问题描述
我在QTP中使用了CustomControl,当我试图监视控件它将生成以下代码时,
I have used CustomControl in QTP when i try to spy the control it will generate the following code,
自定义控件
SwfWindow("Form1").SwfObject("customControl1").Click 72,49
此外,我已经测试了面板和按钮。请参考以下代码, Panel
SwfWindow("Form1").SwfObject("panel1").Click 72,34
Button
SwfWindow("Button").SwfButton("Button").Click "Button"
如何在HP-QTP中实现类似按钮?
How can i achieve like button in HP-QTP?
提前致谢。
Thanks in advance.
Mohanraj G
Mohanraj G
推荐答案
此论坛适用于编码的UI测试问题,您正在使用QTP,其最新版本不称为UFT ,您可以在自己的社区中提出这类问题,在那里您将获得更专业和快速的答案。我找到了适合您的论坛:
This forum is for Coded UI testing issues, and you are using QTP, which the latest version is not called UFT, you may ask this kind of issue on it own community, where you will get more professional and quick answer. I found its forum for you:
https://community.saas.hpe.com/t5/Functional-Testing-UFT-etc/ct -p / sws-Fun_Test
感谢您的理解。
最好的问候,
Fletcher
这篇关于如何检测QTP中的控件?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!