Question

我正在尝试查询以下内容：

列出在其注册的课程中至少有一个A的学生的姓名。（不要将A-视为A。）

我使用以下查询：

    MATCH (studentPerson)<-[:S2P]-(:Student)-[:Taking]->(:Offering)-
    [:Covers]->(studentCourse)

    WHERE studentCourse.Grade = "A"
    RETURN studentPerson.Name as student

但每次我运行它，我都会继续＃34;（没有变化，没有记录）＆＃34;。我也尝试使用EXISTS(studentCourse.Grade = "A")无济于事。

当记录添加到数据库时，它们是从如下文件中添加的：

    match (a:Student), (b:Offering) where a.ID = 419180204 and b.ID = 15 
    create (a) - [r:Taking{EnrollmentID:'Enrollment306',Grade: 'A-'}]-> (b) 
    WITH count(*) as dummy

    match (a:Student), (b:Offering) where a.ID = 449976666 and b.ID = 15 
    create (a) - [r:Taking{EnrollmentID:'Enrollment307',Grade: 'C+'}]-> (b) 
    WITH count(*) as dummy

    match (a:Student), (b:Offering) where a.ID = 477453864 and b.ID = 15 
    create (a) - [r:Taking{EnrollmentID:'Enrollment308',Grade: 'A'}]-> (b) 
    WITH count(*) as dummy

    match (a:Student), (b:Offering) where a.ID = 495490053 and b.ID = 15 
    create (a) - [r:Taking{EnrollmentID:'Enrollment309',Grade: 'A-'}]-> (b)

是否可能由于实体被添加到关系中（＆＃34; r：Take＆＃34;），“提供”节点甚至没有成绩？我是neo4j的全新品牌并且尽力而为，但到目前为止这件事让我很困惑。

感谢任何帮助。

Answer 1

问题是Grade是:Taking关系的属性。因此，在您的查询中，将变量添加到该关系类型，例如(:Student)-[t:Taking]->(:Offering)并使用t.Grade = 'A'进行过滤：

MATCH (studentPerson)<-[:S2P]-(:Student)-[t:Taking]->(:Offering)-
  [:Covers]->(studentCourse)
WHERE t.Grade = `A`
RETURN studentPerson.Name as student

基于来自另一个节点

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