Install and start an application using wsadmin

Copy the application ear-file to /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/',['-appname '])

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 :-)

= Name of server (JVM) where the application is deployed
= Name of the application (not the ear-files name)