Question

我想根据DB中的列值创建节点和关系。 Node1值将来自列＆＃34; App-Name＆＃34;，Node2-＆gt; ＆＃34; Corresponding_App＆＃34; ＆安培;关系 - ＆GT;＆＃34; INTERFACE_NAME＆＃34; 基于列值＆＃34; Interface_Type_Name＆＃34; - ＆gt;传入/传出箭头将被绘制到特定节点。

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);     
    }

对于每一行创建新节点。如何在上面的代码中创建唯一的节点？

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。这样的事情（警告，代码没有测试，但这应该给你的想法）。

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。但至少应该可以使用几千个。如果您的数字非常庞大，请使用LRU缓存，然后确保您按照名称处理结果。

在java类中创建Neo4j Unique节点

1 个答案: