java - 在 java 类中创建 Neo4j Unique 节点

Question

我想根据数据库中的列值创建节点和关系。Node1 值将来自“App-Name”列，Node2 -> “Corresponding_App” & relationship->“Interface_Name”基于列值“Interface_Type_Name”->传入/传出箭头将绘制到特定节点。

节点应该是唯一的，如果已经创建，则不应创建重复的节点。

Transaction tx=graphDb.beginTx();       
    Connection conn = null;
    try{{       
    conn=ConnectionFactory.getConnection();
    Statement stat=conn.createStatement();
    String sql="select * from Fade where App_Int_Id < 22";
    ResultSet rs=stat.executeQuery(sql);        
    String n1 = "",n2="",rel="",type="";
    while(rs.next()){
        n1=rs.getString(1);              
        n2=rs.getString(7);
        rel=rs.getString(3);    
        type=rs.getString(4);           

    Node first=graphDb.createNode();
    first.setProperty("name", n1);
    first.addLabel(DynamicLabel.label(n1));

    Node sec=graphDb.createNode();
    sec.setProperty("name", n2);
    sec.addLabel(DynamicLabel.label(n2));       

    RelationshipType KNOWS = DynamicRelationshipType.withName(rel);

    if(type.equalsIgnoreCase("Incoming")){
        sec.createRelationshipTo(first, KNOWS);     
    }
    else if(type.equalsIgnoreCase("Outgoing")){
        first.createRelationshipTo(sec, KNOWS);     
    }

它为每一行创建新节点。所以如何在上面的代码中创建唯一节点？

Answer 1

如果您的节点在名称方面是唯一的，您可以在简单的 HashMap 中跟踪它们以重用节点而不是创建新节点。像这样的东西(警告，代码未经测试，但这应该给你想法)。

HashMap<String,Node> nodes = new HashMap<String,Node>();

Transaction tx=graphDb.beginTx();       
    Connection conn = null;
    try{{       
    conn=ConnectionFactory.getConnection();
    Statement stat=conn.createStatement();
    String sql="select * from Fade where App_Int_Id < 22";
    ResultSet rs=stat.executeQuery(sql);        
    String n1 = "",n2="",rel="",type="";
    while(rs.next()){
        n1=rs.getString(1);              
        n2=rs.getString(7);
        rel=rs.getString(3);    
        type=rs.getString(4);           

    Node first=nodes.get(n1);

    // Only create it if it didn't already exist.
    if(first == null) { 
       first = graphDb.createNode();
       first.setProperty("name", n1);
       first.addLabel(DynamicLabel.label(n1));

       // Now put it in the map to ensure it doesn't get created again.
       nodes.put(n1, first);
    } 

    Node sec=nodes.get(n2); 

    if(sec == null) { 
       sec = graphDb.createNode();
       sec.setProperty("name", n2);
       sec.addLabel(DynamicLabel.label(n2));       

       // Now put it in the map to ensure it doesn't get created again.
       nodes.put(n2, sec);
    } 

    RelationshipType KNOWS = DynamicRelationshipType.withName(rel);

    if(type.equalsIgnoreCase("Incoming")){
        sec.createRelationshipTo(first, KNOWS);     
    }
    else if(type.equalsIgnoreCase("Outgoing")){
        first.createRelationshipTo(sec, KNOWS);     
    }

最后 - 如果您有大量节点，请不要尝试这样做，因为您的 HashMap 会超出内存。但这至少对几千人来说应该是可行的。如果您的数字非常大，请使用 LRU 缓存，然后确保按名称对您正在处理的结果进行排序。