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Visual Basic dot Net developer for small tool

there is a tool already present in my Desktop computer, which tracks my Production time on the basis of Login time, logout time, if i am idle for more than 10 sec, it starts counting away from system, and it shows active time in the last. i want to by pass this. is it possible? so that if i go idle, it wont track me.

Навыки: .NET, Visual Basic, Программирование на C#

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О работодателе:
( 0 отзыв(-а, -ов) ) Mumbai, India

ID проекта: #26536344

9 фрилансеров(-а) готовы выполнить эту работу в среднем за ₹1331

krishantha972

Dear Client; I have gone through your project description and as per analysis ready to give you complete work. I am working as senior software engineer with over 10 Years’ experience for software development using c#, Больше

₹700 INR за 2 дней(-я)
(4 отзывов(-а))
3.4
rangav12891

Hi, I suggest you to open note and keep a object on the space bar when you are going out. It will be active always.

₹750 INR за 1 день
(3 отзывов(-а))
2.0
₹1300 INR за 3 дней(-я)
(1 отзыв)
1.0
sumitdwivedi565

Hi, I am a C#,.Net and programming expert with more than 7 years of experience. Let me know if you want to hire me for this task.

₹1500 INR за 1 день
(0 отзывов(-а))
0.0
paulelan2017

sir this tool can be modified using vb.net .pls contact me for further dis [login to view URL] with Regards Paul

₹3333 INR за 2 дней(-я)
(0 отзывов(-а))
0.0
Kyrp70

I can make the project if you have the source code if not I can do it in reverse enginering if it is in .NET

₹1050 INR за 15 дней(-я)
(0 отзывов(-а))
0.0
₹1300 INR за 1 день
(0 отзывов(-а))
0.0
Wilsonm3110

Hello, I am a developer with more than 6 years of experience in .Net (C # / VB) with different databases such as Oracle, SQL and MySQL. Please contact me to help you on your project and together review the technical de Больше

₹1000 INR за 7 дней(-я)
(0 отзывов(-а))
0.0
lidoratal

It would be easy to get bypass this with a great idea. Contact me and I will solve the problem for you

₹1050 INR за 7 дней(-я)
(0 отзывов(-а))
0.0