cyhper将两列合并为一个 [英] cyhper combine two columns into a single

问题描述

我找不到类似的帖子，因此，如果您已经知道一个帖子，或者如果我的问题不正确，请告诉我.

I couldn't find a similar post, so if you already know one or if my question is not the proper one, please let me know.

我有这个查询

MATCH
  (t:Taxi {name:'Taxi1813'})<-[:ASSIGNED]-(u2:User)-[rd2:DROP_OFF]->
  (g2:Grid)-[r:TO*1..2]-(g:Grid)<-[rd:DROP_OFF]-(u:User)-[:ASSIGNED]->(t)
WHERE ID(u2) < ID(u) AND rd2.time >= '04:38' AND rd2.time <= '04:42'
WITH DISTINCT u2, g2, u, g, rd2, rd
MATCH p=shortestPath((g2)-[r:TO*1..2]-(g))
WITH rd2, rd,u2, g2, u, g, p, REDUCE(totalTime = 0, x IN RELATIONSHIPS(p) | totalTime + x.time) AS totalTime
WHERE totalTime <= 4
RETURN u2.name, u.name

所以最后我有两列

u2.name  u.name
User179  UserTest
User177  User179

是否存在将两列合并为一个并删除重复项的方法或功能

Is there is a way or function to merge both columns into a single one and remove duplicates

Users
User179
User177
UserTest

有什么建议吗?谢谢

推荐答案

您可以将两个集合合并为一个集合，然后仅返回不同的项目.

You can combine the two collections into a single collection and then return just the distinct items.

WITH ['User179', 'User177'] AS list1
, ['UserTest', 'User179'] AS list2
UNWIND list1 + list2 AS item
RETURN DISTINCT item

或者，如果您使用的是APOC，则可以使用apoc.coll.union().

Alternatively, if you are using APOC you could use apoc.coll.union() instead.

WITH ['User179', 'User177'] AS list1
, ['UserTest', 'User179'] AS list2
RETURN apoc.coll.union(list1,list2)

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