python - 无法让 pywinauto 单击按钮
问题描述
我正在尝试使用pywinauto
运行大量exe
将项目添加到现有应用程序的文件。该exe
文件按预期启动,并出现感兴趣的对话框(如下),但我无法单击对话框中的“下一步”按钮。看起来pywinauto
找不到“下一步”按钮。一个想法如何让他工作?这是我的代码:
import time
import pywinauto
def do_installs():
app = pywinauto.application.Application()
app.start("You're_a_Hoot_Brushes-(lrd)_windows.exe")
# setup wizard starts
# a "Setup" windows appears with a `Next >" button that I want to click
time.sleep(10) # excessive wait for Setup dialog
dlg = app['Setup']
print('dlg:')
dlg.print_control_identifiers()
dlg['Next'].click()
if __name__ == '__main__':
do_installs()
这是输出日志:
C:\Users\John\AppData\Local\Programs\Python\Python38-32\python.exe C:/Users/John/PycharmProjects/installMM_downloads/main.py
dlg:
Control Identifiers:
SunAwtFrame - 'Setup' (L710, T345, R1210, B735)
['Setup', 'SunAwtFrame', 'SetupSunAwtFrame']
child_window(title="Setup", class_name="SunAwtFrame")
Traceback (most recent call last):
File "C:\Users\John\AppData\Local\Programs\Python\Python38-32\lib\site-packages\pywinauto\application.py", line 250, in __resolve_control
ctrl = wait_until_passes(
File "C:\Users\John\AppData\Local\Programs\Python\Python38-32\lib\site-packages\pywinauto\timings.py", line 458, in wait_until_passes
raise err
pywinauto.timings.TimeoutError
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "C:/Users/John/PycharmProjects/installMM_downloads/main.py", line 22, in <module>
do_installs()
File "C:/Users/John/PycharmProjects/installMM_downloads/main.py", line 18, in do_installs
dlg['Next'].click()
File "C:\Users\John\AppData\Local\Programs\Python\Python38-32\lib\site-packages\pywinauto\application.py", line 379, in __getattribute__
ctrls = self.__resolve_control(self.criteria)
File "C:\Users\John\AppData\Local\Programs\Python\Python38-32\lib\site-packages\pywinauto\application.py", line 261, in __resolve_control
raise e.original_exception
File "C:\Users\John\AppData\Local\Programs\Python\Python38-32\lib\site-packages\pywinauto\timings.py", line 436, in wait_until_passes
func_val = func(*args, **kwargs)
File "C:\Users\John\AppData\Local\Programs\Python\Python38-32\lib\site-packages\pywinauto\application.py", line 222, in __get_ctrl
ctrl = self.backend.generic_wrapper_class(findwindows.find_element(**ctrl_criteria))
File "C:\Users\John\AppData\Local\Programs\Python\Python38-32\lib\site-packages\pywinauto\findwindows.py", line 87, in find_element
raise ElementNotFoundError(kwargs)
pywinauto.findwindows.ElementNotFoundError: {'best_match': 'Next', 'top_level_only': False, 'parent': <win32_element_info.HwndElementInfo - 'Setup', SunAwtFrame, 2490776>, 'backend': 'win32'}
Process finished with exit code 1
这是我要单击的带有“下一步”的对话框:
解决方案
解决方案是不要尝试找到按钮。
我能够通过以下方式做我想做的事:
keyboard.send_keys('{ENTER}')
在选择所需选择的情况下按回车键,就像我发布的对话框一样。和:
dlg.click_input(coords=(x, y))
dlg
在其他情况下,在指定坐标(相对于对话框, )处单击鼠标。
推荐阅读
- ruby-on-rails - IP 地址 URL 的 Nginx 配置
- node.js - 如何在 MongoDB 查询中输入 NodeJS 变量作为参数
- javascript - 单击li元素时如何在谷歌地图上触发标记事件
- awk - 如何使用 sed 或 awk 在特定位置删除包含特定字符串的行?
- python - 如何使用 Python 从 Firebase 存储中检索图像?
- ionic-framework - 动态显示离子列表中的离子项
- c++ - 二维数组,一个是已知的,另一个是未知的
- javascript - 如何在 $http GET 方法中设置参数以从下拉 Angularjs 发送数据
- java - Spring Security BadCredentialsException
- javascript - 如果我不返回值,可以使用简洁的箭头函数语法吗?