I have a sewer network represented as a connected pipelines in a GIS shapefile.

the network is very close to a tree graph that each node is only connected with one or two nodes.

I need to reconstruct the layout of this network on another program.

I first used vertices to points feature to specify the start and the end point of each line.

However, this doesn't help too much because I need to specify which node is connected to which node and the feature of the line that is connecting these two nodes like its diameter or its FID for example.

1 Answer 1


There are several ways to achieve this. You could build a coverage dataset, it's attribute table will store the logic of the network. You could download ArcHydro and this too will extract out the logical network.

For the record you should always mention which license version of ArcGIS you have as this dictates the solution.

  • @naaeemmohammad 10.2.2 is the version and ArcMap is the application that you are using. Your license level will be Basic, Standard or Advanced.
    – PolyGeo
    Jun 11, 2015 at 11:01
  • 3
    I was surprised to see a coverage being proposed here - do you think they still have a role to play? I noticed an answer at gis.stackexchange.com/a/1748 which seems to be describing the geodatabase equivalent.
    – PolyGeo
    Jun 11, 2015 at 11:07
  • Just old school I guess! As I understood he just needed a way of extracting the logical connectivity and coverages is just one way. A Geodatabase approach is just as sensible.
    – Hornbydd
    Jun 11, 2015 at 13:19
  • I am really new to gis so I am wondering how to use build coverage I read the help page but it just a description there is no steps on how to apply the coverage so can you please explain how to use the coverage Jun 11, 2015 at 22:28
  • 1
    For someone new to GIS I think you will be better to focus on Geodatabase Topology and Geometric Networks than Coverages which are from the old Esri architecture that began to be superseded from the mid-late 1990s.
    – PolyGeo
    Jun 12, 2015 at 1:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.