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biomotor
Asked:2022-06-16 21:27:25 +0000 UTC2022-06-16 21:27:25 +0000 UTC

如何在树视图中对列表进行排序？

  • 772

我有一个程序代码，它在树视图中显示来自 xml 文件的数据。

from tkinter import *
from tkinter.ttk import Treeview
import xml.etree.ElementTree as ET

class App:

  def __init__(self, root):

    try:
      with open('FileInfo.xml') as f:
        read_data = f.read()
        ansi = read_data.encode('ansi')
      rootNode = ET.fromstring(ansi)

      self.tree = Treeview(root)
      self.tree.pack(expand=True, fill='both')
      self.tree.heading('#0', text=rootNode.get('displayName'), anchor='w')
      self.walk_dict(rootNode)
    except Exception as e:
      print(e)

  def walk_dict(self, d, depth=0, parent=""):
    for child in d:
      if(child.tag == 'folder'):
          item = self.tree.insert(parent, 'end', None, text=child.get('displayName'))

      elif(child.tag == 'element'):
          item = self.tree.insert(parent, 'end', None, text=child.get('file'))

      if child.__len__() > 0:
          self.walk_dict(child, depth + 1, parent=item)

root = Tk()
root.geometry("500x600")
App(root)
root.mainloop()

包含内容的 XML 文件：

<?xml version="1.0"?>
<library displayName="MyLibrary">
    <description />
    <folder displayName="9_">
        <folder displayName="9_">
            <element displayName="9_" file="9_.kdw" />
            <description>Description.</description>
            <element displayName="1_" file="1_.kdw" />
            <element displayName="я_" file="я_.kdw" />
            <element displayName="а_" file="а_.kdw" />
            <element displayName="z_" file="z_.kdw" />
            <element displayName="a_" file="a_.kdw" />
        </folder>
        <folder displayName="1_">
            <element displayName="9_" file="9_.kdw" />
        </folder>
        <folder displayName="я_">
            <element displayName="9_" file="9_.kdw" />
        </folder>
        <folder displayName="а_">
            <element displayName="9_" file="9_.kdw" />
        </folder>
        <folder displayName="z_">
            <element displayName="9_" file="9_.kdw" />
        </folder>
        <folder displayName="a_">
            <element displayName="9_" file="9_" />
        </folder>
    </folder>
    <folder displayName="1_">
    <element displayName="9_" file="9_.kdw" />
    </folder>
    <folder displayName="я_">
        <element displayName="9_" file="9_.kdw" />
    </folder>
    <folder displayName="а_">
        <element displayName="9_" file="9_" />
    </folder>
    <folder displayName="z_">
        <element displayName="9_" file="9_.kdw" />
    </folder>
    <folder displayName="a_">
        <element displayName="9_" file="9_.kdw" />
  </folder>
    <element displayName="9_" file="9_.kdw" />
    <element displayName="1_" file="1_.kdw" />
    <element displayName="я_" file="я_.kdw" />
    <element displayName="а_" file="а_.kdw" />
    <element displayName="z_" file="z_.kdw" />
    <element displayName="a_" file="a_.kdw" />
</library>

有必要对所有目录和文件进行排序。1-999...，az，a-z。

显然排序应该在这里的某个地方......

ansi = read_data.encode('ansi')
....
rootNode = ET.fromstring(ansi)

这是我目前得到的：

在此处输入图像描述

python

1 个回答

  1. Best Answer

    我们写了一个排序方法：

      @staticmethod
  def sort_by_attr(node, tags, attr):
    return sorted(filter(lambda x: x.tag in tags, node), key=lambda child: child.get(attr))

    如果您只想排序，那么只需像这样使用它：

      def walk_dict(self, d, depth=0, parent=""):
    for child in self.sort_by_attr(d, ('folder', 'element'), "displayName"):
       ...

    真正的文件将与文件夹混合。

    如果需要将文件夹移到开头，则如下所示：

      def walk_dict(self, d, depth=0, parent=""):
    for child in self.sort_by_attr(d, ('folder'), "displayName"):
      item = self.tree.insert(parent, 'end', None, text=child.get('displayName'))
      if len(child) > 0:
        self.walk_dict(child, depth + 1, parent=item)
    for child in self.sort_by_attr(d, ('element'), "file"):
      item = self.tree.insert(parent, 'end', None, text=child.get('file'))
    • 1

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