Question

我很难从两个不同的来源收集数据并合并集合，以便最后一个是由'dateCreated'排序的一组对象。

上下文

用户可以分组提问。问题可以是一般性的，也可以与特定的视频游戏相关。如果小组中提出的问题与视频游戏相关，则此问题也会出现在视频游戏的问题页面中。

目前，我有两个一般性问题，一个特定于一个视频游戏。因此，在提问时，我应该有3个问题。

查询

以下是查询：

START group = node(627)
MATCH generalQuestions-[?:GENERAL_QUESTION]->group
WITH group, generalQuestions
MATCH gamesQuestions-[?:GAME_QUESTION]->games<-[:GAMES]-group
WITH (collect(generalQuestions) + collect(gamesQuestions)) as questions
RETURN questions
ORDER BY questions.dateCreated

第一期：使用ORDER BY

Cached(questions of type Collection) expected to be of type Map but it is of type Collection - maybe aggregation removed it?

实现我想要做的事情的正确方法是什么？

第二个问题：结果错误

如果我删除了ORDER BY子句，而不是有3个结果，我得到14 ......：

[
Node[641]{dateCreated:1380892636,dateUpdated:1380892636,title:"GENERAL TITLE 1",type:1,content:"GENERAL CONTENT 1"},
Node[641]{dateCreated:1380892636,dateUpdated:1380892636,title:"GENERAL TITLE 1",type:1,content:"GENERAL CONTENT 1"},
Node[641]{dateCreated:1380892636,dateUpdated:1380892636,title:"GENERAL TITLE 1",type:1,content:"GENERAL CONTENT 1"},
Node[641]{dateCreated:1380892636,dateUpdated:1380892636,title:"GENERAL TITLE 1",type:1,content:"GENERAL CONTENT 1"},
Node[641]{dateCreated:1380892636,dateUpdated:1380892636,title:"GENERAL TITLE 1",type:1,content:"GENERAL CONTENT 1"},
Node[641]{dateCreated:1380892636,dateUpdated:1380892636,title:"GENERAL TITLE 1",type:1,content:"GENERAL CONTENT 1"},
Node[642]{dateCreated:1380892642,dateUpdated:1380892642,title:"GENERAL TITLE 2",type:1,content:"GENERAL CONTENT 2"},
Node[642]{dateCreated:1380892642,dateUpdated:1380892642,title:"GENERAL TITLE 2",type:1,content:"GENERAL CONTENT 2"},
Node[642]{dateCreated:1380892642,dateUpdated:1380892642,title:"GENERAL TITLE 2",type:1,content:"GENERAL CONTENT 2"},
Node[642]{dateCreated:1380892642,dateUpdated:1380892642,title:"GENERAL TITLE 2",type:1,content:"GENERAL CONTENT 2"},
Node[642]{dateCreated:1380892642,dateUpdated:1380892642,title:"GENERAL TITLE 2",type:1,content:"GENERAL CONTENT 2"},
Node[642]{dateCreated:1380892642,dateUpdated:1380892642,title:"GENERAL TITLE 2",type:1,content:"GENERAL CONTENT 2"},
Node[632]{dateCreated:1380889484,dateUpdated:1380889484,title:"GTA5 TITLE",type:2,content:"GTA5 CONTENT"},
Node[632]{dateCreated:1380889484,dateUpdated:1380889484,title:"GTA5 TITLE",type:2,content:"GTA5 CONTENT"}
]

我收集结果的方式有问题吗？

修改

扩展查询以获取gamesQuestion：

gamesQuestions-[:GAME_QUESTION]->()<-[:QUESTIONS]-games-[:INTERESTS]->()<-[:HAS_‌INTEREST_FOR]-interests<-[:INTERESTS]-group

感谢您的帮助，

Answer 1

“Order by”需要节点或关系上的属性。查询中的“问题”是节点的集合而不是节点/关系，您无法使用“排序依据”对集合进行排序，您只能对其属性上的节点或关系进行排序。

为了使用“Order by”，您需要将问题作为一列而不是一个集合返回。根据原始查询中指示的关系，以下查询应将常规和特定游戏问题作为行列返回，并在属性“dateCreated”上对其进行排序，

START group = node(627) 
Match question-[?:GENERAL_QUESTION|GAME_QUESTION]->()<-[:GAMES*0..1]-(group)
Return distinct question
Order by question.dateCreated

对于游戏问题通过关系序列“游戏问题 - [?: GAME_QUESTION] - ＆gt;游戏＆lt; - [：GAMES] -group与群组相关的扩展案例，我有游戏问题 - [：GAME_QUESTION] - ＆gt;（）＆lt; - [：QUESTIONS] -games - [：INTERESTS] - ＆gt;（）＆lt; - [：HAS_ INTEREST_FOR] -interests＆lt; - [：INTERESTS] -group“，你可以简单地扩展上一个查询中的模式如下，

START group = node(627) 
Match question-[:GENERAL_QUESTION|GAME_QUESTION]->()-[*0..4]-(group)
Return distinct question
Order by question.dateCreated

我们的想法是将可以到达组节点的问题与一步或四步相匹配。

另一种选择是在where子句中指定两种模式，

START group = node(627) 
MATCH question-[*]-group
Where question-[:GENERAL_QUESTION]->group or (question-[:GAME_QUESTION]->()<-[:QUESTIONS]-()-[:INTERESTS]->()<-[:HAS_INTERESTS_FOR]-()<-[:INTERESTS]-group)
Return distinct q
Order by q.dateCreated

Neo4j cypher查询：使用ORDER BY和COLLECT（S）

1 个答案: