返回 Neo4J 中每个 session 的任务(列表列表)
Returning tasks for each session in Neo4J (list of lists)
我将 Neo4J 用于我正在构建的导师平台,但我被以下问题难住了:
给定以下节点和属性:
Mentor{ login, ... }
Mentee{ login, ... }
Session{ notes, ... }
Task{ complete, name }
以及以下协会:
// each session has 1 mentor and 1 mentee
(Mentor)<-[:HAS]-(Session)-[:HAS]->(Mentee)
// each task is FOR one person (a Mentor or Mentee)
// each task is FROM one Session
(Session)<-[:FROM]-(Task)-[:FOR]->(Mentor or Mentee)
查询此数据以生成以下形状的 API 响应的最佳方法是什么?同样,这是对数据建模的合理方法吗?也许 coalesce
?
{
mentor: { login: '...', /* ... */ },
mentee: { login: '...', /* ... */ },
sessions: [
{
notes,
/* ... */
mentorTasks: [{ id, name, complete }],
menteeTasks: [{ id, name, complete }]
]
我第一次尝试:
MATCH (mentor:Mentor{ github: "mentorlogin" })
MATCH (session:Session)-[:HAS]->(mentee:Mentee{ github: "menteelogin" })
OPTIONAL MATCH (mentor)<-[:FOR]-(mentorTask:Task)-[:FROM]->(session)
OPTIONAL MATCH (mentee)<-[:FOR]-(menteeTask:Task)-[:FROM]->(session)
RETURN
mentor,
mentee,
session,
COLLECT(DISTINCT mentorTask) as mentorTasks,
COLLECT(DISTINCT menteeTask) as menteeTasks
ORDER BY session.date DESC
但这很糟糕 - 导师和受训者数据被 returned 多次,如果受训者没有会话,它就完全消失了。
这似乎更合适,但我不确定如何折叠任务:
MATCH (mentor:Mentor{ github: "mentorlogin" })
MATCH (mentee:Mentee{ github: "menteelogin })
OPTIONAL MATCH (session:Session)-[:HAS]->(mentee)
OPTIONAL MATCH (mentor)<-[:FOR]-(mentorTask:Task)-[:FROM]->(session)
OPTIONAL MATCH (mentee)<-[:FOR]-(menteeTask:Task)-[:FROM]->(session)
RETURN
mentor,
mentee,
COLLECT(DISTINCT session) as sessions
编辑:工作!感谢 Graphileon 的及时回复。我做了一些修改:
- 更改了
MATCH
声明,因此它 return 是导师和受训者,即使没有会议
- 按日期排序会话(最近的排在最前面)
- return所有节点属性,而不是白名单
MATCH (mentor:Mentor{ github: $mentorGithub })
MATCH (mentee:Mentee{ github: $menteeGithub })
RETURN DISTINCT {
mentor: mentor{ .*, id: toString(id(mentor)) },
mentee: mentee{ .*, id: toString(id(mentee)) },
sessions: apoc.coll.sortMaps([(mentor:Mentor)<-[:HAS]-(session:Session)-[:HAS]->(mentee:Mentee) |
session{
.*,
id: toString(id(session)),
mentorTasks: [
(session)<-[:FROM]-(task:Task)-[:FOR]->(mentor) |
task{ .*, id: toString(id(task)) }
],
menteeTasks: [
(session)<-[:FROM]-(task:Task)-[:FOR]->(mentee) |
task{ .*, id: toString(id(task)) }
]
}
], "date")
} AS result
假设您拥有这些数据:
你可以按照这些思路做一些事情,嵌套 pattern comprehensions
MATCH (mentor:Mentor)<-[:HAS]-(:Session)-[:HAS]->(mentee:Mentee)
RETURN DISTINCT {
mentor: {id:id(mentor), name: mentor.name},
mentee: {id:id(mentee), name: mentee.name},
sessions: [(mentor:Mentor)<-[:HAS]-(session:Session)-[:HAS]->(mentee:Mentee) |
{ id: id(session),
name: session.name,
mentorTasks: [(session)<-[:FROM]-(task:Task)-[:FOR]->(mentor) |
{id:id(task), name: task.name}
],
menteeTasks: [(session)<-[:FROM]-(task:Task)-[:FOR]->(mentee) |
{id:id(task), name: task.name}
]
}
]
} AS myResult
回归
{
"mentor": {
"name": "Mentor Jill",
"id": 211
},
"sessions": [
{
"menteeTasks": [
{
"id": 223,
"name": "Task D"
},
{
"id": 220,
"name": "Task C"
},
{
"id": 219,
"name": "Task B"
}
],
"name": "Session 1",
"id": 208,
"mentorTasks": [
{
"id": 213,
"name": "Task A"
}
]
}
],
"mentee": {
"name": "Mentee Joe",
"id": 212
}
}
请注意,使用模式理解,您可以避免 OPTIONAL 匹配。如果模式理解没有找到任何东西,它 returns []
我将 Neo4J 用于我正在构建的导师平台,但我被以下问题难住了:
给定以下节点和属性:
Mentor{ login, ... }
Mentee{ login, ... }
Session{ notes, ... }
Task{ complete, name }
以及以下协会:
// each session has 1 mentor and 1 mentee
(Mentor)<-[:HAS]-(Session)-[:HAS]->(Mentee)
// each task is FOR one person (a Mentor or Mentee)
// each task is FROM one Session
(Session)<-[:FROM]-(Task)-[:FOR]->(Mentor or Mentee)
查询此数据以生成以下形状的 API 响应的最佳方法是什么?同样,这是对数据建模的合理方法吗?也许 coalesce
?
{
mentor: { login: '...', /* ... */ },
mentee: { login: '...', /* ... */ },
sessions: [
{
notes,
/* ... */
mentorTasks: [{ id, name, complete }],
menteeTasks: [{ id, name, complete }]
]
我第一次尝试:
MATCH (mentor:Mentor{ github: "mentorlogin" })
MATCH (session:Session)-[:HAS]->(mentee:Mentee{ github: "menteelogin" })
OPTIONAL MATCH (mentor)<-[:FOR]-(mentorTask:Task)-[:FROM]->(session)
OPTIONAL MATCH (mentee)<-[:FOR]-(menteeTask:Task)-[:FROM]->(session)
RETURN
mentor,
mentee,
session,
COLLECT(DISTINCT mentorTask) as mentorTasks,
COLLECT(DISTINCT menteeTask) as menteeTasks
ORDER BY session.date DESC
但这很糟糕 - 导师和受训者数据被 returned 多次,如果受训者没有会话,它就完全消失了。
这似乎更合适,但我不确定如何折叠任务:
MATCH (mentor:Mentor{ github: "mentorlogin" })
MATCH (mentee:Mentee{ github: "menteelogin })
OPTIONAL MATCH (session:Session)-[:HAS]->(mentee)
OPTIONAL MATCH (mentor)<-[:FOR]-(mentorTask:Task)-[:FROM]->(session)
OPTIONAL MATCH (mentee)<-[:FOR]-(menteeTask:Task)-[:FROM]->(session)
RETURN
mentor,
mentee,
COLLECT(DISTINCT session) as sessions
编辑:工作!感谢 Graphileon 的及时回复。我做了一些修改:
- 更改了
MATCH
声明,因此它 return 是导师和受训者,即使没有会议 - 按日期排序会话(最近的排在最前面)
- return所有节点属性,而不是白名单
MATCH (mentor:Mentor{ github: $mentorGithub })
MATCH (mentee:Mentee{ github: $menteeGithub })
RETURN DISTINCT {
mentor: mentor{ .*, id: toString(id(mentor)) },
mentee: mentee{ .*, id: toString(id(mentee)) },
sessions: apoc.coll.sortMaps([(mentor:Mentor)<-[:HAS]-(session:Session)-[:HAS]->(mentee:Mentee) |
session{
.*,
id: toString(id(session)),
mentorTasks: [
(session)<-[:FROM]-(task:Task)-[:FOR]->(mentor) |
task{ .*, id: toString(id(task)) }
],
menteeTasks: [
(session)<-[:FROM]-(task:Task)-[:FOR]->(mentee) |
task{ .*, id: toString(id(task)) }
]
}
], "date")
} AS result
假设您拥有这些数据:
你可以按照这些思路做一些事情,嵌套 pattern comprehensions
MATCH (mentor:Mentor)<-[:HAS]-(:Session)-[:HAS]->(mentee:Mentee)
RETURN DISTINCT {
mentor: {id:id(mentor), name: mentor.name},
mentee: {id:id(mentee), name: mentee.name},
sessions: [(mentor:Mentor)<-[:HAS]-(session:Session)-[:HAS]->(mentee:Mentee) |
{ id: id(session),
name: session.name,
mentorTasks: [(session)<-[:FROM]-(task:Task)-[:FOR]->(mentor) |
{id:id(task), name: task.name}
],
menteeTasks: [(session)<-[:FROM]-(task:Task)-[:FOR]->(mentee) |
{id:id(task), name: task.name}
]
}
]
} AS myResult
回归
{
"mentor": {
"name": "Mentor Jill",
"id": 211
},
"sessions": [
{
"menteeTasks": [
{
"id": 223,
"name": "Task D"
},
{
"id": 220,
"name": "Task C"
},
{
"id": 219,
"name": "Task B"
}
],
"name": "Session 1",
"id": 208,
"mentorTasks": [
{
"id": 213,
"name": "Task A"
}
]
}
],
"mentee": {
"name": "Mentee Joe",
"id": 212
}
}
请注意,使用模式理解,您可以避免 OPTIONAL 匹配。如果模式理解没有找到任何东西,它 returns []