python - Robot Framework location and name of keyword -

i want create python library 0 argument function custom robot framework keywords can call. needs know absolute path of file keyword defined, , name of keyword. know how similar test cases using robot.libraries.builtin library , ${suite_source} , ${test name} variables, can't find similar custom keywords. don't care how complicated answer is, maybe have dig guts of robot framework's internal classes , access data somehow. there way this?

thanks janne able find solution.

from robot.running.context import execution_contexts  def resource_locator():     name      = execution_contexts.current.keywords[-1].name     libname   = execution_contexts.current.get_handler(name).libname     resources = execution_contexts.current.namespace._kw_store.resources     path = ""     key in resources._keys:         if resources[key].name == libname:             path = key             break     return {'name': name, 'path': path} 

execution_contexts.current.keywords stack of keywords called, earliest first , recent last, execution_contexts.current.keywords[-1] gets last keyword, keyword called function.

execution_contexts.current.get_handler(name).libname gets name of library in keyword name defined. in case of user defined keywords, filename (not full path) minus extension.

execution_contexts.current.namespace._kw_store.resources dictionary of included resources, key absolute path. because file path key, have search key such value's name name of resource in keyword defined (libname)