条件:
vadas
是否与lop
内的2 hops
相关联vadas
是否与peter
内的3 hops
相关联vadas
是否与does-not-exists
中的1 hops
相关(搜索不会给出任何结果)虚拟搜索预期结果
1
和2
=> [vadas-marko-lop,vadas-marko-lop-peter]1
或3
=> [vadas-marko-lop]我能得到什么
1
和2
gremlin> g.V().has("person", "name", "vadas").as("from")
.select("from").as("to1").repeat(both().as("to1")).times(2).emit().has("software", "name", "lop")
.select("from").as("to2").repeat(both().as("to2")).times(3).emit().has("person", "name", "peter")
.project("a", "b")
.by(select(all, "to1").unfold().values("name").fold())
.by(select(all, "to2").unfold().values("name").fold())
==>[a:[vadas,marko,lop],b:[vadas,marko,lop,peter]]
1
或2
gremlin> g.V().has("person", "name", "vadas").as("nodes")
.union(repeat(both().as("nodes")).times(2).emit().has("software", "name", "lop"),
out().has("x", "y", "does-not-exist").as("nodes"))
.project("a")
.by(select(all, "nodes").unfold().values("name").fold())
==>[a:[vadas,marko,lop]]
那么如何实现这一点我有两种不同的查询格式,有没有办法编写一个可以同时执行这两种查询格式的查询格式?
这没用,这里有什么不对吗?不返回已遍历的节点
g.V().has("person", "name", "vadas").as("nodes")
.or(
repeat(both().as("nodes")).times(2).emit().has("software", "name", "lop"),
repeat(both().as("nodes")).times(3).emit().has("person", "name", "peter")
)
.project("a").by(select(all, "nodes").unfold().values("name").fold())
==>[a:[vadas]]
// Expect paths to be printed here vadas..lop, vadas...peter
我不知道我是否理解你所追求的是什么,但如果你只是需要像查询模板这样的东西,那么这可能会有所帮助:
gremlin> conditions = [
......1> [filter: {has("software", "name", "lop")}, distance: 2],
......2> [filter: {has("person", "name", "peter")}, distance: 3],
......3> [filter: {has("x", "y", "does-not-exist")}, distance: 1]]
==>[filter:groovysh_evaluate$_run_closure1@378bd86d,distance:2]
==>[filter:groovysh_evaluate$_run_closure2@2189e7a7,distance:3]
==>[filter:groovysh_evaluate$_run_closure3@69b2f8e5,distance:1]
gremlin> g = TinkerFactory.createModern().traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
gremlin> g.V().has("person", "name", "vadas").
......1> union(repeat(both().simplePath()).
......2> times(conditions[0].distance).
......3> emit().
......4> filter(conditions[0].filter()).store("x"),
......5> repeat(both().simplePath()).
......6> times(conditions[1].distance).
......7> emit().
......8> filter(conditions[1].filter()).store("x")).
......9> barrier().
.....10> filter(select("x").
.....11> and(unfold().filter(conditions[0].filter()),
.....12> unfold().filter(conditions[1].filter()))).
.....13> path().
.....14> by("name")
==>[vadas,marko,lop]
==>[vadas,marko,lop,peter]
gremlin> g.V().has("person", "name", "vadas").
......1> union(repeat(both().simplePath()).
......2> times(conditions[0].distance).
......3> emit().
......4> filter(conditions[0].filter()).store("x"),
......5> repeat(both().simplePath()).
......6> times(conditions[2].distance).
......7> emit().
......8> filter(conditions[2].filter()).store("x")).
......9> barrier().
.....10> filter(select("x").
.....11> or(unfold().filter(conditions[0].filter()),
.....12> unfold().filter(conditions[2].filter()))).
.....13> path().
.....14> by("name")
==>[vadas,marko,lop]
而更多的抽象应该更清楚地表明两个查询只有一步(and
vs or
):
apply = { condition ->
repeat(both().simplePath()).
times(condition.distance).
emit().
filter(condition.filter()).store("x")
}
verify = { condition ->
unfold().filter(condition.filter())
}
// condition 1 AND 2
g.V().has("person", "name", "vadas").
union(apply(conditions[0]),
apply(conditions[1])).
barrier().
filter(select("x").
and(verify(conditions[0]),
verify(conditions[1]))).
path().
by("name")
// condition 1 OR 3
g.V().has("person", "name", "vadas").
union(apply(conditions[0]),
apply(conditions[2])).
barrier().
filter(select("x").
or(verify(conditions[0]),
verify(conditions[2]))).
path().
by("name")