Question

在我的社交媒体应用程序中，我有一个neo4j数据库用于FRIEND，POSTED，LIKE，COMMENT，SHARE这样的关系

节点

(User{user_id})
(Post{title, text, created, picture})

边

(User)-[:FRIEND]->(User)
(User)-[:POSTED]->(Post) 
(Post)-[:PINNED]->(User)
(User)-[:LIKE]->(Post)
(User)-[:SHARE]->(Post)

我正在寻找一个返回所有帖子中所有数据的查询由我张贴或给我打字或张贴并钉在了我的朋友身上包括每个帖子的LIKE关系和SHARE关系的数量。除此之外，我想将每个具有POSTED关系的用户的user_id返回到其中一个帖子。所有这些ORDER BY都创建了DESC。

我正在寻找的结果将是那样的

{ 
    {
        user_id:56,
        text:'This is a test',
        title:'This is a title',
        created:'2016.12.01.18.00.00'
        post_picture:'',
        likecount:9,
        sharecount:3
    },
    {
        user_id:52,
        text:'This is a test',
        title:'This is a title',
        created:'2016.12.01.17.00.00'
        post_picture:'',
        likecount:12,
        sharecount:4
    }
}

这不起作用

MATCH (post)-[:POSTED]-(user:User)-[:FRIEND]-(friend:User)
WHERE (user{user_id:'97'})-[:POSTED]-(post)
OR (user{user_id:'97'})-[:PINNED]-(post)
OR (friend{user_id:'97'})-[:FRIEND]-(user)-[:PINNED]-(post) 
AND(friend{user_id:'97'})-[:FRIEND]-(user)-[:POSTED]-(post)
OPTIONAL MATCH (post)-[liked:LIKE]-(b:User)
OPTIONAL MATCH (post)-[share:SHARE]-(d:User)
WITH post, user, count(liked) as likecount, 
count(share) as sharecount
RETURN DISTINCT
user.user_id as user_id,
post.title as title,
post.text as text,
post.picture as post_picture,
post.created as created,
likecount,
sharecount,
ORDER BY created DESC

有没有办法在一个查询中获得所有这些？

Answer 1

此查询应该有效：

MATCH (me:User {user_id: 123})
OPTIONAL MATCH (me)-[:POSTED]->(p1)
WITH me, COLLECT(p1) AS posts
OPTIONAL MATCH (p2)-[:PINNED]->(me)
WITH me, posts + COLLECT(p2) AS posts
OPTIONAL MATCH (me)-[:FRIEND]-(friend)-[:POSTED]->(p3)-[:PINNED]->(friend)
WITH me, posts + COLLECT(p3) AS posts
UNWIND posts AS post
WITH DISTINCT post
OPTIONAL MATCH (poster)-[:POSTED]->(post)
RETURN post, COUNT(()-[:LIKE]->(post)) AS likes, COUNT(()-[:SHARE]->(post)) AS shares, COLLECT(DISTINCT poster.user_id) AS posters
ORDER BY post.created DESC;

它收集符合posts集合中每个条件的帖子，过滤掉重复项，查找发布帖子的用户，并按所需顺序返回结果。

Answer 2

什么不起作用？你有什么回来的？这将是一个无效的密码查询，因为在ORDERing之前有一个悬空的逗号。

这是您的密码查询的已清理版本。

MATCH (post)-[:POSTED]-(user:User)-[:FRIEND]-(friend:User)
WHERE (user{user_id:'97'})-[:POSTED]-(post)
OR (user{user_id:'97'})-[:PINNED]-(post)
OR (friend{user_id:'97'})-[:FRIEND]-(user)-[:PINNED]-(post) 
AND(friend{user_id:'97'})-[:FRIEND]-(user)-[:POSTED]-(post)
OPTIONAL MATCH (post)-[liked:LIKE]-(b:User)
OPTIONAL MATCH (post)-[share:SHARE]-(d:User)
WITH post, user, count(liked) as likecount, 
count(share) as sharecount
RETURN DISTINCT
user.user_id as user_id,
post.title as title,
post.text as text,
post.picture as post_picture,
post.created as created,
likecount,
sharecount,
ORDER BY created DESC

如果这不起作用，您是否可以提供测试数据库或cypher语句来创建您要操作的图形？在为您的问题找到一个好的解决方案时，这将非常有用。

Neo4J分别计算节点的关系

2 个答案: