2

I have one Java application and I want that it should run as window services. I have chosen YAJSW for this. I have run its GetConfig.bat with my Java process ID and it has generated it config file. I have placed the User name and password in the config file and run the runConsole.bat. My jar file run successfully and I have seen in the log file of JAJSW which I have print out in my application. Now when I run installService.bat file. It successfully generate service but when I go the window services and run the that window service it show the error : "error 1053: The service did not respond to the start or control request in the timely fashion"

Can someone help me, how can I solve this problem.

Ashish Aggarwal
  • 3,018
  • 2
  • 23
  • 46
Malik Ehtasham
  • 343
  • 2
  • 11
  • 25

3 Answers3

1

Did you enter a username and/or password under wrapper.app.account or wrapper.ntservice.account (with corresponding password) in the wrapper configuration file?

If so, try commenting them out, maybe you have a problem with the user-account.

Graphfoto
  • 123
  • 6
  • I have run along the same problem, have not filled in the username or password, any other suggestions? – MRK187 Oct 26 '15 at 09:06
1

Maybe the issue is related to have both, Java 7 and Java 8. Please check this answer https://stackoverflow.com/a/37047073/1666929

Community
  • 1
  • 1
jfuentes
  • 477
  • 5
  • 8
0

Set the path to the java.exe in your service

Firstly get the path to the bin folder where java.exe is in. Then open your service and copy the path executable > java.exe....

sc config <servicename> binPath="FULLPATHTOBIN"+"\"+"PATHTOEXECUTABLE"
MRK187
  • 1,545
  • 2
  • 13
  • 20