Question

条件：

vadas是否已连接到lop中的2 hops
vadas是否已连接到peter中的3 hops
vadas已连接到does-not-exists中的1 hops （搜索不会给出任何结果）

虚拟搜索具有预期结果

条件1和2 => [vadas-marko-lop，vadas-marko-lop-peter]
条件1或3 => [vadas-marko-lop]

我能得到的

条件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]]

条件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]]

所以，要实现这一点，我有两种不同的查询格式，有没有办法编写一种可以同时实现这两种查询格式的查询格式？

这没有用，这里有什么问题吗？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

Answer 1

我不知道我是否了解您的要求，但是如果您只需要查询模板之类的东西，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]

更多一点的抽象应该更清楚地表明这两个查询只有一步之遥（and与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")

TinkerPop：通用查询，用于合并和过滤多个遍历

1 个答案: