我想知道如何在Java控制台中返回节点名而不是节点ID。
控制台中显示以下输出:
所需的输出应如下所示:
只是没有所有信息,只有节点名称(等于Airportnames)。
我的Java代码如下所示:
package com.routesNeo4j;
import org.neo4j.driver.v1.*;
import java.util.ArrayList;
import java.util.List;
/**
* Created by e on 11.06.17.
*/
public class Neo4JRouting implements AutoCloseable, Neo4J_Connector {
static Driver driver;
public Neo4JRouting(String startAirport, String destinationAirport, StatementResult shortestPath) {
driver = GraphDatabase.driver("bolt://ec2-13-58-101-13.us-east-2.compute.amazonaws.com:7687",
AuthTokens.basic("neo4j", "Einloggen_123"));
try(Session session = driver.session()) {
shortestPath = session.run("MATCH (a:" + startAirport.toLowerCase() + "), (b:" + destinationAirport.toLowerCase()
+ "), p = allShortestPaths((a)-[r*1..4]-(b)) UNWIND rels(p) AS rel RETURN nodes(p), sum(rel.weight) " +
"AS weight ORDER BY sum(rel.weight)");
List<Record> storeList = storeList(shortestPath);
while (shortestPath.hasNext()) {
System.out.println(shortestPath.next().toString());
}
System.out.println(storeList);
} catch (Exception e) {
e.printStackTrace();
}
}
public List<Record> storeList(StatementResult statementResult) {
List<Record> list = new ArrayList<>();
while (statementResult.hasNext()) {
list.add(statementResult.next());
}
return list;
}
@Override
public Driver runDriver(String user, AuthToken basicAuthToken) throws Exception {
return null;
}
@Override
public void close() throws Exception {
}
}
我期待着你的回答。非常感谢!
答案 0 :(得分:0)
您返回的每一行都包含节点列表和权重。这就是您在查询中提出的问题,以及您获得的内容。所以你必须&#34;解包&#34;这会产生你想要的格式。
一些代码片段来显示我的意思:
StatementResult vResult = vSession.run(aCypher);
while (vResult.hasNext()) {
Record vRecord = vResult.next();
vMutator.pushNode("row");
for (Pair <String, Value> vListEntry : vRecord.fields()) {
process_listentry(vSession, vMutator, vListEntry.key(), vListEntry.value());
}
vMutator.popNode(); // row
}
然后在process_listentry中:
private void process_listentry(Session vSession, IHDSMutator vMutator, String vKey, Value vValue) {
...
else if (vValue.type().equals(vSession.typeSystem().NODE())){
vMutator.pushNode(vKey);
vMutator.addNode("_id", Long.toString(vValue.asNode().id()));
for (String lLabel : vValue.asNode().labels()) {
vMutator.addNode("_label", lLabel);
}
for (String lKey : vValue.asNode().keys()) {
Value lValue = vValue.asNode().get(lKey);
process_listentry(vSession, vMutator, lKey, lValue);
}
vMutator.popNode();
}
... 但它确实取决于你在查询中提出的要求,因此必须解压...
希望这有帮助, 汤姆