2

我有以下 Cypher 查询。它返回一个球员列表和每个球员参加的所有联赛的列表。现在对于每个返回的玩家，我想创建Person NodeEntity而不是使用NodeProxy. 想知道这样做的有效方法是什么。

String q = "START t=node({teamId}) MATCH player-[:PLAYED_WITH_TEAM]->t-[:CONTESTED_IN]->league WITH player AS player, league.startDate AS startDate, league.name AS leagueName ORDER BY startDate RETURN player,  collect(leagueName) AS leagueNames";

Map<String, Object> params = Maps.newHashMap();
params.put("teamId", selectedTeam);

Result<Map<String, Object>> result = template.query(q, params);

final List<Player> players = new ArrayList<Player>();

result.handle(new Handler<Map<String, Object>>()
{
    @Override
    public void handle(Map<String, Object> value)
    {       
        players.add((Player) value.get("player"));
    }
});

例外

SEVERE: Servlet.service() for servlet [appServlet] in context with path [/avl] threw exception [Request processing failed; nested exception is java.lang.ClassCastException: org.neo4j.kernel.impl.core.NodeProxy cannot be cast to com.aravind.avl.domain.Player] with root cause
java.lang.ClassCastException: org.neo4j.kernel.impl.core.NodeProxy cannot be cast to com.aravind.avl.domain.Player
    at com.aravind.avl.controller.RegistrationController$1.handle(RegistrationController.java:103)
4

1 回答 1

3

您应该使用Neo4jOperations接口中的convert方法 将返回的对象转换为其正确的类；这是一个例子：

neo4jOperations.convert(value.get("player"), Player.class);

Neo4jOperations对象由 Spring Data Neo4j 基础设施使用@Autowired注解注入。

于 2013-06-13T12:19:03.720 回答