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  1. PO
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    <p>This is the query that should return the collection of the nodes that have the relationship [:CONTAINS] to the nodes "midNode" on the path with the specified starting and ending point. It groups the desired "searchedNode" by the path. </p> <pre><code>START startNode=node(3), endNode=node(5) Match path=startNode-[:Connect*]-&gt;midNode-[:Connect*]-&gt;endNode, searchedNode-[:CONTAINS]-&gt;midNode Return distinct path, collect(searchedNode) as searchedNodes </code></pre> <p>Update: It is to be noted that the identifier 'midNode' represents every node on the path between the two specified starting and ending nodes because of the variable length relationships [:Connect*] that connects it to the starting and ending points. Concretely, it represents the node('m1') one step away from the starting node as in this case startNode-[:Connect]->m1-[:Connect]->...->endNode; and 'm2' that is two steps away from the starting node as in this case startNode-[:Connect]-()-[:Connect]->m2-[:Connect]->...->endNode. As a matter of fact, the matched paths would be a collection of paths each of which corresponds to one node between the startNode and the endNode. Since we only want one path for all of the middle nodes on the path, the DISTINCT is used to remove the duplicate paths, and the searchedNode would be grouped by the key that is the path, so the collection of the searchedNode would be the collection of all the nodes that are related to all the nodes on the path between the startNode and the endNode. </p>
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    1. POFind node with relation to all nodes in collection
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    1. USLisa Li
    1. USLisa Li
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    1. COThank You for your reply but this means that if one node have relationship with midNode [:CONTAINS] then it will be returned. The next midNode doesn't need to have such relation with this searchedNode. I want to find the nodes that have relations to all those nodes on the path, not to just one midNode.
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      1. USMarcin Kapusta
    2. CONo, the identifier 'midNode' represents every node on the path not just single one. I have updated the answer to make it clearer. Give it a try and see if it gets what you want.
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      1. USLisa Li
 

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