You don’t need to specify ‘java’ in the AppParameters because it’s already specified as the Application

I followed this instructions to run RC as service.

I was able to create registory entery by name SeleniumRC. I don’t know what is -Xrs -jar selenium-server.jar is used for so I changed it to java -jar selenium-server.jar command.

I hope its okay.

After this I am not sure how to start SeleniumRC as service.

Can some one help?

1. Copy “JavaService.exe” to selenium-server-1.0.3\ directory as “selenium.exe”
2. Run the following batch in the same directory (edit paths and SEL_PARAMS if needed):
========
@echo off
set JAVA_HOME=c:\java\jre6
set JVMDIR=%JAVA_HOME%\bin\client
set JSBINDIR=%CD%
set JSEXE=%JSBINDIR%\selenium.exe
set SVC_NAME=Selenium4445

set SEL_PARAMS=”-port 4445 -log %JSBINDIR%\logs\%SVC_NAME%.log -ensureCleanSession -trustAllSSLCertificates”
set SEL_JAR=%JSBINDIR%\selenium-server.jar
set SEL_START=org.openqa.selenium.server.SeleniumServer

echo . About to remove the previously installed service if any:
pause
%JSEXE% -uninstall %SVC_NAME%

echo . Using following version of JavaService executable:
%JSEXE% -version
echo .

echo Installing service… Press Control-C to abort
pause
echo .

%JSEXE% -install %SVC_NAME% %JVMDIR%\jvm.dll -Djava.class.path=%SEL_JAR% -start “%SEL_START%” -params “%SEL_PARAMS%”
echo .
pause

%JSEXE% -queryconfig %SVC_NAME%
========

Voila!

PS: Draw your attention to how double quotes are used. Each substring in SEL_PARAMS should be passed to SeleniumServer.class as a separate parameter and should not be combined to the single string. Otherwise nothing will work.
See the corresponding registry key below:
========
REGEDIT4

[HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Services\Selenium4445\Parameters]
“JavaService Version”=”2,0,10,0″
“JVM Library”=”c:\\java\\jre6\\bin\\client\\jvm.dll”
“JVM Option Count”=dword:00000001
“JVM Option Number 0″=”-Djava.class.path=C:\\Java\\selenium\\selenium-server-1.0.3\\selenium-server.jar”
“Start Class”=”org.openqa.selenium.server.SeleniumServer”
“Start Method”=”main”
“Shutdown Timeout”=dword:00007530
“Overwrite Files Flag”=dword:00000000
“Startup Sleep”=dword:00000000
“Start Param Count”=dword:00000006
“Start Param Number 0″=”-port”
“Start Param Number 1″=”4445″
“Start Param Number 2″=”-log”
“Start Param Number 3″=”C:\\Java\\selenium\\selenium-server-1.0.3\\logs\\Selenium4445.log”
“Start Param Number 4″=”-ensureCleanSession”
“Start Param Number 5″=”-trustAllSSLCertificates”
========

I ran into the same problem (the parameters) and solved it by using a BAT-file containing the whole command line:

Windows Registry Editor Version 5.00
[HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Services\SeleniumRC\Parameters]
“Application”=”C:\\Path\To\My\BAT\File”

I must admit I haven’t tried starting it on anything other than the default port - but I would have thought it should work.

The first thing to try would be to start the service from the command line using the arguments you’re setting in the registry; so, in your case:

cd “C:\\Program Files\\selenium-server-1.0.1″
java -Xrs -jar selenium-server.jar -port 8888

If that doesn’t start correctly, it should give you an error message with the cause.

HTH -

Have you tried starting the SeleniumRC service on a port other than the default (4444)? I’m trying to pass in the parameter -port 8888 without any luck. I tried adding it to the registry key:

AppParameters”=”-Xrs -jar selenium-server.jar -port 8888″

But it doesn’t seem to get picked up. (How can I view console output?)

Neither does it when I add it to the service properties Start parameters. Then the system tells me the service was started and stopped because it may not have had any work to do.