I notice there is a method to get the current apps dir from python. There is also a getCurrentApp() available in javascript, but I need the current app from Python.
Does anyone have any suggestions?
Which type of python file? We tend to use python's __file__
attribute. I assume that you have imported os and cherrypy at least.
For example, in the file $SPLUNK_HOME/etc/apps/TA-opseclea/appserver/controllers/:
app_name = __file__.split('.')[-2]
If I was implementing a controller method where a request was routed to me:
@route('/:app/:action=edit/:name')
@expose_page(must_login=True, methods=['GET'])
def edit(self, app, action, name, **kwargs):
host_app = cherrypy.request.path_info.split('/')[3]
For a scripted input in an app's .\bin directory:
app_name = __file__.split(os.sep)[-3]
Which type of python file? We tend to use python's __file__
attribute. I assume that you have imported os and cherrypy at least.
For example, in the file $SPLUNK_HOME/etc/apps/TA-opseclea/appserver/controllers/:
app_name = __file__.split('.')[-2]
If I was implementing a controller method where a request was routed to me:
@route('/:app/:action=edit/:name')
@expose_page(must_login=True, methods=['GET'])
def edit(self, app, action, name, **kwargs):
host_app = cherrypy.request.path_info.split('/')[3]
For a scripted input in an app's .\bin directory:
app_name = __file__.split(os.sep)[-3]
I am writing a custom search command. Looks like the last option you presented may be the one. I will give it a shot. Thanks