Get the first or last node from a list

Neo4j Graph Platform Cypher
1

First of all I must admit I could not find a suitable title for the problem, thus could not find results in search. Trying to explain it.
I have a situation where i store few translations of a node like following.

(n1)-[:HAS_TRANSLATION { code: "en" }]->(:Translation { text: "some string" })
(n1)-[:HAS_TRANSLATION { code: "esp" }]->(:Translation { text: "another string" })
(n2)-[:HAS_TRANSLATION { code: "en" }]->(:Translation { text: "another string" })
(n3)-[:HAS_TRANSLATION { code: "tr" }]->(:Translation { text: "another string" })
(n3)-[:HAS_TRANSLATION { code: "zh" }]->(:Translation { text: "another string" })

Now i have a situation where i need to search the nodes by translation code such that if i search by code: "en" I will get n1 and n2 (from above example) i do like

match (n)-[:HAS_TRANSLATION { code: "en" }]->(tr:Translation)
return tr.text as name

but I also intend to show other nodes as well such that following list is formed

  • n1 - en - textstring
  • n2 - en - textstring
  • n3 - tr - textstring

I understand I can use apoc.case for this but what I am struggling is
If I use match (n)-[:HAS_TRANSLATION]->(tr:Translation) for n3 i will get 2 non "en" nodes, but i want to pick one of them only.... can i write a query for that so that it gives one node for n3 -> either the tr or the esp one... remember i will not know what translation i have for a node befor hand.
Thanks in advance...

2

Hi Mithun,

I saw your insert I have few questions

  1. Why there is no label for n
  2. Please write on your second requirement

Regards
vivek

3

Those were examples.... and there is no second requirement... just wanted a help to understand how to achieve the scenario of getting all nodes irrespecitive of language choice... So to clarify more
I have

(n1)-[tr:HAS_TRANSLATION { code: "en" }]->(:Translation { text: "some string" })
(n1)-[tr:HAS_TRANSLATION { code: "esp" }]->(:Translation { text: "another string" })
(n2)-[tr:HAS_TRANSLATION { code: "en" }]->(:Translation { text: "another string" })
(n3)-[tr:HAS_TRANSLATION { code: "tr" }]->(:Translation { text: "another string" })
(n3)-[tr:HAS_TRANSLATION { code: "zh" }]->(:Translation { text: "another string" })

here n1, n2, n3 are example nodes

now if i search with tr.code==="en" I want to get
n1, n2 in en
and n3 in either tr or esp

I know its really tough to explain... but thanks for trying to help.

4

It sounds like you only have specific exceptions that you also want to consider.

I think you want to use a WHERE clause here, using a list of codes you're interested in finding:

match ()-[r:HAS_TRANSLATION]->(tr:Translation)
where r.code in ["en", "tr"]
return tr.text as name

If you have many :HAS_TRANSLATION relationships, you might want to also consider fulltext indexes, which can be used over relationships.

5

Thank you once again, but seems like my question is not clear.....
first of all i do not know what are the languages that can come... each node has several translations.... i will pass one translation code in my query... -> here it is "en"... and i want back all nodes with "en" translation PLUS also those nodes who do not have "en" but atleast have one translation in any code...

So if i have a data like

n1 has english, turkish translation,
n2 has english translation
n3 has spanish, german translation

Then i want output like

n1 -> with english translation
n2 -> with english translation
n3 -> with either one of spanish or german translation

Hopefully this makes my question more clear.

6

Ah, okay, so this can be solved with a sorting operation so the desired type of translation comes first, then taking the first one per text.

Something like this:

MATCH (n)-[r:HAS_TRANSLATION]->(tr:Translation)
WITH n, tr
ORDER BY r.code = 'en' DESC
WITH n, collect(tr)[0] as tr
RETURN tr.text as name
2 Likes
7

Thank u.. this is what I was looking for.. and once again sorry to make it so complicated in my question :slight_smile:
Wish I could have written a better title for this topic so that in future people can readily find this solution.