基于性别关系不同的Neo4j密码查询
在家谱学中，我们用DNA来寻找匹配物。Y-DNA发现父系匹配。执行以下操作的neo4j查询(其中RN是个人的唯一标识符)：
MATCH (n{RN:1}) match p=n-[r:father*..22]->m return m.RN as RN,m.fullname as FullName,m.sex as Sex,m.bd as BD,m.dd as DD,length(p) as generation,case when left(m.bd,4)>'1930' and rtrim(m.dd)='' then 'Y' else 'N' end as mtDNA_Candidate, reduce(srt2 ='|', q IN nodes(p)| srt2 + q.RN + '|') AS PathOrder order by generation desc,PathOrder desc
或者我们用线粒体DNA来匹配母系：
`MATCH (n{RN:1}) match p=n-[r:mother*..22]->m return m.RN as RN,m.fullname as FullName,m.sex as Sex,m.bd as BD,m.dd as DD,length(p) as generation,case when left(m.bd,4)>'1930' and rtrim(m.dd)='' then 'Y' else 'N' end as mtDNA_Candidate, reduce(srt2 ='|', q IN nodes(p)| srt2 + q.RN + '|') AS PathOrder order by generation desc,PathOrder desc`我的问题与X染色体DNA有关.父亲只给他的女儿一个X染色体，母亲给她所有的孩子一个染色体。因此，我需要一个密码查询，得到所有母亲的，但只有父亲的，当有一个女儿在最近的时间世代。如果后世有个儿子，我就把父亲排除在外。我在节点中有一个属性“性别”，值为M或F。出生日期并不总是已知的，因此不能用来确定方向性。
我试过了，但有个错误：
`MATCH (n{RN:1}) match p=n-[r:mother*..22|father*..1]->m return m.RN as RN,m.fullname as FullName,m.sex as Sex,m.bd as BD,m.dd as DD,length(p) as generation,case when left(m.bd,4)>'1930' and rtrim(m.dd)='' then 'Y' else 'N' end as mtDNA_Candidate, reduce(srt2 ='|', q IN nodes(p)| srt2 + q.RN + '|') AS PathOrder order by generation desc,PathOrder desc`回答 1
Stack Overflow用户
发布于 2015-12-14 21:47:16
已更新
[r:mother*..22|father*..1]语法是非法的。Cypher查询中的关系最多只能有一个变量长度规范，并且它必须在关系类型之后。(旁白:还请注意，[:father*..1]与[:father]是相同的)。
这个似乎在逻辑上等价的查询对您有用吗？
MATCH pf=(n { RN:1 })-[:father]->()
MATCH pm=n-[:mother*..22]->()
WITH [pf] + COLLECT(pm) AS paths
UNWIND paths AS p
WITH LENGTH(p) AS generation, NODES(p) AS ancestors
WITH generation, ancestors, LAST(ancestors) AS m
RETURN m.RN AS RN, m.fullname AS FullName, m.sex AS Sex, m.bd AS BD, m.dd AS DD, generation,
CASE WHEN left(m.bd,4)>'1930' AND rtrim(m.dd)='' THEN 'Y' ELSE 'N' END AS mtDNA_Candidate,
reduce(srt2 ='|', q IN ancestors | srt2 + q.RN + '|' ) AS PathOrder
ORDER BY generation DESC, PathOrder DESC;https://stackoverflow.com/questions/34276166
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