0)
entry_2_problem_1.insert(1.0, 'Here')
entry_2_problem_1.bind("<Tab>", focus_next_widget)
entry_2_problem_1.bind('<FocusIn>', text_resizer)
entry_2_problem_1.bind('<FocusOut>', text_resizer)

entry_3 = iText(stdheight_mult=1, height=1, bd=2, font='futura', relief='flat', highlightcolor='#3A3A3A', highlightbackground='#3A3A3A', highlightthickness=2, bg="#D9D9D9", fg="#000716")
entry_3.place(x=70.0, y=121.0, width=102.0)
entry_3.bind("<Tab>", focus_next_widget)
entry_3.bind('<FocusIn>', text_resizer)
entry_3.bind('<FocusOut>', text_resizer)
# The following line solves the stacking problem, but creates a focus order one
# entry_2_problem_1.lift()

entry_4 = iText(width=4, bd=2, font='futura', relief='flat', highlightcolor='#3A3A3A', highlightbackground='#3A3A3A', highlightthickness=2, bg="#D9D9D9", fg="#000716")
entry_4.place(x=6.0, y=165.0, height=43.0)
entry_4.bind("<Tab>", focus_next_widget)
entry_4.bind('<FocusIn>', text_resizer)
entry_4.bind('<FocusOut>', text_resizer)

# Second problematic entry
entry_5_problem_2 = iText(width=4, bd=2, font='futura', relief='flat', highlightcolor='#3A3A3A', highlightbackground='#3A3A3A', highlightthickness=2, bg="#D9D9D9", fg="#000716")
entry_5_problem_2.place(x=6.0, y=213.0, height=43.0)
entry_5_problem_2.insert(1.0, 'Here')
entry_5_problem_2.bind("<Tab>", focus_next_widget)
entry_5_problem_2.bind('<FocusIn>', text_resizer)
entry_5_problem_2.bind('<FocusOut>', text_resizer)

entry_6 = iText(stdheight_mult=1, height=1, bd=2, font='futura', relief='flat', highlightcolor='#3A3A3A', highlightbackground='#3A3A3A', highlightthickness=2, bg="#D9D9D9", fg="#000716")
entry_6.place(x=70.0, y=218.0, width=102.0, height=34.0)
entry_6.bind("<Tab>", focus_next_widget)
entry_6.bind('<FocusIn>', text_resizer)
entry_6.bind('<FocusOut>', text_resizer)
# The following line solves the stacking problem, but creates a focus order one
# entry_8_problem_2.lift()

window.mainloop()

此外，还有一些当前和所需输出的照片，涉及堆叠问题。

Current stacking-situation (with GOOD focus order)

Desired stacking-situation (but has a BAD focus order behavior)

Answer 1

我无法发表评论(由于代表阈值)，因此我无法提出澄清问题，但我相信您想要的行为是按 Tab 键前进 iText 条目的每一行按照 tkinter 默认行为，按照您声明的顺序排列框，直到您 lift() 一个对象。

最简单的答案

最简单的方法是在小部件获得焦点时而不是在声明它们时使用.lift()，以便您获得焦点的小部件始终高于其他小部件当你使用它时:

def text_resizer(event):
    if event.widget == event.widget.focus_get():
        event.widget.lift() # Always lift the current widget when it gains focus
        if not event.widget.stdheight == None:event.widget.configure(height=int(event.widget.stdheight*event.widget.stdheight_mult))
        if not event.widget.stdwidth == None: event.widget.configure(width=int(event.widget.stdwidth*event.widget.stdwidth_mult))
    else:
        if not event.widget.stdheight == None:event.widget.configure(height=event.widget.stdheight)
        if not event.widget.stdwidth == None: event.widget.configure(width=event.widget.stdwidth)

在此应用程序中，默认输入框在未展开时没有任何重叠，这是有效的。这还假设您只想使用事件系统循环浏览元素，而不是按特定顺序单击它们。

只要不是这种情况，请尝试以下操作:

更加动态的方法

但是，如果您的布局更紧凑，并且有许多重叠的输入框(实际上是任何小部件)，那么当它们的相对层发生变化时，可能会导致它们在视觉上变得困惑。为了避免这种情况，您可以将对输入框对象的引用放入可迭代对象中，并编写一个函数，按照可迭代对象中的顺序对它们进行排序。在此示例中，我搭载了您的 focusOut 回调:

def text_resizer(event):
    if event.widget == event.widget.focus_get():
        event.widget.lift() # Always lift the current widget when it gains focus
        if not event.widget.stdheight == None:event.widget.configure(height=int(event.widget.stdheight*event.widget.stdheight_mult))
        if not event.widget.stdwidth == None: event.widget.configure(width=int(event.widget.stdwidth*event.widget.stdwidth_mult))
    else:
        if not event.widget.stdheight == None:event.widget.configure(height=event.widget.stdheight)
        if not event.widget.stdwidth == None: event.widget.configure(width=event.widget.stdwidth)
        for widget in window.widget_order:
            window.widget_order[widget].lift() # Sort the widgets back into their default order to restore the appearance

将小部件添加到可迭代的位置，如下所示:

window.widget_order[len(window.widget_order)] = entry1 # Add the widget to the widget order list

这将出现与您最初遇到的相同问题，因此跟踪当前聚焦的小部件并更改将焦点从 tkinter 的 focusNext() 切换的方法将很有趣，这可能是您首先选择的解决方案:

window = Tk()
window.geometry("300x300")
window.widget_order = {}    # Store the desired visual ordering of widgets
window.currentFocus = 0     # Store the current focus number to use as a key for the widget dict

def focus_next_widget(event):
    window.currentFocus = (window.currentFocus + 1) % len(window.widget_order) # Calculate the next focus value, wrapping based on dict length
    window.widget_order[window.currentFocus].focus() # Focus the next widget
    return("break")

完整的代码将如下所示:

from tkinter import Tk, Text

def focus_next_widget(event):
    window.currentFocus = (window.currentFocus + 1) % len(window.widget_order) # Calculate the next focus value, wrapping based on dict length
    window.widget_order[window.currentFocus].focus() # Focus the next widget
    return("break")

class iText(Text):
    def __init__(self, stdwidth_mult=2.5, stdheight_mult=3, **kwargs):
        super(iText, self).__init__(**kwargs)

        self.stdwidth = kwargs.get('width')
        self.stdheight = kwargs.get('height')
        self.stdwidth_mult = stdwidth_mult
        self.stdheight_mult = stdheight_mult

def text_resizer(event):
    if event.widget == event.widget.focus_get():
        event.widget.lift() # Always lift the current widget when it gains focus
        if not event.widget.stdheight == None:event.widget.configure(height=int(event.widget.stdheight*event.widget.stdheight_mult))
        if not event.widget.stdwidth == None: event.widget.configure(width=int(event.widget.stdwidth*event.widget.stdwidth_mult))
    else:
        if not event.widget.stdheight == None:event.widget.configure(height=event.widget.stdheight)
        if not event.widget.stdwidth == None: event.widget.configure(width=event.widget.stdwidth)
        for widget in window.widget_order:
            window.widget_order[widget].lift() # Sort the widgets back into their default order to restore the appearance

window = Tk()
window.geometry("300x300")
window.widget_order = {}    # Store the desired visual ordering of widgets
window.currentFocus = 0     # Store the current focus number to use as a key for the widget dict

entry1 = iText(width=4, bd=2, font='futura', relief='flat', highlightcolor='#3A3A3A', highlightbackground='#3A3A3A', highlightthickness=2, bg="#D9D9D9", fg="#000716")
entry1.insert(1.0, "e1")
entry1.place(x=6.0, y=68.0, height=43.0)
entry1.bind("<Tab>", focus_next_widget)
entry1.bind('<FocusIn>', text_resizer)
entry1.bind('<FocusOut>', text_resizer)
window.widget_order[len(window.widget_order)] = entry1 # Add the widget to the widget order list

# First problematic entry
entry_2_problem_1 = iText(width=4, bd=2, font='futura', relief='flat', highlightcolor='#3A3A3A', highlightbackground='#3A3A3A', highlightthickness=2, bg="#D9D9D9", fg="#000716")
entry_2_problem_1.place(x=6.0, y=116.0, height=43.0)
entry_2_problem_1.insert(1.0, "e2")
entry_2_problem_1.bind("<Tab>", focus_next_widget)
entry_2_problem_1.bind('<FocusIn>', text_resizer)
entry_2_problem_1.bind('<FocusOut>', text_resizer)
window.widget_order[len(window.widget_order)] = entry_2_problem_1 # Add the widget to the widget order list

entry_3 = iText(stdheight_mult=1, height=1, bd=2, font='futura', relief='flat', highlightcolor='#3A3A3A', highlightbackground='#3A3A3A', highlightthickness=2, bg="#D9D9D9", fg="#000716")
entry_3.place(x=70.0, y=121.0, width=102.0)
entry_3.insert(1.0, "e3")
entry_3.bind("<Tab>", focus_next_widget)
entry_3.bind('<FocusIn>', text_resizer)
entry_3.bind('<FocusOut>', text_resizer)
window.widget_order[len(window.widget_order)] = entry_3 # Add the widget to the widget order list

entry_4 = iText(width=4, bd=2, font='futura', relief='flat', highlightcolor='#3A3A3A', highlightbackground='#3A3A3A', highlightthickness=2, bg="#D9D9D9", fg="#000716")
entry_4.place(x=6.0, y=165.0, height=43.0)
entry_4.insert(1.0, "e4")
entry_4.bind("<Tab>", focus_next_widget)
entry_4.bind('<FocusIn>', text_resizer)
entry_4.bind('<FocusOut>', text_resizer)
window.widget_order[len(window.widget_order)] = entry_4 # Add the widget to the widget order list

# Second problematic entry
entry_5_problem_2 = iText(width=4, bd=2, font='futura', relief='flat', highlightcolor='#3A3A3A', highlightbackground='#3A3A3A', highlightthickness=2, bg="#D9D9D9", fg="#000716")
entry_5_problem_2.place(x=6.0, y=213.0, height=43.0)
entry_5_problem_2.insert(1.0, 'e5')
entry_5_problem_2.bind("<Tab>", focus_next_widget)
entry_5_problem_2.bind('<FocusIn>', text_resizer)
entry_5_problem_2.bind('<FocusOut>', text_resizer)
window.widget_order[len(window.widget_order)] = entry_5_problem_2 # Add the widget to the widget order list

entry_6 = iText(stdheight_mult=1, height=1, bd=2, font='futura', relief='flat', highlightcolor='#3A3A3A', highlightbackground='#3A3A3A', highlightthickness=2, bg="#D9D9D9", fg="#000716")
entry_6.place(x=70.0, y=218.0, width=102.0, height=34.0)
entry_6.insert(1.0, 'e6')
entry_6.bind("<Tab>", focus_next_widget)
entry_6.bind('<FocusIn>', text_resizer)
entry_6.bind('<FocusOut>', text_resizer)
window.widget_order[len(window.widget_order)] = entry_6 # Add the widget to the widget order list

window.mainloop()

这还允许您划分所有小部件订单代码，以便更轻松地更改订单；只需将它们以新的顺序插入到字典中即可。对我来说这是有道理的，尽管其他人可能有所不同。

我使用了字典，但我相信列表也同样有效。