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我想在增加时highScore根据用户更新 a 。score如果新增加score的大于highScore，则设置highScore= score。

// initial data
g.addV("player")
  .property(id, 1)
  .property(single, "score", 0)
  .property(single, "highScore", 0)

// increment score by 1 and set highScore if required
g.V(1)
  .sack(assign)
  .by("score")
  .sack(sum)
  .by(__.constant(1))
  .property(single, "score", sack())
  .choose(
    __.values("highScore").is(lt(__.values("score"))),
    __.property(single, "highScore", __.values("score")))
  )

它似乎在lt(__.values("score")). 它将其解析为遍历而不是值。

com.amazon.neptune.tinkerpop.structure.NeptuneGraph$NeptuneGraphTraversal 不能转换为 java.lang.Integer

如何将当前值传递score到该谓词中？我试过添加.value(),.iterate()和.next()

4

1 回答 1

1

这种使用方法where()似乎有效：

gremlin> g.V(1).as('a').
......1>   sack(assign).
......2>     by("score").
......3>   sack(sum).
......4>     by(__.constant(1)).
......5>   property(single, "score", sack()).
......6>   choose(where('a', lt('a')).by('highScore').by('score'),
......7>          __.property(single, "highScore", sack()))
==>v[1]
gremlin> g.V().valueMap()
==>[score:[1],highScore:[1]]
gremlin> g.V(1).as('a').
......1>   sack(assign).
......2>     by("score").
......3>   sack(sum).
......4>     by(__.constant(1)).
......5>   property(single, "score", sack()).
......6>   choose(where('a', lt('a')).by('highScore').by('score'),
......7>          __.property(single, "highScore", sack()))
==>v[1]
gremlin> g.V().valueMap()
==>[score:[2],highScore:[2]]
gremlin> g.V().property('highScore',10)
==>v[1]
gremlin> g.V().valueMap()
==>[score:[2],highScore:[10]]
gremlin> g.V(1).as('a').
......1>   sack(assign).
......2>     by("score").
......3>   sack(sum).
......4>     by(__.constant(1)).
......5>   property(single, "score", sack()).
......6>   choose(where('a', lt('a')).by('highScore').by('score'),
......7>          __.property(single, "highScore", sack()))
==>v[1]
gremlin> g.V().valueMap()
==>[score:[3],highScore:[10]]
于 2019-01-11T01:24:52.437 回答