nebula2.6.2 怎样获取节点全部属性并排序

问答 问题
1

1.希望获取节点全部属性并排序，下面语句前面排序后，在fetch * 查询全部属性但是前面排序字段结果后面全部属性中无效
GO FROM 34580 OVER Singer_song YIELD Singer_song._dst AS Singer_song | FETCH PROP ON Thing $-.Singer_song YIELD Thing.Thing_popularity AS Thing_popularity | ORDER BY $-.Thing_popularity DESC |limit 10 |fetch prop on * $-.VertexID
2.用match 实现但是性能不高
MATCH (v:Thing)-[e:Singer_song]->(v2)
WHERE id(v) == 34580
RETURN id(v2),v2.Thing_popularity as Thing_popularity ,v2
ORDER BY Thing_popularity desc LIMIT 3

2

你的需求是 做一跳查询, 对查询结果的每个vertex查询所有属性, 然后按照某个属性排序吗?

3

是的先一条查出目标节点ID，再根据目标节点的字段排序，然后取前10条，在通过vid fetch * 查出节点全部字段，但是最后一个fetch 没有保留上一次vid 的排序，就是问这个问题怎么处理，下面我用match 也可实现了，但是查询性能太慢

1 个赞
4

需要从前面一个 fetch 的 yield 子句透传全部属性。这是我本地测试:

(root@nebula) [nba]> GO FROM "Tim Duncan" OVER like YIELD like._dst AS dst | FETCH PROP ON * $-.dst YIELD player.age AS age,vertex AS allProperties | ORDER BY $-.age | LIMIT 3 | YIELD $-.allProperties AS allProperties
+-----------------------------------------------------------+
| allProperties                                             |
+-----------------------------------------------------------+
| ("Tony Parker" :player{age: 36, name: "Tony Parker"})     |
| ("Manu Ginobili" :player{age: 41, name: "Manu Ginobili"}) |
+-----------------------------------------------------------+
Got 2 rows (time spent 4378/4719 us)
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5

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