如果后续匹配失败，则WITH中的Neo4j值将消失

我遇到了这个问题，我有一个使用

WITH

MATCH (u1:user {username: "pewpewlasers"}), (u2:user {username: "pewpew2"})
MERGE (u1)-[y:KNOWS]->(u2)
ON CREATE SET y.connected = 1
WITH y, has(y.connected) AS connected 
REMOVE y.connected
RETURN connected

这很有效。当

知道

关系被创建时，

已连接

为

true

else

false

。所以这个密码返回

true

或

false

。现在我想扩展这个

MATCH (u1:user {username: "pewpewlasers"}), (u2:user {username: "pewpew2"})
MERGE (u1)-[y:KNOWS]->(u2)
ON CREATE SET y.connected = 1
WITH y, has(y.connected) AS connected 
REMOVE y.connected
WITH y, connected
MATCH (x)-[zz:LIKES]->(y)    <------------- this match
WITH zz, connected
RETURN zz, connected

---

可选匹配解决匹配问题时，它将失败，并出现merge giving

null

错误。有什么建议吗？

我想这样做可能会奏效

MATCH (u1:user {username: "pewpewlasers"}), (u2:user {username: "pewpew2"})
MERGE (u1)-[y:KNOWS]->(u2)
ON CREATE SET y.connected = 1
WITH y, has(y.connected) AS connected 
REMOVE y.connected
WITH y, connected
OPTIONAL MATCH (x)-[zz:LIKES]->(yy)
FOREACH(ignoreMe IN CASE WHEN yy IS null THEN [] ELSE [1] END |
    MERGE x-[:LOVES]->(yy))
WITH zz, connected
RETURN zz, connected

---

可选匹配允许保留所有初始行，如果没有关系，则“zz”的值为空。

您的解决方案是有意义的。然而，我错过了一个细节。如果您有时间，请查看编辑。谢谢。谢谢你的回答，但这给了我一个错误：类型不匹配：y已经定义了冲突的类型关系（预期节点）啊yeh，因为y是前面定义的。在这种情况下，我们可以称第二个为“yy”。如果现在答案是好的，请接受。

MATCH (u1:user {username: "pewpewlasers"}), (u2:user {username: "pewpew2"})
MERGE (u1)-[y:KNOWS]->(u2)
ON CREATE SET y.connected = 1
WITH y, has(y.connected) AS connected 
REMOVE y.connected
WITH y, connected
OPTIONAL MATCH (x)-[zz:LIKES]->(yy)
FOREACH(ignoreMe IN CASE WHEN yy IS null THEN [] ELSE [1] END |
    MERGE x-[:LOVES]->(yy))
WITH zz, connected
RETURN zz, connected