How to return all the nodes with a common property

[vasu_sethia]

How to return all the nodes and relationships with a common property without forming cartesian product.

[david_allen]

Abstract out the common property values into separate nodes, and do something like this:

MATCH (n:Stuff)
WITH n, distinct(n.color)
MERGE (c:Color { name: n.color })
MERGE (n)-[:COLOR]->(c)

Now you've got an enhanced graph where everything is linked to its color, no cartesian product, and you can navigate your graph by that commonality

[vasu_sethia]

Hey, thanks for the reply.Suppose we have one node labeled as Filetype with a property of lets say file_name and then I have another node labeled as FileSize which also has a property named as file_name and few other properties.Then I have a other node with label as FileFormat and common property as file_name, this node has a relationship with FileSize and FileSize has relationship with FileType.So how do I extract the files with based on the common property of files_name and their relationships with each other?

[david_allen]

What have you tried? Please post your code and describe what it does and why that isn't right.

[vasu_sethia]

MATCH (a:FileType),(b:FileName),(c:FileSize) where a.md5='84c82835a5d21bbcf75a61706d8ab549' and b.md5='84c82835a5d21bbcf75a61706d8ab549' and c.md5='84c82835a5d21bbcf75a61706d8ab549' return a,b,c
FileName has a relationship (x) with FileType label and FileSize has a relationship (y) with FileName label node.They all have a common property of md5 hash .So how do I return all the nodes with the common property.

[david_allen]

There are two ways to approach this. Match "1 at a time" and use WITH, or abstract the MD5 out into a separate node. The second approach is better if you need to find matching MD5s frequently, which I'm guessing you do.

Approach 1 (which is similar to what you're doing)

WITH "84c82835a5d21bbcf75a61706d8ab549" as val
MATCH (a:FileType { md5: val })
WITH a
MATCH (b:FileName { md5: val })
WITH a, b
MATCH (c:FileSize { md5: val })

RETURN a, b, c;

The second approach is what I already suggested, and I think it's probably better for you:

MATCH (anyNode)
WHERE anyNode.md5 = "84c82835a5d21bbcf75a61706d8ab549"
MERGE (hash:MD5 { value: "84c82835a5d21bbcf75a61706d8ab549" })
MERGE (anyNode)-[:HASH]->(hash);

If you do this second thing, then you can find all commonalities really easily:

MATCH (hash:MD5 { value: "Whatever you're looking for" })<-[:HASH]-(stuff)
RETURN stuff

[Joel]

First I need to say that I think a separate node as suggested is the best answer. A graph approach.

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but if these options didn't do what you need, I'd mention secondary labels

neo4j allows multiple labels, and I believe very few people use that capability. It will mangle your apoc.meta.graph output, but using multiple labels on nodes can be very useful in a few situations.

For example, one could add an additional label to all nodes with the md5 property (e.g. MD5), then put an index on that label/property combination, it would enable returning all three nodes with this cypher

match(a:MD5 {md5:"84c82835a5d21bbcf75a61706d8ab549})
return a

Caveats: Depending on your specific situation, there are likely better ways to model it (as proposed earlier). Technically this would pull back all nodes listed in that index with a matching md5 value... (if you cross label more that those three...)