CODEWITHSUNDEEP
pythonflaskneo4jdockerpy2neo
Tom Rijntjes
Tom Rijntjes

Reputation: 654

Flask API breaks at long Cypher query

I'm building an API in Flask that talks to a neo4j db. One particularly large query (15 minutes+) breaks the API. Breaking means the Docker container in which it runs stops without logging the request. The trouble is I can't reproduce the error when I run the API locally.

What I think I need is a way to run the query using py2neo, then return an arbitrary response without waiting for the query to finish.

def post(self):
    g.cypher.run("MATCH a-[r]-b SET r.cost = "
    "CASE WHEN r.cost <1 THEN 0.01*exp(4.60517*(r.costx+0.01)) ELSE r.cost END " 
    "SET r.costx = "
    "CASE WHEN r.costx < 1 THEN r.costx + 0.01 ELSE r.costx END "
    "RETURN r")
    return make_response(jsonify({'success': 'all relationship costs increased'}), 200)

I'm really not an ops guy so any broader insights to this conundrum are most welcome.

Upvotes: 0

Views: 154

Answers (1)

Michael Hunger
Michael Hunger

Reputation: 41676

How much data do you have in that database? You might be better off running start r=rel(*) ...

why do you return r in the first place?

I would batch your query and add this condition:

START r=rel(*)
WITH r
WHERE r.cost < 1 OR r.costx < 1
WITH r
SKIP {batchSize} LIMIT 100000
SET r.cost = CASE WHEN r.cost < 1 THEN 0.01*exp(4.60517*(r.costx+0.01)) ELSE r.cost END 
WITH r WHERE r.costx < 1
SET r.costx = r.costx + 0.01

or better run it in two passes:

START r=rel(*)
WITH r
WHERE r.cost < 1
WITH r
SKIP {batchSize} LIMIT 100000
SET r.cost = 0.01*exp(4.60517*(r.costx+0.01))

and same for cost.x

Upvotes: 1

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