我正在尝试使用identify_element()方法实现treeview中的拖放,以指示我需要在哪里插入项。
当触发<ButtonRelease-1>事件时,我调用identify_element(),它返回单词“填充”,而不是treeview项。我知道它为什么不像它应该做的那样在x/y弦处返回项。
import os
import tkinter as tk
import tkinter.ttk as ttk
from tkinter import messagebox
class App(tk.Tk):
def __init__(self, path):
super().__init__()
self.title("Ttk Treeview")
abspath = os.path.abspath(path)
self.nodes = {}
self.tree = ttk.Treeview(self)
self.tree.heading("#0", text=abspath, anchor=tk.W)
ysb = ttk.Scrollbar(self, orient=tk.VERTICAL, command=self.tree.yview)
xsb = ttk.Scrollbar(self, orient=tk.HORIZONTAL, command=self.tree.xview)
self.tree.configure(yscroll=ysb.set, xscroll=xsb.set)
self.tree.grid(row=0, column=0, sticky=tk.N + tk.S + tk.E + tk.W)
ysb.grid(row=0, column=1, sticky=tk.N + tk.S)
xsb.grid(row=1, column=0, sticky=tk.E + tk.W)
self.rowconfigure(0, weight=1)
self.columnconfigure(0, weight=1)
self.tree.bind("<<TreeviewOpen>>", self.open_node)
def button_release(e):
print('Results =', self.tree.identify_element(e.x, e.y))
self.tree.bind("<ButtonRelease-1>", button_release)
self.populate_node("", abspath)
def populate_node(self, parent, abspath):
try:
for entry in os.listdir(abspath):
entry_path = os.path.join(abspath, entry)
node = self.tree.insert(parent, tk.END, text=entry, open=False)
if os.path.isdir(entry_path):
self.nodes[node] = entry_path
self.tree.insert(node, tk.END)
except:
messagebox.showerror('Access Denied', f'Sorry you do not have access privileges for, {abspath}.')
def open_node(self, _):
item = self.tree.focus()
abspath = self.nodes.pop(item, False)
if abspath:
children = self.tree.get_children(item)
self.tree.delete(children)
self.populate_node(item, abspath)
if __name__ == "__main__":
app = App(path="/")
app.mainloop()我希望identify_element()的输出应该返回一个treeview项。
发布于 2019-09-28 23:06:25
identify_element被设计成返回单击了小部件的哪个部分(即: treeview布局中定义的元素之一),而不是您单击的项目。
在tcl/tk中,您将使用.tree identify item而不是.tree identify element。奇怪的是,tkinter没有提供类似的identify_item方法。但是,它确实公开了泛型identify方法,该方法可以作为它的第一个参数传递给"item",以标识光标下的树项id。
print('Results =', self.tree.identify("item", e.x, e.y))https://stackoverflow.com/questions/58150792
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