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jeff_inflection · 12-09-2020

I want to return all nodes/relationships until I find the first node that has a certain relationship in an arbitrary number of iterations. Example Given: X(1)-[foo]->X(2)-[foo]->X(3)-[bar]->X(4)-[foo]->X(5) I want this returned: X(1)-[foo]->X(2)-[fo...

jeff_inflection · 12-09-2020

OK, here's the second answer: MATCH (f:X {Id:1}), x=(f)-[:foo|bar*1..]->(t:X) WITH x, relationships(x) AS r WITH x, startNode(r[size(r)-1]) AS p1, startNode(r[size(r)-2]) AS c1 WHERE NOT (c1)-[:bar]->(p1) RETURN x Whew! Now that's a lot of Cyphering...

jeff_inflection · 12-09-2020

Figured it out the first question myself after all: MATCH (f:X {Id:1}), x=(f)-[:foo|bar*1..]->(t:X) WITH x, startNode(last(relationships(x))) AS s WHERE NOT ()-[:bar]->(s) RETURN x Someone can still answer whether (f-1) is possible somehow... These ...

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