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问题:
So I know when you created nodes neo4j has a UUID for each node. I know you can access a particular node by that UUID by accessing the ID. For example:
START n=node(144) RETURN n;
How would I get the last node that was created? I know I could show all nodes and then run the same command in anotehr query with the corresponding ID, but is there a way to do this quickly? Can I order nodes by id and limit by 1? Is there a simpler way? Either way I have not figured out how to do so through a simple cypher query.
回答1:
Every time not guaranteed that a new node always has a larger id than all previously created nodes,
So Better way is to set created_at property which stores current time-stamp while creating node. You can use timestamp() function to store current time stamp
Then,
Match (n) Return n Order by n.created_at desc Limit 1
回答2:
Please be aware that Neo4j's internal node id is not a UUID. Nor is it guaranteed that a new node always has a larger id than all previously created nodes. The node id is (multiplied with some constant) the offset of the node's location within a store file. Due to space reclaiming a new node might get a lower id number.
BIG FAT WARNING: Do not take any assumption on node ids.
Depending on your requirements you could organize all nodes into a linked list. There is one "magic" node having a specific label, e.g. References that has always a relationship to the latest created node:
CREATE (entryPoint:Reference {to:'latest'}) // create reference node
When a node from your domain is created, you need to take multiple actions:
- remove the
latest relationships if existing - create your node
- connect your new node to the previously latest node
- create the reference link
.
MATCH (entryPoint:Reference {to:'latest'})-[r:latest]->(latestNode) CREATE (domainNode:Person {name:'Foo'}), // create your domain node (domainNode)-[:previous]->(latestNode), // build up a linked list based on creation timepoint (entryPoint)-[:latest]->(domainNode) // connect to reference node DELETE r //delete old reference link
回答3:
I finally found the answer. The ID() function will return the neo4j ID for a node:
Match (n) Return n Order by ID(n) desc Limit 1;
