Neo4j-Cypher: Remove a node from a path and maintain the link to all nodes of the path










1















I want to remove one node from a path, without jeopardizing the original path nodes.



This is my test db:



Test DB



I want to remove node (2) from the path, but I want nodes 1, 3, 4, and 5 to stay linked in the path.



Is there a way to do it in one query? So far I have the following:



MATCH p = (:Connect)-[:to*]-(:Connect)
WITH nodes(p) AS connectNodes
UNWIND connectNodes AS connectNode
WITH distinct connectNode
WHERE connectNode.connectID = 2
DETACH DELETE (connectNode)


This will remove node 2 and unlinkes the path



Unlinked graph



How can maintain the link between the nodes of the original path without node 2?



EDIT:



I solved it by modifying the accepted answer's response



//Make sure node (n) belongs to the path
MATCH (n:Connect cID:2)-[:to*]-(:Connect cID:5)
//get attached nodes, if any, ignoring directions
OPTIONAL MATCH (oa:connect)-[:to]-(n)-[:to]-(ob:connect)
//make sure nothing is duplicated
WHERE oa.cID <> ob.cID
//Use FOREACH to mimic if/else. Only merge oa to ob if they exist. Query fails without it
FOREACH (_ IN case when oa IS NOT NULL then [true] else end |
MERGE (oa)-[:to created: 1542103211]-(ob)
)
//Get n, and get all connected nodes to it, and delete the relationship(s)
WITH n
OPTIONAL MATCH (n)-[r:to]-(:Connect) DELETE r









share|improve this question
























  • What should your preferred solution look like? Scenario I: 5->4->1 and 3->1, Scenario II: 5->4<-1 and 3->1 Scenario III: 5->4<-3 and 3->1 or Scenario IV: 5->4->3 and 3->1?

    – ThirstForKnowledge
    Nov 12 '18 at 14:41











  • @ThirstForKnowledge direction doesn't matter, as long as I can get all nodes of the path.

    – Tarek Deeb
    Nov 13 '18 at 7:33















1















I want to remove one node from a path, without jeopardizing the original path nodes.



This is my test db:



Test DB



I want to remove node (2) from the path, but I want nodes 1, 3, 4, and 5 to stay linked in the path.



Is there a way to do it in one query? So far I have the following:



MATCH p = (:Connect)-[:to*]-(:Connect)
WITH nodes(p) AS connectNodes
UNWIND connectNodes AS connectNode
WITH distinct connectNode
WHERE connectNode.connectID = 2
DETACH DELETE (connectNode)


This will remove node 2 and unlinkes the path



Unlinked graph



How can maintain the link between the nodes of the original path without node 2?



EDIT:



I solved it by modifying the accepted answer's response



//Make sure node (n) belongs to the path
MATCH (n:Connect cID:2)-[:to*]-(:Connect cID:5)
//get attached nodes, if any, ignoring directions
OPTIONAL MATCH (oa:connect)-[:to]-(n)-[:to]-(ob:connect)
//make sure nothing is duplicated
WHERE oa.cID <> ob.cID
//Use FOREACH to mimic if/else. Only merge oa to ob if they exist. Query fails without it
FOREACH (_ IN case when oa IS NOT NULL then [true] else end |
MERGE (oa)-[:to created: 1542103211]-(ob)
)
//Get n, and get all connected nodes to it, and delete the relationship(s)
WITH n
OPTIONAL MATCH (n)-[r:to]-(:Connect) DELETE r









share|improve this question
























  • What should your preferred solution look like? Scenario I: 5->4->1 and 3->1, Scenario II: 5->4<-1 and 3->1 Scenario III: 5->4<-3 and 3->1 or Scenario IV: 5->4->3 and 3->1?

    – ThirstForKnowledge
    Nov 12 '18 at 14:41











  • @ThirstForKnowledge direction doesn't matter, as long as I can get all nodes of the path.

    – Tarek Deeb
    Nov 13 '18 at 7:33













1












1








1








I want to remove one node from a path, without jeopardizing the original path nodes.



This is my test db:



Test DB



I want to remove node (2) from the path, but I want nodes 1, 3, 4, and 5 to stay linked in the path.



Is there a way to do it in one query? So far I have the following:



MATCH p = (:Connect)-[:to*]-(:Connect)
WITH nodes(p) AS connectNodes
UNWIND connectNodes AS connectNode
WITH distinct connectNode
WHERE connectNode.connectID = 2
DETACH DELETE (connectNode)


This will remove node 2 and unlinkes the path



Unlinked graph



How can maintain the link between the nodes of the original path without node 2?



EDIT:



I solved it by modifying the accepted answer's response



//Make sure node (n) belongs to the path
MATCH (n:Connect cID:2)-[:to*]-(:Connect cID:5)
//get attached nodes, if any, ignoring directions
OPTIONAL MATCH (oa:connect)-[:to]-(n)-[:to]-(ob:connect)
//make sure nothing is duplicated
WHERE oa.cID <> ob.cID
//Use FOREACH to mimic if/else. Only merge oa to ob if they exist. Query fails without it
FOREACH (_ IN case when oa IS NOT NULL then [true] else end |
MERGE (oa)-[:to created: 1542103211]-(ob)
)
//Get n, and get all connected nodes to it, and delete the relationship(s)
WITH n
OPTIONAL MATCH (n)-[r:to]-(:Connect) DELETE r









share|improve this question
















I want to remove one node from a path, without jeopardizing the original path nodes.



This is my test db:



Test DB



I want to remove node (2) from the path, but I want nodes 1, 3, 4, and 5 to stay linked in the path.



Is there a way to do it in one query? So far I have the following:



MATCH p = (:Connect)-[:to*]-(:Connect)
WITH nodes(p) AS connectNodes
UNWIND connectNodes AS connectNode
WITH distinct connectNode
WHERE connectNode.connectID = 2
DETACH DELETE (connectNode)


This will remove node 2 and unlinkes the path



Unlinked graph



How can maintain the link between the nodes of the original path without node 2?



EDIT:



I solved it by modifying the accepted answer's response



//Make sure node (n) belongs to the path
MATCH (n:Connect cID:2)-[:to*]-(:Connect cID:5)
//get attached nodes, if any, ignoring directions
OPTIONAL MATCH (oa:connect)-[:to]-(n)-[:to]-(ob:connect)
//make sure nothing is duplicated
WHERE oa.cID <> ob.cID
//Use FOREACH to mimic if/else. Only merge oa to ob if they exist. Query fails without it
FOREACH (_ IN case when oa IS NOT NULL then [true] else end |
MERGE (oa)-[:to created: 1542103211]-(ob)
)
//Get n, and get all connected nodes to it, and delete the relationship(s)
WITH n
OPTIONAL MATCH (n)-[r:to]-(:Connect) DELETE r






neo4j path cypher graph-databases






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 13 '18 at 17:42









Tezra

5,05121042




5,05121042










asked Nov 12 '18 at 12:50









Tarek DeebTarek Deeb

1328




1328












  • What should your preferred solution look like? Scenario I: 5->4->1 and 3->1, Scenario II: 5->4<-1 and 3->1 Scenario III: 5->4<-3 and 3->1 or Scenario IV: 5->4->3 and 3->1?

    – ThirstForKnowledge
    Nov 12 '18 at 14:41











  • @ThirstForKnowledge direction doesn't matter, as long as I can get all nodes of the path.

    – Tarek Deeb
    Nov 13 '18 at 7:33

















  • What should your preferred solution look like? Scenario I: 5->4->1 and 3->1, Scenario II: 5->4<-1 and 3->1 Scenario III: 5->4<-3 and 3->1 or Scenario IV: 5->4->3 and 3->1?

    – ThirstForKnowledge
    Nov 12 '18 at 14:41











  • @ThirstForKnowledge direction doesn't matter, as long as I can get all nodes of the path.

    – Tarek Deeb
    Nov 13 '18 at 7:33
















What should your preferred solution look like? Scenario I: 5->4->1 and 3->1, Scenario II: 5->4<-1 and 3->1 Scenario III: 5->4<-3 and 3->1 or Scenario IV: 5->4->3 and 3->1?

– ThirstForKnowledge
Nov 12 '18 at 14:41





What should your preferred solution look like? Scenario I: 5->4->1 and 3->1, Scenario II: 5->4<-1 and 3->1 Scenario III: 5->4<-3 and 3->1 or Scenario IV: 5->4->3 and 3->1?

– ThirstForKnowledge
Nov 12 '18 at 14:41













@ThirstForKnowledge direction doesn't matter, as long as I can get all nodes of the path.

– Tarek Deeb
Nov 13 '18 at 7:33





@ThirstForKnowledge direction doesn't matter, as long as I can get all nodes of the path.

– Tarek Deeb
Nov 13 '18 at 7:33












2 Answers
2






active

oldest

votes


















5














This is the simplest way to do it would be to match the paths that would be broken, and then create the new links in the same cypher that deletes the node.



// Match the target node for deletion
MATCH (nid:2)
// Match broken paths, if any
OPTIONAL MATCH (a)-[ra]->(n)-[rb]->(b)
// Create new link to replace destroyed ones
CREATE (a)-[r:to]->(b)
// Copy properties over, if any
SET r+=ra, r+=rb
// Remove target node
DETACH DELETE n
// If you want to keep the node and just disconnect it, replace last line with
// OPTIONAL MATCH (n)-[r:to]-() DELETE r


There are APOC functions you can use to create a relation with a dynamic type if you want to copy the type from one of the removed relationships.






share|improve this answer

























  • after modifying this query, it worked (please see the edited OP). But it fails on edge nodes. It only works on nodes that are connected to more than 1 node. In my original graph, it fails on 1, 3, and 5. How can I make it dynamic enough to work on any node?

    – Tarek Deeb
    Nov 13 '18 at 16:31











  • @TarekDeeb detach delete was suppose to clean up the edges, if you want to keep the node, and just remove it from all graphs, you can replace detach delete with OPTIONAL MATCH (n)-[r:to]-() DELETE r

    – Tezra
    Nov 13 '18 at 16:51











  • This will just remove the relationship. I need to maintain the link to the original path. So if I have (1)-(2)-(3) and remove (2), your solution works, and I'll get (1)-(3). But If I want to remove (3), I want to still be left with (1)-(2). The removed node will not be deleted, but only not belonging to the original path.

    – Tarek Deeb
    Nov 13 '18 at 16:55











  • @TarekDeeb With the replace I gave you, (1)-(2)-(3) target 3 will result in (1)-(2) and (3), I don't understand what else you want to happen (your description sounds like that is what you wanted)

    – Tezra
    Nov 13 '18 at 17:02











  • I'm sorry I just saw your edit in your response. I modified the query to work on both cases and all is well. Check the modified query in my OP. Many thanks for your help!

    – Tarek Deeb
    Nov 13 '18 at 17:19


















1














An additional opportunity would be the removal of node "2" and the usage of the APOC procedure "Redirect relationship to". You can find a detailed description and illustrating images in the procedure documentation.






share|improve this answer






















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    This is the simplest way to do it would be to match the paths that would be broken, and then create the new links in the same cypher that deletes the node.



    // Match the target node for deletion
    MATCH (nid:2)
    // Match broken paths, if any
    OPTIONAL MATCH (a)-[ra]->(n)-[rb]->(b)
    // Create new link to replace destroyed ones
    CREATE (a)-[r:to]->(b)
    // Copy properties over, if any
    SET r+=ra, r+=rb
    // Remove target node
    DETACH DELETE n
    // If you want to keep the node and just disconnect it, replace last line with
    // OPTIONAL MATCH (n)-[r:to]-() DELETE r


    There are APOC functions you can use to create a relation with a dynamic type if you want to copy the type from one of the removed relationships.






    share|improve this answer

























    • after modifying this query, it worked (please see the edited OP). But it fails on edge nodes. It only works on nodes that are connected to more than 1 node. In my original graph, it fails on 1, 3, and 5. How can I make it dynamic enough to work on any node?

      – Tarek Deeb
      Nov 13 '18 at 16:31











    • @TarekDeeb detach delete was suppose to clean up the edges, if you want to keep the node, and just remove it from all graphs, you can replace detach delete with OPTIONAL MATCH (n)-[r:to]-() DELETE r

      – Tezra
      Nov 13 '18 at 16:51











    • This will just remove the relationship. I need to maintain the link to the original path. So if I have (1)-(2)-(3) and remove (2), your solution works, and I'll get (1)-(3). But If I want to remove (3), I want to still be left with (1)-(2). The removed node will not be deleted, but only not belonging to the original path.

      – Tarek Deeb
      Nov 13 '18 at 16:55











    • @TarekDeeb With the replace I gave you, (1)-(2)-(3) target 3 will result in (1)-(2) and (3), I don't understand what else you want to happen (your description sounds like that is what you wanted)

      – Tezra
      Nov 13 '18 at 17:02











    • I'm sorry I just saw your edit in your response. I modified the query to work on both cases and all is well. Check the modified query in my OP. Many thanks for your help!

      – Tarek Deeb
      Nov 13 '18 at 17:19















    5














    This is the simplest way to do it would be to match the paths that would be broken, and then create the new links in the same cypher that deletes the node.



    // Match the target node for deletion
    MATCH (nid:2)
    // Match broken paths, if any
    OPTIONAL MATCH (a)-[ra]->(n)-[rb]->(b)
    // Create new link to replace destroyed ones
    CREATE (a)-[r:to]->(b)
    // Copy properties over, if any
    SET r+=ra, r+=rb
    // Remove target node
    DETACH DELETE n
    // If you want to keep the node and just disconnect it, replace last line with
    // OPTIONAL MATCH (n)-[r:to]-() DELETE r


    There are APOC functions you can use to create a relation with a dynamic type if you want to copy the type from one of the removed relationships.






    share|improve this answer

























    • after modifying this query, it worked (please see the edited OP). But it fails on edge nodes. It only works on nodes that are connected to more than 1 node. In my original graph, it fails on 1, 3, and 5. How can I make it dynamic enough to work on any node?

      – Tarek Deeb
      Nov 13 '18 at 16:31











    • @TarekDeeb detach delete was suppose to clean up the edges, if you want to keep the node, and just remove it from all graphs, you can replace detach delete with OPTIONAL MATCH (n)-[r:to]-() DELETE r

      – Tezra
      Nov 13 '18 at 16:51











    • This will just remove the relationship. I need to maintain the link to the original path. So if I have (1)-(2)-(3) and remove (2), your solution works, and I'll get (1)-(3). But If I want to remove (3), I want to still be left with (1)-(2). The removed node will not be deleted, but only not belonging to the original path.

      – Tarek Deeb
      Nov 13 '18 at 16:55











    • @TarekDeeb With the replace I gave you, (1)-(2)-(3) target 3 will result in (1)-(2) and (3), I don't understand what else you want to happen (your description sounds like that is what you wanted)

      – Tezra
      Nov 13 '18 at 17:02











    • I'm sorry I just saw your edit in your response. I modified the query to work on both cases and all is well. Check the modified query in my OP. Many thanks for your help!

      – Tarek Deeb
      Nov 13 '18 at 17:19













    5












    5








    5







    This is the simplest way to do it would be to match the paths that would be broken, and then create the new links in the same cypher that deletes the node.



    // Match the target node for deletion
    MATCH (nid:2)
    // Match broken paths, if any
    OPTIONAL MATCH (a)-[ra]->(n)-[rb]->(b)
    // Create new link to replace destroyed ones
    CREATE (a)-[r:to]->(b)
    // Copy properties over, if any
    SET r+=ra, r+=rb
    // Remove target node
    DETACH DELETE n
    // If you want to keep the node and just disconnect it, replace last line with
    // OPTIONAL MATCH (n)-[r:to]-() DELETE r


    There are APOC functions you can use to create a relation with a dynamic type if you want to copy the type from one of the removed relationships.






    share|improve this answer















    This is the simplest way to do it would be to match the paths that would be broken, and then create the new links in the same cypher that deletes the node.



    // Match the target node for deletion
    MATCH (nid:2)
    // Match broken paths, if any
    OPTIONAL MATCH (a)-[ra]->(n)-[rb]->(b)
    // Create new link to replace destroyed ones
    CREATE (a)-[r:to]->(b)
    // Copy properties over, if any
    SET r+=ra, r+=rb
    // Remove target node
    DETACH DELETE n
    // If you want to keep the node and just disconnect it, replace last line with
    // OPTIONAL MATCH (n)-[r:to]-() DELETE r


    There are APOC functions you can use to create a relation with a dynamic type if you want to copy the type from one of the removed relationships.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 13 '18 at 17:04

























    answered Nov 12 '18 at 15:23









    TezraTezra

    5,05121042




    5,05121042












    • after modifying this query, it worked (please see the edited OP). But it fails on edge nodes. It only works on nodes that are connected to more than 1 node. In my original graph, it fails on 1, 3, and 5. How can I make it dynamic enough to work on any node?

      – Tarek Deeb
      Nov 13 '18 at 16:31











    • @TarekDeeb detach delete was suppose to clean up the edges, if you want to keep the node, and just remove it from all graphs, you can replace detach delete with OPTIONAL MATCH (n)-[r:to]-() DELETE r

      – Tezra
      Nov 13 '18 at 16:51











    • This will just remove the relationship. I need to maintain the link to the original path. So if I have (1)-(2)-(3) and remove (2), your solution works, and I'll get (1)-(3). But If I want to remove (3), I want to still be left with (1)-(2). The removed node will not be deleted, but only not belonging to the original path.

      – Tarek Deeb
      Nov 13 '18 at 16:55











    • @TarekDeeb With the replace I gave you, (1)-(2)-(3) target 3 will result in (1)-(2) and (3), I don't understand what else you want to happen (your description sounds like that is what you wanted)

      – Tezra
      Nov 13 '18 at 17:02











    • I'm sorry I just saw your edit in your response. I modified the query to work on both cases and all is well. Check the modified query in my OP. Many thanks for your help!

      – Tarek Deeb
      Nov 13 '18 at 17:19

















    • after modifying this query, it worked (please see the edited OP). But it fails on edge nodes. It only works on nodes that are connected to more than 1 node. In my original graph, it fails on 1, 3, and 5. How can I make it dynamic enough to work on any node?

      – Tarek Deeb
      Nov 13 '18 at 16:31











    • @TarekDeeb detach delete was suppose to clean up the edges, if you want to keep the node, and just remove it from all graphs, you can replace detach delete with OPTIONAL MATCH (n)-[r:to]-() DELETE r

      – Tezra
      Nov 13 '18 at 16:51











    • This will just remove the relationship. I need to maintain the link to the original path. So if I have (1)-(2)-(3) and remove (2), your solution works, and I'll get (1)-(3). But If I want to remove (3), I want to still be left with (1)-(2). The removed node will not be deleted, but only not belonging to the original path.

      – Tarek Deeb
      Nov 13 '18 at 16:55











    • @TarekDeeb With the replace I gave you, (1)-(2)-(3) target 3 will result in (1)-(2) and (3), I don't understand what else you want to happen (your description sounds like that is what you wanted)

      – Tezra
      Nov 13 '18 at 17:02











    • I'm sorry I just saw your edit in your response. I modified the query to work on both cases and all is well. Check the modified query in my OP. Many thanks for your help!

      – Tarek Deeb
      Nov 13 '18 at 17:19
















    after modifying this query, it worked (please see the edited OP). But it fails on edge nodes. It only works on nodes that are connected to more than 1 node. In my original graph, it fails on 1, 3, and 5. How can I make it dynamic enough to work on any node?

    – Tarek Deeb
    Nov 13 '18 at 16:31





    after modifying this query, it worked (please see the edited OP). But it fails on edge nodes. It only works on nodes that are connected to more than 1 node. In my original graph, it fails on 1, 3, and 5. How can I make it dynamic enough to work on any node?

    – Tarek Deeb
    Nov 13 '18 at 16:31













    @TarekDeeb detach delete was suppose to clean up the edges, if you want to keep the node, and just remove it from all graphs, you can replace detach delete with OPTIONAL MATCH (n)-[r:to]-() DELETE r

    – Tezra
    Nov 13 '18 at 16:51





    @TarekDeeb detach delete was suppose to clean up the edges, if you want to keep the node, and just remove it from all graphs, you can replace detach delete with OPTIONAL MATCH (n)-[r:to]-() DELETE r

    – Tezra
    Nov 13 '18 at 16:51













    This will just remove the relationship. I need to maintain the link to the original path. So if I have (1)-(2)-(3) and remove (2), your solution works, and I'll get (1)-(3). But If I want to remove (3), I want to still be left with (1)-(2). The removed node will not be deleted, but only not belonging to the original path.

    – Tarek Deeb
    Nov 13 '18 at 16:55





    This will just remove the relationship. I need to maintain the link to the original path. So if I have (1)-(2)-(3) and remove (2), your solution works, and I'll get (1)-(3). But If I want to remove (3), I want to still be left with (1)-(2). The removed node will not be deleted, but only not belonging to the original path.

    – Tarek Deeb
    Nov 13 '18 at 16:55













    @TarekDeeb With the replace I gave you, (1)-(2)-(3) target 3 will result in (1)-(2) and (3), I don't understand what else you want to happen (your description sounds like that is what you wanted)

    – Tezra
    Nov 13 '18 at 17:02





    @TarekDeeb With the replace I gave you, (1)-(2)-(3) target 3 will result in (1)-(2) and (3), I don't understand what else you want to happen (your description sounds like that is what you wanted)

    – Tezra
    Nov 13 '18 at 17:02













    I'm sorry I just saw your edit in your response. I modified the query to work on both cases and all is well. Check the modified query in my OP. Many thanks for your help!

    – Tarek Deeb
    Nov 13 '18 at 17:19





    I'm sorry I just saw your edit in your response. I modified the query to work on both cases and all is well. Check the modified query in my OP. Many thanks for your help!

    – Tarek Deeb
    Nov 13 '18 at 17:19













    1














    An additional opportunity would be the removal of node "2" and the usage of the APOC procedure "Redirect relationship to". You can find a detailed description and illustrating images in the procedure documentation.






    share|improve this answer



























      1














      An additional opportunity would be the removal of node "2" and the usage of the APOC procedure "Redirect relationship to". You can find a detailed description and illustrating images in the procedure documentation.






      share|improve this answer

























        1












        1








        1







        An additional opportunity would be the removal of node "2" and the usage of the APOC procedure "Redirect relationship to". You can find a detailed description and illustrating images in the procedure documentation.






        share|improve this answer













        An additional opportunity would be the removal of node "2" and the usage of the APOC procedure "Redirect relationship to". You can find a detailed description and illustrating images in the procedure documentation.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 13 '18 at 21:11









        ThirstForKnowledgeThirstForKnowledge

        6551112




        6551112



























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