Question

首先让我指出我是Neo4j的新手，如果有更好的方法，请告诉我。

其次，我怎样才能找到：

// QUERY
if(!empty($_SESSION['uid'])) {
    $uid = $_SESSION['uid'];
    $query = 'MATCH (cu:User)
    WHERE cu.id = "'.$uid.'"
    WITH cu
    MATCH (p:Painting)<-[:PAINTED]-(u:User)-[:LIVES_IN]->(l:Location) 
    WHERE (p.slug) = "'.$slug.'"
        RETURN p,u,l, 
        EXISTS((cu)-[:LIKES]->(p)) as liked,
        EXISTS((cu)-[:FOLLOWS]->(u)) as followed,
        SIZE((:User)-[:LIKES]->(p)) as total_likes, 
        SIZE((:User)-[:FOLLOWS]->(u)) as total_follows
    LIMIT 1';    
} else {
    $query = 'MATCH (p:Painting)<-[:PAINTED]-(u:User)-[:LIVES_IN]->(l:Location) WHERE (p.slug) = "'.$slug.'"
        RETURN p,u,l, 
        SIZE((:User)-[:LIKES]->(p)) as total_likes, 
        SIZE((:User)-[:FOLLOWS]->(u)) as total_follows
        LIMIT 1';
}

所以现在我根据用户的可用性运行两个查询。我必须想象必须有一个更好/更有效的方法来做到这一点，但至少它现在有效。

Answer 1

MATCH (p:Painting)<-[:PAINTED]-(u:User)-[:LIVES_IN]->(l:Location) WHERE (p.slug) = "blah-blah-blah"
WITH p, u, l
OPTIONAL MATCH (ul:User)-[likes:LIKES]->(p)
OPTIONAL MATCH (uf:User)-[follows:FOLLOWS]->(u)
RETURN p,u,l,ul,uf LIMIT 1

这会返回正确的图表，我无法弄清楚为什么我很难验证类似/跟随连接。

唉，我今天尝试过很多东西，我不知道上下是什么。我认为OPTIONAL MATCHES现在实际上已经处理了总值。我想我遇到的问题是验证现有用户是否喜欢或关注过。

Answer 2

听起来有一个当前用户的概念，您可能会通过ID或名称或类似物获得，并且您希望看到他们是否喜欢这幅画并跟随用户为其“slug”定义的这幅特定画作“属性（这是一个独特的价值，还是与多幅画相匹配？）

我会对这个进行一次尝试，并提供一些改进，以便更好地获得喜欢和跟随的数量。

MATCH (currentUser:User)
WHERE currentUser.ID = 123
WITH currentUser
MATCH (p:Painting)<-[:PAINTED]-(painter:User)-[:LIVES_IN]->(l:Location) 
WHERE (p.slug) = "blah-blah-blah"
    RETURN p,painter,l, 
    SIZE( (:User)-[:LIKES]->(p) ) as painting_likes, 
    SIZE( (:User)-[:FOLLOWS]->(painter) ) as painter_follows, 
    EXISTS( (currentUser)-[:LIKES]->(p) ) as current_user_liked,
    EXISTS( (currentUser)-[:FOLLOWS]->(painter) ) as current_user_followed
LIMIT 1

Neo4j - 来自Node的全部关系

2 个答案: