如何用Gremlin递归查询树形结构并获取顶点多属性

Hey there! I totally get what you're trying to do—using the tree() step to generate a hierarchical structure but only pulling specific vertex properties instead of the entire vertex objects. The tree().by('name') trick works great for single properties, but for multiple attributes, we can leverage Gremlin's flexible step modifiers to shape exactly what we need.

Here are a couple of straightforward, practical approaches:

1. Use project() to Explicitly Define Your Desired Properties #

This is the most flexible and readable method, as it lets you specify exactly which properties to include and how to label them in the output.

Example query:

// Replace 'root-id' with your actual root vertex ID, adjust edge labels/properties as needed
g.V('root-id')
  .repeat(out('child')) // Traverse your tree structure (tweak edge direction/label if required)
  .tree()
  .by(
    project('name', 'id', 'status') // Define the keys for your output map
      .by('name') // Map the vertex's 'name' property to the 'name' key
      .by(id()) // Include the vertex's unique ID (use label() here if you need the vertex type too)
      .by('status') // Add another custom property you want to retrieve
  )

This returns a tree where each node is a clean map containing only the name, id, and status properties you specified—no extra vertex metadata cluttering up the result.

2. Use valueMap() for a Quick Multi-Property Grab (With Cleanup) #

If you want a faster way to pull multiple properties without explicitly defining each key, valueMap() works well—just note you'll need to clean up the default list values (since Gremlin supports multi-valued properties):

g.V('root-id')
  .repeat(out())
  .tree()
  .by(
    valueMap('name', 'id', 'type') // Pull only the properties you list here
      .unfold()
      .group() // Convert list values to single values (ideal for single-valued attributes)
  )

This gives you a similar map structure, but keep in mind: if any of your properties are multi-valued, this will collapse them to a single entry. Stick with this only if you're working exclusively with single-valued attributes.

Key Takeaway #

The by() modifier in tree() accepts any valid Gremlin step, so you can use steps like project(), valueMap(), or even custom projections to shape your node data exactly how you want it. project() is almost always the best choice for clarity and control when selecting multiple properties.

内容的提问来源于stack exchange，提问作者flq