3

我很难从两个不同的来源收集数据并合并这些集合，以便最后一个是一组按“dateCreated”排序的对象。

语境

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和一个特定于一个视频游戏的问题。因此，在提取问题时，我应该有 3 个问题。

询问

这是查询：

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

第一个问题：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：错误的结果

如果我删除 ORDER BY 子句，而不是 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？

编辑

扩展查询以获取 gamesQuestion ：

gamesQuestions-[:GAME_QUESTION]->()<-[:QUESTIONS]-games-[:INTERESTS]->()<-[:HAS_‌​INTEREST_FOR]-interests<-[:INTERESTS]-group

谢谢你的帮助，

4

1 回答 1

3

“Order by”需要节点或关系上的属性。查询中的“问题”是节点集合而不是节点/关系，您不能使用“排序依据”对集合进行排序，只能根据节点或关系的属性对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，以下查询应将一般和特定游戏问题作为行列返回，并按属性“dateCreated”对它们进行排序，

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“gamesQuestions-[?:GAME_QUESTION]->games<-[:GAMES]-group 与组相关的扩展案例，我有 gamesQuestions-[:GAME_QUESTION]->() <-[:QUESTIONS]-games-[:INTERESTS]->()<-[:HAS_‌​INTEREST_FOR]-interests<-[:INTERESTS]-group"，您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

这个想法是用一个步骤或 4 个步骤来匹配可以到达组节点的问题。

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
于 2013-10-04T16:47:18.207 回答