使用neo4j进行规则引擎的开发-cypher语句

1.申请人之前有多少个逾期的进件

match (p:Person)-[h:HAS_APPLICATION]->(a:APPLICATION) where a.status='OVER_DUE' and p.person_id='123456' return count(a)

2.申请人的一度关系中有多少个触碰黑名单

match (p:Person)-[]-(p1:Person)-[h:HAS_PHONE]-(b:Black) where p.Person_id='123456' return count(b)

3.申请人的二度关系中有多少个触碰黑名单

match (p:Person)-[]-(p1:Person)-[]-(p2:Person)-[h:HAS_PHPNE]-(b:Black) where p.Person_id='123456' return count(b)