密码-匹配两个不同的可能路径并返回两者 [英] Cypher - matching two different possible paths and return both
问题描述
我有一个数据集，在这里以示例的方式表示：
http：/ /console.neo4j.org/?id=3dq78v
I have a a data set that I have represented here as an example: http://console.neo4j.org/?id=3dq78v
我想要做的是针对图形中的每个Z节点（该示例只有一个但我有很多）我想捕获一组涵盖所有关联的A，B，C和D节点和关系的属性。
What I am trying to do is for every Z node in my graph (the example only has one but I have many) I want to capture a set of properties that cover all all of the associated A, B, C and D nodes and relationships.
我遇到了试图做到这一点的两个问题。首先是C节点连接到B节点或A节点。第二个D节点连接到C节点，但并不总是存在。
I have come across two issues trying to do this. The first is that the C nodes are either connected to the to the B nodes OR the A nodes. Second the D nodes are connected to C nodes but not always present.
我要输出一个看起来像这样的表：
I am looking to output a table that looks something like this:
Z.prop | A.prop | B.prop | （A-B rel）.prop | c.prop | （c-d rel）.prop | d.prop
Z.prop | A.prop | B.prop | (A-B rel).prop | c.prop | (c-d rel).prop | d.prop
我尝试了很多使用OPTIONAL MATCH和WITH的组合，但我无法做到。我使用OPTIONAL MATCH尝试过的所有内容都会找到该匹配项（如果存在），然后不允许我维护与该可选匹配项不匹配的项目（这是我应该做的）。我可以分享我尝试过的更具体的查询是否有帮助。
I have tried a lot of combinations of using OPTIONAL MATCH and using WITH but I can not get it. Everything I have tried with OPTIONAL MATCH will find that match if it exists and then not allow me to maintain the items that did not match this optional match (Which I realize is what it is supposed to do). I can share more specific queries I have tried if that will help.
任何洞察力都很好！
编辑：使用Neo4j 2.0.3版本
编辑：使用较小的更正更新了控制台链接
using version Neo4j 2.0.3 updated console link with a small correction
这是我尝试过的查询。我了解了为什么这行不通，但是也许您可以看到我的逻辑，我希望可选匹配项在发现某些匹配项时不删除不匹配的节点。我想要那些不匹配的东西。
Here is a query I have tried. I realize why this doesn't work, but maybe you can see my 'logic' I want the optional match to not remove nodes if they dont match if it finds some matches. I want the ones that dont match and do.
MATCH (z:Z)-[:has]->(a:A)
OPTIONAL MATCH (a)-[:has*1..2]->(c:C)
OPTIONAL MATCH (a)-[:has]->(b:B)-[:CONTAINS]->(c)
OPTIONAL MATCH (c)-[cd:knows]->(d:D)
RETURN z.name, a.name, b.name, c.name, d.name, cd.score;
编辑：我正在尝试使用以下4个查询的结果，但获取结果带有1个查询
What I am trying to do is use the results from the 4 queries below but get the results with 1 query
#1
MATCH (z:Z)-[:has]->(a:A)
MATCH (A)-[mr:has]->(b:B)-[:has]->(c:C)
MATCH (c)-[ds:knows]->(d:d)
RETURN a.name, b.name, mr.order, c.name, d.name, ds.score;
#2
MATCH (z:Z)-[:has]->(a:A)
MATCH (A)-[mr:has]->(b:B)-[:has]->(c:C)
WHERE NOT (c)-[:knows]->(:d)
RETURN a.name, mr.order, b.name, c.name;
#3
MATCH (z:Z)-[:has]->(a:A)
MATCH (A)-[:has]->(c:C)
WHERE NOT (c)-[:knows]->(:d)
RETURN a.name, c.name;
#4
MATCH (z:Z)-[:has]->(a:A)
MATCH (A)-[:has]->(c:C)
MATCH (c)-[ds:knows]->(d:d)
RETURN a.name, c.name, d.name, ds.score;
由于某些结果，这些方法并没有达到我的预期。例如，我期望查询3仅返回：
These don't work the way I was expecting though because some results. For example I was expecting query 3 to only return:
A2 C4 d4 8
A2 C4 d5 6
A2 C4 d6 9
编辑-精确输出我的目标是：
EDIT - EXACT output I am aiming for:
a.name mr.order b.name c.name d.name d.score
A1 1 B3
A1 2 B1 C2
A1 2 B1 C5
A1 2 B1 C1 d1 1
A1 2 B1 C1 d3 4
A1 2 B1 C1 d2 3
A2 1 B4
A2 2 B2 C3
A2 C4 d4 8
A2 C4 d5 6
A2 C4 d6 9
这与我要查找的11行中的9行匹配，它错过了带有B3和B4的行
This matches 9 of the 11 lines I am looking for, it misses the lines with B3 and B4
MATCH (z:Z)-[:has]->(a:A)
WITH a, z
MATCH (a)-[*1..2]-(c:C)
OPTIONAL MATCH (a)-[mr:has]->(b:B)-[:has]-(c:C)
WITH a, b, c, z, mr
OPTIONAL MATCH (c)-[cd:knows]->(d:d)
RETURN z.name, a.name, mr.order, b.name, c.name, d.name, cd.score;
推荐答案
不确定作为输出的期望，但这可能会起作用：
Not completely sure about what you would expect as the ouput, but this one might work:
MATCH (z:Z)-[:has]->(a:A)
WITH a, z
MATCH (b:B), (a)-[*1..2]-(c:C)
WHERE (a)-[:has]->(b)-[:CONTAINS]->(c) OR (a)-[:has*1..2]->(c)
WITH a, b, c, z
OPTIONAL MATCH (c)-[cd:knows]->(d:d)
RETURN z.name, a.name, b.name, c.name, d.name, cd.score;
从那里可以对其进行优化。
From there you could optimize it.
编辑
添加后，我认为唯一的实现方法是使用两个查询的 UNION
MATCH (a:A)-[mr:has]->(b:B)
OPTIONAL MATCH (b)-->(c:C)
OPTIONAL MATCH (c)-[cd]->(d:d)
RETURN a.name, mr.order, b.name, c.name, d.name, cd.score
UNION
MATCH (a:A)-[mr:has]->(c:C)
OPTIONAL MATCH (c:C)-[cd]->(d:d)
OPTIONAL MATCH (b:B)-->(c:C)
RETURN a.name, mr.order, b.name, c.name, d.name, cd.score
请注意，在第二个查询中，可选匹配（b：B）->（c：C）仅用于生成 UNION 是可能的，因为Neo4j不允许只返回一个空字符串而不是 b.name 。好消息是 UNION 将删除重复项，因此会得到预期的结果。
Note that in the second query after the UNION, the OPTIONAL MATCH (b:B)-->(c:C) is only there to make the UNION possible, since Neo4j doesn't allow to just return an empty string instead of b.name. The good news is that UNION will remove duplicates, therefore resulting in the expected result.
a.name mr.order b.name c.name d.name cd.score
A1 1 B3
A1 2 B1 C5
A1 2 B1 C1 d1 1
A1 2 B1 C1 d2 4
A1 2 B1 C1 d3 3
A1 2 B1 C2
A2 1 B4
A2 2 B2 C3
A2 C4 d4 8
A2 C4 d5 6
A2 C4 d6 9
Showing 1 to 11 of 11 entries
查询花费了9毫秒并返回了11行。
Query took 9 ms and returned 11 rows.
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