pywinauto：迭代窗口中的所有控件

Question

我正在尝试编写一个通用测试脚本来查找新软件版本中的错误。我的想法是迭代窗口中的控件并与每个控件进行交互，记录引起的任何错误，并在软件崩溃时重新启动软件。

我正在寻找一种动态查找控件标识符的方法，有点像print_control_identifiers()但输出是一个列表或类似的结构，我可以对其进行迭代。

在关于控件标识符的GitHub 问题上提到了这一点：

可以通过使用.children()（仅直接子级）和.descendants() （整个子树作为普通列表）来遍历层次结构

我假设我可以迭代Application对象的descendants()列表并为每个对象调用相关的交互方法，但是我不知道如何获取此列表。我以为我可以做这样的事情，但我没有取得任何成功：

def test(application):
    for child in application.descendants():
        #interact with child control

software = Application(backend='uia').start(cmd_line=FILE_PATH)
test(software)

AttributeError：既没有找到 GUI 元素（包装器）也没有找到包装器方法“后代”（拼写错误？）

---

编辑

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我通过查看代码找到了print_control_identifiers方法：

class Application(object):

    def print_control_identifiers(self, depth=None, filename=None):
        """
        Prints the 'identifiers'
        Prints identifiers for the control and for its descendants to
        a depth of **depth** (the whole subtree if **None**).
        .. note:: The identifiers printed by this method have been made
               unique. So if you have 2 edit boxes, they won't both have "Edit"
               listed in their identifiers. In fact the first one can be
               referred to as "Edit", "Edit0", "Edit1" and the 2nd should be
               referred to as "Edit2".
        """
        if depth is None:
            depth = sys.maxsize
        # Wrap this control
        this_ctrl = self.__resolve_control(self.criteria)[-1]

        # Create a list of this control and all its descendants
        all_ctrls = [this_ctrl, ] + this_ctrl.descendants()

        # Create a list of all visible text controls
        txt_ctrls = [ctrl for ctrl in all_ctrls if ctrl.can_be_label and ctrl.is_visible() and ctrl.window_text()]

        # Build a dictionary of disambiguated list of control names
        name_ctrl_id_map = findbestmatch.UniqueDict()
        for index, ctrl in enumerate(all_ctrls):
            ctrl_names = findbestmatch.get_control_names(ctrl, all_ctrls, txt_ctrls)
            for name in ctrl_names:
                name_ctrl_id_map[name] = index

        # Swap it around so that we are mapped off the control indices
        ctrl_id_name_map = {}
        for name, index in name_ctrl_id_map.items():
            ctrl_id_name_map.setdefault(index, []).append(name)

这表明它.descendants()不是Application类的方法，而是属于控件。看来我错了。是否可以创建我自己的版本，print_control-identifiers()该版本返回可以迭代的控制对象列表？

Answer 1

列出顶级窗口的正确方法是application.windows()。然后您可以调用.descendants()每个列出的窗口。在大多数情况下，应用程序只有一个顶层窗口。特别是对于backend="uia"新对话框来说，它们也是主窗口的子窗口（对于backend="win32"每个对话框都是一个顶级窗口）。