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我有一个密码查询，我正在尝试编写等效的遍历描述，但我被顺序卡住了。

start n=node:EduNodes(title='bachelor of science')     
match p=n-[:JOB_CHANGE|EDUCATION|CAREER_ENTRY|EDUCATION_CONTINUED*1..4]->m     
with p, relationships(p) as rels, n, m     
return p,n,m,     
reduce(weight=0, x in rels: weight + x.weight) as totalWeight     
order by totalWeight asc

这个查询很耗内存而且很慢，遍历描述很快：

TraversalDescription td = Traversal.traversal().evaluator(Evaluators.excludeStartPosition())
        .uniqueness(Uniqueness.NONE).depthFirst()
        .evaluator(Evaluators.includingDepths(1, 4))
        .relationships(RelationshipTypes.JOB_CHANGE, Direction.OUTGOING)
        .relationships(RelationshipTypes.EDUCATION, Direction.OUTGOING)
        .relationships(RelationshipTypes.CAREER_ENTRY, Direction.OUTGOING)
        .relationships(RelationshipTypes.EDUCATION_CONTINUED, Direction.OUTGOING);

我可以在遍历描述中也有密码查询的最后两行（减少顺序）吗？之后我可以自己循环执行..

Traverser traverser = td.traverse(start);
List<PathWrapper> orderedTraversalResult = new ArrayList<>();
for (Path path : traverser) {
    orderedTraversalResult.add(new PathWrapper(path));
}
Collections.sort(orderedTraversalResult);
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1 回答 1

0

您可以提供自己PathEvaluator的能够处理分支状态的。最好的资源是 Neo4j 本身的单元测试，请查看https://github.com/neo4j/neo4j/blob/master/community/kernel/src/test/java/org/neo4j/kernel/impl/traversal /TestBranchState.java。

于 2014-09-10T15:22:04.853 回答