如何优化Gremlin查询以输出指定结构的配方层级JSON？

我是Gremlin新手，若有表述不当之处敬请谅解！我拥有一个结构较为简单的配方与食材层级，希望编写Gremlin查询以输出如下格式的JSON：

{
  "id": "sandwich-a",
  "name": "Sandwich A",
  "composedOf": [
    {
      "id": "tomato",
      "name": "Tomato",
      "weight": 40,
      "composedOf": []
    },
    {
      "id": "sauce-a",
      "name": "Sauce A",
      "weight": 2.8,
      "composedOf": [
        {
          "id": "salt",
          "name": "Salt",
          "weight": null,
          "composedOf": []
        },
        {
          "id": "pepper",
          "name": "Pepper",
          "weight": null,
          "composedOf": []
        }
      ]
    },
    {
      "id": "sauce-b",
      "name": "Sauce B",
      "weight": null,
      "composedOf": [
        {
          "id": "water",
          "name": "Water",
          "weight": null,
          "composedOf": []
        },
        {
          "id": "lemon-juice",
          "name": "Lemon Juice",
          "weight": null,
          "composedOf": []
        }
      ]
    }
  ]
}

其中，id和name是顶点的属性；weight是边的属性，可能不存在；composedOf是通过标签为composed-of的边连接的顶点集合。

我目前编写的查询如下：

g.V()
    .has("id", "new-york-ciabatta")
    .project("id", "name", "composedOf")
        .by("id")
        .by("name")
        .by(
            repeat(
                out("composed-of")
                .outE().as('e')
                .inV().as('v')
                .simplePath()
                )
            .emit()
            .tree())

该查询可正常运行，但会返回大量冗余数据，请问如何修改以输出上述指定格式的JSON？

解决方案 #

你需要用递归嵌套的project结构构建目标JSON，同时处理weight属性缺失的情况，保证叶子节点的composedOf为空数组。以下是优化后的查询：

g.V().has("id", "new-york-ciabatta").
  project("id", "name", "composedOf").
    by("id").
    by("name").
    by(outE("composed-of").
        project("id", "name", "weight", "composedOf").
          by(inV().values("id")).
          by(inV().values("name")).
          by(coalesce(values("weight"), constant(null))).
          by(inV().
              project("id", "name", "composedOf").
                by("id").
                by("name").
                by(outE("composed-of").
                    project("id", "name", "weight", "composedOf").
                      by(inV().values("id")).
                      by(inV().values("name")).
                      by(coalesce(values("weight"), constant(null))).
                      by(inV().fold()).
                    fold()).
              fold()).
        fold())

关键说明： #

• 匹配层级结构：通过多层project对应JSON的嵌套层级，每个子节点复用类似结构，确保composedOf的嵌套关系正确。
• 处理weight属性：用coalesce(values("weight"), constant(null))实现“有属性则返回值，无则返回null”的逻辑。
• 空数组兜底：对无后续子节点的顶点，用fold()将空的边集合转为空数组，保证composedOf始终是数组类型。
• 消除冗余：相比tree()返回的全路径数据，这种方式仅保留当前节点的直接子节点，逐层构建结构，不会产生冗余内容。

如果你的配方层级深度不固定，可以用repeat()实现通用递归，无需手动嵌套多层project：

g.V().has("id", "new-york-ciabatta").
  repeat(
    project("id", "name", "weight", "composedOf").
      by(values("id")).
      by(values("name")).
      by(coalesce(__.inE("composed-of").values("weight"), constant(null))).
      by(outE("composed-of").inV().fold())
  ).emit().
  tree().
  unfold().select(values).unfold()

手动嵌套适合层级固定的简单结构，repeat方式则适配深度不确定的场景。

内容的提问来源于stack exchange，提问作者zXynK