Install and start an application using wsadmin

opy the application ear-file to <WAS_PROFILE_ROOT>/installableApps

1. Open wsadmin (with jython)
1.1 Go to "WebSphere_Profile_Root"/bin
1.2 Execute

wsadmin.sh(.bat) -lang jython

1.3 Enter UserID/Password (if promted)

2. Define needed cell and node variables
2. Execute

cell=AdminConfig.list('Cell')
cellname=AdminConfig.showAttribute(cell,'name')
nodes=AdminConfig.list('Node',cell)
nodename=AdminConfig.showAttribute(nodes,'name')
-----------------------------------
(Alternative:
e.g.: nodename=NAME_OF_NODE)

3. Install application
3.1 Execute
AdminApp.install('../installableApps/<NAME_OF_EAR_FILE>',['-appname <NAME_OF_APPLICATION>'])

4. Save changes
4.1 Execute

AdminConfig.save()

5. Define application manager variable

5.1 Execute

appManager=AdminControl.queryNames
('type=ApplicationManager,cell='+cellname+',
node='+nodename+',process=,*')

5.2 To print out variable execute

print appManager

6. Assign application to variable
6.1 Execute

app=AdminControl.queryNames
('type=Application,cell='+cellname+',node='+nodename+',
process=,J2EEName=,*')

6.2 To print out variable execute

print app

7. Define application name to variable
7.1 Execute

appName=AdminControl.getAttributes(app,'name')

7.2 To print out variable execute

print appName

8. Start application
8.1 Execute

AdminControl.invoke(appManager, 'startApplication',appName)

Hint:
It is easier to use administrative console :-)
<APP_SERVER_NAME>= Name of server (JVM) where the application is deployed
<NAME_OF_APPLICATION> = Name of the application (not the ear-files name)