使用pywinauto按标题选择窗口

Question

使用pywinauto，我试图关闭软件中出现的弹出窗口，因为它是有条件的

def get_window():
    app = pywinauto.application.Application(backend="uia")
    app.connect(path='gofer.exe')
    #app.Properties.print_control_identifiers()
    trade = app.window(best_match='Warning:')
    # trade.wrapper_object().close
    print(trade)
    if trade == 'Warning:':
        print("You see the Window")
        # press enter key to displace
        # start next action
    else:
        print("Naw, No window bro")
        # Log to file 
        pass

印刷品（贸易）的输出为：

<pywinauto.application.WindowSpecification object at 0x0000019B8296DBA8>

所以我知道它至少起作用了，但不是去我想去的地方。根据spy++，警告是一个弹出的窗口，标题为“警告”

但是，我无法打印窗口数据。。。尽管该窗口是一个弹出窗口，但如果有区别的话，它不是一个toast弹出窗口。这是一个dlg窗口

属性打印一个dict，该dict仅引用主程序和对话框窗口中的提示，但从不指定属性。即使在搜索主程序标题时，我也无法正确执行此操作

Answer 1

这就是我用来识别弹出窗口的方法。基本上，您希望创建一个对话框（dlg），表示作为弹出窗口父窗口的窗口：

app = pywinauto.application.Application(backend="uia")
app.connect(path='gofer.exe')

# Using regular expression to create a dialog of the gofer.exe app
# I am assuming the title will match "*Gofer*" eg: "Gofer the Application"
dlg = app.window(title_re=".*Gofer.*")

# Now I am going to identify the title of the popup window:
dlg.print_ctrl_ids()

# If you did the last step correctly, the output will look something like:
#Control Identifiers:
#Dialog - 'Gofer The Application'    (L688, T518, R1065, B1006)
#[u'Dialog', u'Gofer Dialog']
#child_window(title="Warning: ", control_type="Window")
   #|
   #| Image - ''    (L717, T589, R749, B622)
   #| [u'', u'0', u'Image1', u'Image0', 'Image', u'1']
   #| child_window(auto_id="13057", control_type="Image")
   #|
   #| Image - ''    (L717, T630, R1035, B632)
   #| ['Image2', u'2']
   #| child_window(auto_id="13095", control_type="Image")
   #|


# Now using the same title and control type for the popup that we identified
# We check to see if it exists as follows:

if dlg.child_window(title="Warning:", control_type="Window").exists():
    print("Bro, you got a pop-up bro...")

else:
    print("No popup Bro...")