Run Java application at Windows startup

I have a JAR file containing a Java application. How can I make it start with Windows, without needing user interaction?

8 Answers 8

Create a .bat file and put this inside:

javaw -Xmx200m -jar C:\Path\to\jarfile\TheJar.jar

Then put the .bat file into the windows startup folder.

One more thing: There’s a difference between using java and javaw. While java is better when you are debugging an application, the application prints text or something like that, javaw is better when you don’t need that. Why? Because java runs java program using a console that shows all that application prints (println’s, exception stacktraces and so on) while javaw doesn’t run on console.

it’s simple as you have to put shortcut in

Windows 7 C:\users\All Users\Start Menu\Programs\Startup (Admin) or User home directory(%userProfile%)

Windows 10 : In Run shell:startup

in it’s property -> shortcut -> target — > java.exe -jar D:\..\runJar.jar

NOTE: This will run only after you login

With Admin Right

sc create serviceName binpath= «java.exe -jar D:\..\runJar.jar» Will create windows service

if you get timeout use cmd /c D:\JAVA7

1\jdk1.7.0_51\bin\java.exe -jar d:\jenkins\jenkins.war but even with this you’ll get timeout but in background java.exe will be started. Check in task manager

In some restricted environment as I was in corporate environment

ERROR:

The service did not respond to the start or control request in a timely fashion

1\jdk1.7.0_51\bin\java.exe -jar d:\jenkins\jenkins.war

This will give you an error if you run manually but will run in background.

NOTE: This will run at windows logon start-up(before sign-in, Based on service ‘ Startup Type ‘)

Regedit

Note: Edit Advanced User only

To Run for Current User Only

To Run for All Users

Create a String with Name and Path using above command