How to return nodes between a node and the top of a tree in neo4j -

How to return nodes between a node and the top of a tree in neo4j -

i trying larn neo4j , going start basic employee / employer illustration employee has manager unless @ top of tree.

my construction looks employee->managedby->manager->managedby->manager->managedby->manager. each manger can have multiple employees or managers under them.

what able list of managers between employee , main boss(ceo/president/whatever).

initially started off query this

match(baseemployee {name: 'josh'})-[:managedby*0..]-(managers) homecoming managers.name;

this seems not show manager , his/her manager , on seems show manage when want show list like

josh, boss, bossboss, bossbossboss, ceo

after searching , luck managed closer using next query

match p=(baseemployee {name: 'josh'})-[:managedby*0..]-(managers) not(managers-[:managedby]->()) homecoming p;

i learning neo4j best guess of happening here path myself first manager doesn't have managedby relationship. issue returns path , prefer have list of managers.

is possible without doing p= query?

you must add together direction of relationship path query! otherwise explore downwards josh.

and if start varlength query @ 0 0.. homecoming josh too. default 1.. can leave off.

you can homecoming nodes of path nodes(p).

match p=(baseemployee {name: 'josh'})-[:managedby*]->(managers) not(managers-[:managedby]->()) homecoming nodes(p);

if want have 1 row per node can 1 of two: either unwind collection rows.

match p=(baseemployee {name: 'josh'})-[:managedby*]->(managers) not(managers-[:managedby]->()) unwind nodes(p) n homecoming n;

or homecoming lastly node of each path root.

match p=(baseemployee {name: 'josh'})-[:managedby*]->(managers) homecoming managers;

neo4j