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我的例子得到我的微博和我的朋友们的微博:Neo4j查询:如何在标签中构建多个标签?
MATCH
(me:User{user_id: "346"})-[:POSTS]->(t:Tweet),
(me)-[:FOLLOWS]->(following:User)-[:POSTS]->(t1:Tweet)
RETURN
t, t1
SKIP 10 LIMIT 10
问:我该如何合并T和T1到结果集。基本上,我想建立json响应如下:
[
{
"tweet_id": "504597",
"message": "Commodi consequatur qui libero.",
"location": "25909 Hermann Village",
"user": {
"user_id": "346",
"user_name": "Madaline.Mayer60346",
"full_name": "Conor Hyatt",
"avatar_url": "http://lorempixel.com/640/480"
}
},
{
"tweet_id": "504261",
"message": "Atque hic ut velit.",
"location": "42920 Esmeralda Lakes",
"user": {
"user_id": "347",
"user_name": "Madaline",
"full_name": "Conor Test",
"avatar_url": "http://lorempixel.com/640/480"
}
}
]
感谢您的反馈意见。首先,你的更新代码工作。但它对响应数据太慢(我的推特样本数据库> 3GB)。是否有其他更新来提高性能? 另外,感谢APOC库 –
性能问题:在156480毫秒内返回10行。 –
尝试用这个“MATCH(me:User {user_id:”346“})替换这一行”MATCH(me:User {user_id:“346”}) - [:POSTS] - >我MATCH(我) - [:POSTS] - >(t:鸣叫)“ – Evgen