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我正在评估 RedisGraph，并试图复制我在 Neo4j 中的结果。

图表和查询来自我的另一个问题，我试图达到相同的结果：路径f1 <- (friend) m2 <- (sibling) b3 <- (coworker) d4（并且没有p*节点）

节点：

CREATE (a:temporary {name: 'm2'});
CREATE (a:temporary {name: 'f1'});
CREATE (a:temporary {name: 'b3'});
CREATE (a:temporary {name: 'd4'});

CREATE (a:temporary {name: 'p5'});
CREATE (a:temporary {name: 'p6'});
CREATE (a:temporary {name: 'p7'});

CREATE (a:temporary {name: 'k1'});
CREATE (a:temporary {name: 'k2'});

和关系：

MATCH (a) WHERE a.name = 'b3' MATCH (b) where b.name = 'm2' CREATE (a)-[:KNOWS {type: 'sibling'}]->(b);
MATCH (a) WHERE a.name = 'm2' MATCH (b) where b.name = 'f1' CREATE (a)-[:KNOWS {type: 'friend'}]->(b);
MATCH (a) WHERE a.name = 'd4' MATCH (b) where b.name = 'b3' CREATE (a)-[:KNOWS {type: 'coworker'}]->(b);

MATCH (a) WHERE a.name = 'p7' MATCH (b) where b.name = 'p6' CREATE (a)-[:KNOWS {type: 'friend'}]->(b);
MATCH (a) WHERE a.name = 'p6' MATCH (b) where b.name = 'p5' CREATE (a)-[:KNOWS {type: 'coworker'}]->(b);

MATCH (a) WHERE a.name = 'k1' MATCH (b) where b.name = 'b3' CREATE (a)-[:HAS]->(b);
MATCH (a) WHERE a.name = 'k1' MATCH (b) where b.name = 'd4' CREATE (a)-[:HAS]->(b);
MATCH (a) WHERE a.name = 'k1' MATCH (b) where b.name = 'm2' CREATE (a)-[:HAS]->(b);
MATCH (a) WHERE a.name = 'k1' MATCH (b) where b.name = 'f1' CREATE (a)-[:HAS]->(b);

MATCH (a) WHERE a.name = 'k2' MATCH (b) where b.name = 'p5' CREATE (a)-[:HAS]->(b);
MATCH (a) WHERE a.name = 'k2' MATCH (b) where b.name = 'p7' CREATE (a)-[:HAS]->(b);
MATCH (a) WHERE a.name = 'k2' MATCH (b) where b.name = 'p6' CREATE (a)-[:HAS]->(b);

原始的 Cypher 查询：

match p=(a:temporary)-[:KNOWS*0..]->(b) 
where not ()-[:KNOWS]->(a) and not (b)-[:KNOWS]->() 
and exists { match (k)-[:HAS]->(a) where k.name='k1' } 
return p

给出以下错误，关于 中的子路径EXISTS：

Invalid input '(': expected ':', ',' or '}' line: 1, column: 112, offset: 111 errCtx: ... (b)-[:KNOWS]->() and exists { match (k)-[:HAS]->(a) where k.name='k1' } r... errCtxOffset: 40

我已经像这样重写了它（带有附加MATCH条款：这是否等同于上述内容？

GRAPH.QUERY test "match p=(a:temporary)-[:KNOWS*0..]->(b) where not ()-[:KNOWS]->(a) and not (b)-[:KNOWS]->() with a, p match (k)-[:HAS]->(a) where k.name='k1'  return p"
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1 回答 1

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它是等价的，重写：

MATCH (k)-[:HAS]->(a:temporary) WHERE k.name='k1'
MATCH p = (a)-[:KNOWS*0..]->(b) 
WHERE not ()-[:KNOWS]->(a) AND not (b)-[:KNOWS]->() 
RETURN p
于 2021-04-12T09:42:46.390 回答