How do I get with cypher the minimum value of array with properties?
MATCH (n)-[r]->(m) RETURN n,m,min(r.timestamps)
Above query does not work.
r has an array with timestamps r.timestamps
How to get the lowest value of timestamps?
You can use unwind:
MATCH (n)-[r]->(m)
UNWIND r.timestamps as timestampts
RETURN n, m, min(timestampts)
I found an answer like this, but it looks ugly
MATCH
(h1)-[r]-(h2)
RETURN h1, h2,
reduce(minTimestamp = 999999999999999999, t IN r.timestamps | CASE WHEN minTimestamp < t THEN minTimestamp ELSE t END)
Related
i have a node like this: (:a{name:'',lname:'',listNumb:[1,3,9]})
i want to select when 1 is in the array
this is what itried to do
match (:a ) where a.listNumb=1 return a ; or match (:a {listNumb=1})
You can use IN to check if a value is present in the array/list.
MATCH (n:a)
WHERE 1 IN n.listNumb
RETURN n;
Match(csav:CSAVHierarchy) with csav
Match(cx:CXCustomerHierarchy) with cx
Optional Match(csav)-[:CSAVCustomerHasChild]->(csa:CSAVHierarchy) where csa._type='CXCustomer' OR csa._type='CXCustomerBU'
Optional Match(cx)-[:CXCustomerHasChild]->(cxc:CXCustomerHierarchy) where cxc._type='CXCustomer' OR cxc._type='CXCustomerBU'
return
CASE
WHEN csa.ssid = cxc.ssid and csa.elementLabel = cxc.elementLabel
THEN "yes"
ELSE "No" END As result
with this query its giving cartesian issue and i want to carry forward both the nodes data for further use.
where I m lacking?
You can use the Apoc plugin (see https://neo4j-contrib.github.io/neo4j-apoc-procedures):
Match(n:Person{ssid:"1234"}) with collect(n) as nodes CALL apoc.refactor.mergeNodes(nodes) YIELD node RETURN node
This query may do what you want. It returns the unique cxc and csa pairs that pass all your tests.
MATCH (csa:CSAVHierarchy)
WHERE
(:CSAVHierarchy)-[:CSAVCustomerHasChild]->(csa) AND
csa._type='CXCustomer' OR csa._type='CXCustomerBU'
MATCH (cxc:CXCustomerHierarchy)
WHERE
(:CXCustomerHierarchy)-[:CXCustomerHasChild]->(cxc) AND
csa.ssid = cxc.ssid AND
csa.elementLabel = cxc.elementLabel AND
(cxc._type='CXCustomer' OR cxc._type='CXCustomerBU')
RETURN cxc, csa
For better performance, you should also create indexes on :CSAVHierarchy(_type) and :CXCustomerHierarchy(_type).
This the solution I came up with
MATCH(cxc:CXCustomerHierarchy)-[:_properties]->(auditnode)-->(spoke)
where cxc._type='CXCustomer' OR cxc._type='CXCustomerBU' AND spoke.start_date <= 1554272198875 <= spoke.end_date AND spoke.status = "Confirmed"
with cxc
OPTIONAL MATCH (cxc)<-[r:CXCustomerHasChild]-(parent) with cxc
MATCH(csav:CSAVHierarchy)-[:_properties]->(auditnode)-->(spoke) with cxc,csav
where csav._type='CXCustomer' OR csav._type='CXCustomerBU' AND spoke.start_date <= 1554272198875 <= spoke.end_date AND spoke.status = "Confirmed"
OPTIONAL MATCH (csav)<-[r:CSAVCustomerHasChild]-(parent) with csav,cxc
return
CASE
WHEN csav.sourceSystemId <> cxc.sourceSystemId , csav.elementLabel <> cxc.elementLabel
THEN csav.elementLabel
ELSE "SIMILAR DATA " END As result
I have a String property in my nodes where the length of the String isn't fix.
Now i must search the right node by this property but i get a fixed length value from another System. For Example my Node has the Value '0123' but I get the Information '000123' for searching.
I need a function like left padding with Zeros and this in the Where Clause like
MATCH (a:LABEL) where leftPad(a.property, 6, '0') = '000123' return a
LIMIT 1
Is something like this possible with a good Performance?
You could do this:
MATCH (a:LABEL)
WHERE SUBSTRING('00000', 0, SIZE(a.property)) + a.property = '000123'
RETURN a
LIMIT 1;
Or, if all the characters are numeric, then you could do this:
MATCH (a:LABEL)
WHERE TOINT(a.property) = TOINT('000123')
RETURN a
LIMIT 1;
However, it would be even better if you could just store the property value as an integer in the first place, and also compare it to an integer, which would be the fastest. This might be very easy to do, depending on your situation.
MATCH (a:LABEL)
WHERE a.property = 000123
RETURN a
LIMIT 1;
Try it with reduce:
MATCH (a:LABEL)
WHERE REDUCE(lp='', n in RANGE(0,5-size(a.name)) | lp+'0')+a. a.property = '000123'
RETURN a
or try it with regular expression:
MATCH (a:LABEL)
WHERE a.property =~ '(0){0,3}123'
RETURN a
I would like to return for a given node-id related nodes and their relationships props
For example:
-> defines a bi direction relationship with property timestamp
1234->777
777->1234
1234->999
999->1234
1234->888
888->1234
1234,777,888,999 are node-ids
When I execute this:
final PreparedStatement ps = conn.prepareStatement("start a = node(1234) match (a)-[k:nearby*]->(b) where a<>b return DISTINCT b, k");
ResultSet rs = ps.executeQuery();
while (rs.next()) {
Map result = (Map<String, Object>) rs.getObject("b");
System.out.println(result.toString());
}
} catch (SQLException e) {
e.printStackTrace();
logger.error("Error returning userId=" + userIdInput, e);
}
return null;
}
I get:
{userId=777}
{userId=999}
{userId=888}
{userId=888}
{userId=999}
{userId=999}
{userId=777}
{userId=888}
{userId=888}
{userId=777}
{userId=888}
{userId=777}
{userId=999}
{userId=999}
{userId=777}
How I do get distinct results only (777,888,999)
How to retrieve the relationship props of 1234 to the dest node? I expect to get the timestamp prop which defined on each relationship
Thank you,
ray.
I'm not sure what language you're using so I'll focus on the Cypher. Firstly I would replace the START query with a MATCH with a WHERE on ID(a):
MATCH (a)-[k:nearby*]->(b)
WHERE ID(a) = 1234 AND a<>b
RETURN DISTINCT b, k
Secondly I'm pretty sure you don't need the a<>b because Cypher paths won't loop back on the same nodes:
MATCH (a)-[k:nearby*]->(b)
WHERE ID(a) = 1234
RETURN DISTINCT b, k
Lastly, and to your question, I suspect the reason that you're getting duplicates is because you have multiple relationships. If so you can return the result node and an array of the relationships like so:
MATCH (a)-[k:nearby*]->(b)
WHERE ID(a) = 1234
RETURN collect(b), k
That should return you node/relationship objects (with properties on both). Depending on your language/library you might get Maps or you might get objects wrapping the data
If your library doesn't return the start/end nodes for relationships for you, you can do something like this:
MATCH (a)-[k:nearby*]->(b)
WHERE ID(a) = 1234
RETURN collect({rel: b, startnode: startnode(b), endnode: endnode(b)}), k
Hopefully that helps!
You get non distinct results, because you return both b and k
If you only want to get distinct b's use:
MATCH (a)-[k:nearby*]->(b)
WHERE ID(a) = 1234 AND a<>b
RETURN DISTINCT b
You should also use parameters!
MATCH (a)-[k:nearby*]->(b)
WHERE ID(a) = {1} AND a<>b
RETURN DISTINCT b
ps.setInt(1,1234);
I have a set of nodes with a property, myproperty = "James" I want to update that property from (myproperty) to (name).
Is there any way to this with Cypher?
Solved by myself, heres what I did:
MATCH (n:term)
SET n.name = n.label
REMOVE n.label
RETURN n
You can change your old column name by using following query
MATCH (n:term)
SET n.name = n.myproperty
REMOVE n.myproperty
RETURN n
match (a:employee {employeeId:123,location:1})
set a.newProperty=a.oldProperty
remove a.oldProperty
return a;