Tkinter 将 Enter 绑定到焦点下一个条目

Question

我正在尝试创建一个函数，将 enter 绑定到“插入”框架中的下一个输入字段。除了在最后一个输入字段的情况下，它将返回所有输入的文本字段。我尝试使用迭代器来获取下一个输入字段无济于事。我是 Python 和 Tkinter 的新手。我的主要问题围绕着“getinsert”函数，我怎样才能专注于下一个输入字段？有某种方式来获取当前焦点并为其添加 1？另外，我如何检查它是否在最后一个字段上，以便我可以调用“getinsert”函数？

任何帮助将不胜感激

import tkinter as tk
from tkinter import messagebox

master = tk.Tk()

insertnames = ["Title", "Location", "Tags", "Section", "Blah"]
insertentries = {}
insertlabels = {}
main = tk.Frame(master)
insert = tk.Frame(master)
currentnames = iter(insertnames[0])

def mainmenu():
    main.pack()
    insertbutton.pack(fill = 'x')

def insertcall():
    main.forget()
    insert.pack()
    insertform()
    insertentries["Title"].bind('<Return>', insertfocus)
    insertentries["Location"].bind('<Return>', insertfocus)
    insertentries["Tags"].bind('<Return>', getinsert)

def getinsert(event):
    for entry in insertentries:
        print (insertentries[entry].get())
        insertentries[entry].delete(0, 'end')
    insert.forget()
    mainmenu()

def insertfocus(event):
    currentnames.next()
    insertentries[currentnames].focus()

def insertform():
    i = 0

    for name in insertnames:
        e = tk.Entry(insert)
        e.grid(row = i, column = 1)
        insertentries[name] = e

        lab = tk.Label(insert, text = name)
        lab.grid(row = i, column = 0)
        insertlabels[name] = lab

        i += 1

insertbutton = tk.Button(main, text = "Insert", command = insertcall)
mainmenu()

tk.mainloop()

Answer 1

您正在复制用于在一组小部件之间移动的 tab 键的功能。这使用虚拟事件<<NextWindow>> 和<<PrevWindow>>。所以要做的是让你的<Return> 处理程序生成这个事件。然后对于您要转到特定小部件的最后一个条目。我已经创建了一个这样做的例子。您可以将这样的东西与小部件集合一起使用，或者如果您深入了解 tk_focusNext 函数正在执行的操作以计算出下一个要转到的小部件，则可以安排检查包含框架中的小部件。

编辑：通过从python生成遍历事件来模拟tk::TabToWindow函数

import tkinter as tk

class MyEntry(tk.Entry):
    """Entry widget with Return bound to provide tabbing functionality"""
    def __init__(self, master, **kw):
        tk.Entry.__init__(self, master, **kw)
        self.next_widget = None
        self.bind("<Return>", self.on_ret)
    def on_ret(self, ev):
        if self.next_widget:
            self.event_generate('<<TraverseOut>>')
            self.next_widget.focus()
            self.next_widget.event_generate('<<TraverseIn>>')
        else:
            self.event_generate('<<NextWindow>>')
        return "break"
    def set_next(self, widget):
        """Override the default next widget for this instance"""
        self.next_widget = widget

def add_entry(parent, row, **kwargs):
    widget = MyEntry(parent, **kwargs)
    widget.grid(row=row, column=0, sticky=tk.NSEW)
    return widget

def main(args=None):
    root = tk.Tk()

    frame = tk.LabelFrame(root, text="Entries", width=200, height=200)

    entries = []
    for row in range(4):
        e = add_entry(frame, row)
        entries.append(e)
    e.set_next(entries[0])

    frame.grid_columnconfigure(0, weight=1)
    frame.grid_rowconfigure(0, weight=0)

    button = tk.Button(root, text="Exit", command=root.destroy)

    frame.grid(row=0, column=0, sticky=tk.NSEW)
    button.grid(row=1, column=0, sticky=tk.SE)
    root.grid_columnconfigure(0, weight=1)
    root.grid_rowconfigure(0, weight=1)

    entries[0].focus_set()
    root.mainloop()

if __name__ == '__main__':
    main()

Answer 2

我知道这不是最好的选择，但一种冗长的方法是硬编码。对字典有更多经验的人可能可以循环执行，但我不能:)

import tkinter as tk

insertnames = ["Title", "Location", "Tags", "Section", "Blah"]
obj = []

root = tk.Tk()

firstFrame = tk.Frame(root)
entryFrame = tk.Frame(root)

def mainmenu():
    firstFrame.pack()
    insertButton.pack(fill='x')

def insertcall():
    firstFrame.forget()
    entryFrame.pack()

    title = tk.Entry(entryFrame)
    title.bind('<Return>', lambda event: focus(title))
    title.grid(row=0, column=1)
    obj.append(title)
    tk.Label(entryFrame, text='Title').grid(row=0, column=0)

    location = tk.Entry(entryFrame)
    location.bind('<Return>', lambda event: focus(location))
    location.grid(row=1, column=1)
    obj.append(location)
    tk.Label(entryFrame, text='Location').grid(row=1, column=0)

    tags = tk.Entry(entryFrame)
    tags.bind('<Return>', lambda event: focus(tags))
    tags.grid(row=2, column=1)
    obj.append(tags)
    tk.Label(entryFrame, text='Tags').grid(row=2, column=0)

    section = tk.Entry(entryFrame)
    section.bind('<Return>', lambda event: focus(section))
    section.grid(row=3, column=1)
    obj.append(section)
    tk.Label(entryFrame, text='Section').grid(row=3, column=0)

    blah = tk.Entry(entryFrame)
    blah.bind('<Return>', lambda event: focus(blah))
    blah.grid(row=4, column=1)
    obj.append(blah)
    tk.Label(entryFrame, text='Blah').grid(row=4, column=0)

def focus(entryobj):
    i = 0
    for item in obj: #for each try and get next item unless it gives an error (then it must be the last entry so put it to 0)
        try:
            if item == entryobj:
                obj[i+1].focus_set()
                break
        except IndexError:
            obj[0].focus_set()
        i += 1

insertButton = tk.Button(firstFrame, text='Insert', command=insertcall)
insertButton.pack()

mainmenu()

root.mainloop()

Answer 3

您可以将相关条目小部件的 键绑定到如下函数：

def entry_next(self, event):         
    event.widget.tk_focusNext().focus()
    return("break")