Neo4j密碼查詢基於性別的不同關係

Question

在系譜中，我們使用DNA來查找匹配。 Y-DNA發現父系匹配。這是否一個Neo4j的查詢（其中RN是一個人的唯一標識符）是：

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

或者我們使用的母系匹配的線粒體DNA：

`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` 

我的問題是相關到X染色體DNA。一位父親只給他的女兒一個X染色體，一位母親給所有的孩子一個。因此，我需要一個密碼查詢來獲取所有母親的信息，但只有在最近的一代中有一個女兒的時候才能得到父親的信息。如果後代有兒子，我就排除父親。我在用M和F的值的出生日期並不總是已知的節點屬性「性」，因此它不能被用來確定方向性

我想這一點，但得到一個錯誤：

`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` 

Answer 1

[增訂]

的[r:mother*..22|father*..1]語法是非法的。 Cypher查詢中的關係最多隻能有一個可變長度規範，並且在之後的關係類型必須爲。 （另外：請注意，[:father*..1]與[:father]相同）。

此查詢似乎在邏輯上等同於您嗎？

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;