Python 3.x 获取TreeView元素的名称

如何获取treeView元素的名称

treeView = addDevWin.TreeView
node2= treeView.GetItem([u'Parent', node1'])
node2.select()
level3nodes = node2.children()

# Output:
#[<pywinauto.controls.common_controls._treeview_element object at 0x068C8750>, <pywinauto.controls.common_controls._treeview_element object at 0x068C82F0>, <pywinauto.controls.common_controls._treeview_element object at 0x068D2750>, <pywinauto.controls.common_controls._treeview_element object at 0x068D2870>, <pywinauto.controls.common_controls._treeview_element object at 0x068D2770>]

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使用：Windows 10，Python3。

您需要方法。treeView对象的项\u文本。win32后端中的树视图项是一种虚拟项，没有通常的win32 API句柄。这就是您需要的原因。获取项目…

这是有效的解决方案：

treeElemList = [u'Parent', node1']
treeView = devWin.TreeView
node2= treeView.GetItem(treeElemList)
node2.select()
level3nodes = node2.children()
for element in level3nodes:
    if 'substring' in element.text():
        treeElemList .append(element.text())
        break

node3= treeView.GetItem(treeElemList)
node3.select()

treeElemList = [u'Parent', node1']
treeView = devWin.TreeView
node2= treeView.GetItem(treeElemList)
node2.select()
level3nodes = node2.children()
for element in level3nodes:
    if 'substring' in element.text():
        treeElemList .append(element.text())
        break

node3= treeView.GetItem(treeElemList)
node3.select()