如何基于Neo4J节点标签返回集合

Question

我不确定如何编写一个neo4j查询，该查询在最短路径中找到包括标签在内的所有节点，然后按标签将其分组为输出。

编辑：ShortestPath包括标签 -> files-> subdirs-> sourcedir之间的链接，我想要所有这些链接。

此neo4j语法在某个地方是错误的，但这基本上是我试图返回我的Java Spring Code

MATCH (a:Tag),(b:File),(c:SourceDir),
path = shortestPath((a)-[*]->(c))
WHERE a.name=~('(?i).*'+"READ"+'.*')
FOREACH (n IN NODES(path) |
WHERE label(n) = "SubDir" as dir |
WHERE label(n) = "SourceDir" as src |
WHERE label(n) = "File" as file |
WHERE label(n) = "Tag" as tag)
RETURN COLLECT(dir) as dirs, src as source, COLLECT(file) as file, COLLECT(tag) as tags

Java语法部分看起来像这样，由Neo4J查询和结果组成

 @Query("MATCH (a:Tag),(b:File),(c:SourceDir), "
             + "path = shortestPath((a)-[*]->(c)) "
             + "WHERE a.name=~('(?i).*'+{search}+'.*') "
             + "FOREACH (n IN NODES(path) | "
             + "WHERE label(n) = 'SubDir' as dir | "
             + "WHERE label(n) = 'SourceDir' as src | "
             + "WHERE label(n) = 'File' as file | "
             + "WHERE label(n) = 'Tag' as tag) "
             + "RETURN COLLECT(dir) as dirs, src as source, COLLECT(file) as files, COLLECT(tag) as tags)")
  public List<ResultData> findByTag(@Param("search") String search);
@QueryResult
public class ResultData
{
   Set<File> files;
   Set<Directory> dirs;
   Source source;
   Set<Tag> tags;
}

有什么适当的neo4j查询给我预期的结果？

Answer 1

这可能对您有用：

MATCH (a:Tag)
WHERE a.name=~('(?i).*'+"READ"+'.*')
MATCH p = shortestPath((a)-[*]->(c:SourceDir))
RETURN
  [n IN NODES(p) WHERE 'SubDir'    IN LABELS(n)] AS dirs,
  [n IN NODES(p) WHERE 'SourceDir' IN LABELS(n)] AS srcs,
  [n IN NODES(p) WHERE 'File'      IN LABELS(n)] AS files,
  [n IN NODES(p) WHERE 'Tag'       IN LABELS(n)] AS tags;

注意：一个节点可以具有多个标签，因此上述查询将检查每个标签的每个节点。