Gremlin查询中Select无法迭代问题求助

问题出在你的mergeV查询未正确将当前遍历的group_id作为匹配条件，导致每次操作都指向同一个顶点，最终返回重复ID。

错误原因 #

原查询中mergeV([(label): 'Peoplegroup']).property('group_id', select('gid'))仅指定了顶点标签，未将group_id纳入匹配逻辑。Gremlin无法区分不同的group_id，只会重复操作同一个顶点（或创建重复顶点但返回同一ID，取决于图数据库的实现）。

修正后的查询 #

方式一：将group_id纳入mergeV的匹配Map #

g.inject(
  [
    [stock_height:111, sh_id:"sh123", inputs: [house: "inputhouse123ia"], outputs: [[house: "outputhouse123oa", group_id: "gid-123a"],[house: "outputhouse123ob", group_id: "gid-123b"]]],
    [stock_height:111, sh_id:"sh1234", inputs: [house: "inputhouse1234ia"], outputs: [[house: "outputhouse1234oa", group_id: "gid-1234a"],[house: "outputhouse1234ob", group_id: "gid-1234b"]]]
  ]).
  unfold().as('sh').
  select('outputs').
  unfold().as('output').
  select('group_id').as('gid').
  mergeV([label: 'Peoplegroup', group_id: select('gid')]).id()

方式二：使用mergeV+on语法（Gremlin 3.5+） #

g.inject(
  [
    [stock_height:111, sh_id:"sh123", inputs: [house: "inputhouse123ia"], outputs: [[house: "outputhouse123oa", group_id: "gid-123a"],[house: "outputhouse123ob", group_id: "gid-123b"]]],
    [stock_height:111, sh_id:"sh1234", inputs: [house: "inputhouse1234ia"], outputs: [[house: "outputhouse1234oa", group_id: "gid-1234a"],[house: "outputhouse1234ob", group_id: "gid-1234b"]]]
  ]).
  unfold().as('sh').
  select('outputs').
  unfold().as('output').
  select('group_id').as('gid').
  mergeV(label, 'Peoplegroup').on('group_id', select('gid')).id()

效果说明 #

修正后，mergeV会根据每个遍历到的group_id值查找对应顶点：

• 若顶点已存在，返回其ID
• 若顶点不存在，创建新顶点并返回新ID

最终会得到4个独立的顶点ID（若所有group_id都是首次出现）。

内容的提问来源于stack exchange，提问作者Kevin Boughton