如何在文本小部件中从树视图打开文件 python tkinter

Question

我正在创建一个代码编辑器，想要一个侧栏，其中包含用户打开的文件夹文件以及如何通过从文本小部件中的树视图中选择文件来打开文件

我已经为文件创建了侧栏，但仍然遇到了一些如何打开文件的问题从文本小部件中的树视图

这是我的代码

from tkinter import*
from tkinter import ttk
import os
from tkinter.filedialog import askdirectory
def Open():
    for i in tree.get_children():
        tree.delete(i)
    path = askdirectory()
    abspath = os.path.abspath(path)
    root_node = tree.insert('', 'end', text=abspath, open=True)
    process_directory(root_node,abspath)
def process_directory( parent, path):

        for p in os.listdir(path):
            abspath = os.path.join(path, p)
            isdir = os.path.isdir(abspath)
            oid = tree.insert(parent, 'end', text=p, open=False)
            if isdir:
                process_directory(oid, abspath)
def Open_file_from_list_box(value):
    global file
    file = tree.selection()
    filepath = os.path.join(value,file)
    root.title(filepath + "                                                                                                                                            Code Editor")
    editor.delete(1.0,END)
    with open(filepath,"r") as f:
        editor.insert(1.0,f.read())
root = Tk()
root.geometry("1550x850+0+0")
Button(root,text="Open",command=Open).pack()
frame = Frame(root)
tree = ttk.Treeview(frame)
tree.pack(expand=True,fill=Y)
path = "."
tree.heading('#0', text=path, anchor='w')
abspath = os.path.abspath(path)
root_node = tree.insert('', 'end', text=abspath, open=True)
process_directory(root_node, abspath)
frame.pack(side=LEFT,fill=Y)
frame = Frame(root)
frame.pack(side=LEFT,expand=True,fill=BOTH)
editor = Text(frame,font="Consolas 15")
editor.pack(expand=True,fill=BOTH)

tree.bind("<<TreeviewSelect>>",lambda event=None:Open_file_from_list_box(path))

root.mainloop()

Answer 1

您只需在 Open_file_from_list_box() 中获取所选项目 iid：

def Open_file_from_list_box(value):
    global file
    file = tree.selection() # return item iid, not the file name
    ...

它应该是：

def Open_file_from_list_box(value):
    global file
    item_id = tree.selection()
    file = tree.item(item_id, 'text') # get the filename from 'text' option
    ...