2014-05-07 21 views
0

我有以下java代码片段,演示了这个问题。我收到的错误也包含在下面。 它正确地拉正确的设置,但我无法打印。 我正在使用org.neo4j.graphdb.Node节点。这是错误的课程吗? 如果不是,我如何从ExecutionEngine获取结果movieid,avgrating和movie_title?如何从ExecutionResult中分别提取结果?

Java代码的

GraphDatabaseService db = new GraphDatabaseFactory().newEmbeddedDatabase(DB_PATH); 
ExecutionEngine engine = new ExecutionEngine(db); 

String cypherQuery =  "MATCH (n)-[r:RATES]->(m) \n" 
          + "RETURN m.movieid as movieid, avg(toFloat(r.rating)) as avgrating, m.title as movie_title \n" 
          + "ORDER BY avgrating DESC \n" 
          + "LIMIT 20;"; 

ExecutionResult result = engine.execute(cypher); 

for (Map<String, Object> row : result) { 
    Node x = (Node) row.get("movie_title"); 
     for (String prop : x.getPropertyKeys()) { 
      System.out.println(prop + ": " + x.getProperty(prop)); 
     } 
    } 

错误

Exception in thread "main" java.lang.ClassCastException: java.lang.String cannot be cast to org.neo4j.graphdb.Node 
    at beans.RecommendationBean.queryMoviesWithCypher(RecommendationBean.java:194) 
    at beans.RecommendationBean.main(RecommendationBean.java:56) 

回答

2
Node x = (Node) row.get("movie_title"); 

...看起来是罪魁祸首。

在您的Cypher语句中,您将m.title返回为movie_title,即您返回节点属性(在本例中为字符串),并且在违规行中尝试将该字符串结果作为节点。

如果您希望Cypher返回可迭代的一系列节点,请尝试返回m(整个节点),而不是仅返回单个属性和聚合,例如,

"...RETURN m AS movie;" 
... 
Node x = (Node) row.get("movie"); 

等等