时间戳

首页 > 解决方案 > 滚动 tkinter 树视图时如何移动弹出窗口？

python - 滚动 tkinter 树视图时如何移动弹出窗口？

问题描述

我有一个带有垂直滚动条的 tkinter 树视图。为了使它（看起来）可编辑，当用户双击树视图的单元格时，我创建了一个弹出条目。但是，滚动树视图时，我无法使弹出窗口移动。

import tkinter  as tk
from tkinter import ttk

class EntryPopup(ttk.Entry):
    def __init__(self, parent, itemId, col, **kw):
        super().__init__(parent, **kw)
        self.tv = parent
        self.iId = itemId
        self.column = col
        self['exportselection'] = False

        self.focus_force()
        self.bind("<Return>", self.onReturn)

    def saveEdit(self):
        self.tv.set(self.iId, column=self.column, value=self.get())
        print("EntryPopup::saveEdit---{}".format(self.iId))

    def onReturn(self, event):
        self.tv.focus_set()        
        self.saveEdit()
        self.destroy()


class EditableDataTable(tk.Frame):
    def __init__(self, parent):
        tk.Frame.__init__(self, parent)

        self.parent = parent
        self.tree = None
        self.entryPopup = None
        columns = ("Col1", "Col2")
        # Create a treeview with vertical scrollbar.
        self.tree = ttk.Treeview(self, columns=columns, show="headings")
        self.tree.grid(column=0, row=0, sticky='news')
        self.tree.heading("#1", text="col1")
        self.tree.heading("#2", text="col2")

        self.vsb = ttk.Scrollbar(self, orient="vertical", command=self.tree.yview)
        self.tree.configure(yscrollcommand=self.vsb.set)
        self.vsb.grid(column=1, row=0, sticky='ns')

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

        self.entryPopup = None
        self.curSelectedRowId = ""

        col1 = []
        col2 = []
        for r in range(50):
            col1.append("data 1-{}".format(r))
            col2.append("data 2-{}".format(r))

        for i in range(min(len(col1),len(col2))):
            self.tree.insert('', i, values=(col1[i], col2[i]))

        self.tree.bind('<Double-1>', self.onDoubleClick)

    def createPopup(self, row, column):
        x,y,width,height = self.tree.bbox(row, column)

        # y-axis offset
        pady = height // 2
        self.entryPopup = EntryPopup(self.tree, row, column)
        self.entryPopup.place(x=x, y=y+pady, anchor='w', width=width)


    def onDoubleClick(self, event):
        rowid = self.tree.identify_row(event.y)
        column = self.tree.identify_column(event.x)
        self.createPopup(rowid, column)

root = tk.Tk()

for row in range(2):
    root.grid_rowconfigure(row, weight=1)
root.grid_columnconfigure(0, weight=1)

label = tk.Label(root, text="Double-click to edit and press 'Enter'")
label.grid(row=0, column=0, sticky='news', padx=10, pady=5)

dataTable = EditableDataTable(root)
dataTable.grid(row=1, column=0, sticky="news", pady=10, padx=10)

root.geometry("450x300")
root.mainloop()

要重现该问题，请双击树视图。当编辑框打开时，移动鼠标指针悬停在树视图上。现在使用鼠标滚轮滚动。树视图移动，但弹出编辑框不移动。

标签： pythontkinterscrolltreeviewpopupwindow

解决方案

我之前做过类似的事情，将一个函数绑定到mousewheel并重新计算悬停小部件的所有新 x 和 y 位置。

class EditableDataTable(tk.Frame):
    def __init__(self, parent):
        tk.Frame.__init__(self, parent)

        self.parent = parent
        self.tree = None
        self.entryPopup = None
        self.list_of_entries = []

        ...

        self.tree.bind("<MouseWheel>", self._on_mousewheel)

    def _on_mousewheel(self, event):
        if self.list_of_entries:
            def _move():
                for i in self.list_of_entries:
                    try:
                        iid = i.iId
                        x, y, width, height = self.tree.bbox(iid, column="Col2") #hardcoded to col2
                        i.place(x=x, y=y+height//2, anchor='w', width=width)
                    except ValueError:
                        i.place_forget()
                    except tk.TclError:
                        pass
            self.master.after(5, _move)

    def createPopup(self, row, column):
        x,y,width,height = self.tree.bbox(row, column)
        # y-axis offset
        pady = height // 2
        self.entryPopup = EntryPopup(self.tree, row, column)
        self.list_of_entries.append(self.entryPopup)
        self.entryPopup.place(x=x, y=y+pady, anchor='w', width=width)

请注意，这仅适用于第二列，但其余部分应该很容易实现。

推荐阅读