I'm trying to do a query to fin all possible paths that correspond to the pattern "(Order) - [ORDERS] -> (Product) - [PART_OF] -> (Category)" and would like to get the whole path (ie all 3 nodes and 2 relationships as their appropriate classes).
The method i used below only let me have 1 column of data (number of orders: 2155). If I tried it once more (the 2nd for loop), the number of row i'd get is 0(number of products: 0). Is there a way to save all the results as nodes and relationships or do I have to query the command 5 times over? Please help!
String query = "MATCH (o:Order)-[:ORDERS]->(p:Product)-[:PART_OF]->(cate:Category) return o,p,cate";
try( Transaction tx = db.beginTx();
Result result = db.execute(query) ){
Iterator<Node> o_column = result.columnAs( "o" );
int i = 0;
for ( Node node : Iterators.asIterable( o_column ) )
{
i++;
}
System.out.println("number of orders: " + i);
i = 0;
Iterator<Node> p_column = result.columnAs( "p" );
for ( Node node : Iterators.asIterable( p_column ) )
{
i++;
}
System.out.println("number of products: " + i);
tx.success();
}
If you do this :
MATCH path=(o:Order)-[:ORDERS]->(p:Product)-[:PART_OF]->(cate:Category) return path
You can process path in your loop and unpack that. Takes a bit of exploring but all the information is in there.
Hope that helps.
Regards, Tom
我在下面的代码中找到了解决此问题的方法,在该代码中,我将使用id()
将返回值更改为节点ID,然后使用GraphDatabaseService.getNodeByID(long)
:
String query = "MATCH (o:Order)-[:ORDERS]->(p:Product)-[:PART_OF]->(cate:Category) return id(o), id(p), id(cate)";
int nodeID = Integer.parseInt(column.getValue().toString()); Node node = db.getNodeById(nodeID);
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