[question] How to check whether two SyntaxNodes are the same?

[FFdhorkin]

I'm attempting to extract attributes of a certain type from a webpage, then reconstruct a JS object showing where they appeared in the original tree.

For example, I want to go from this input:

<div data-test='a'>
    <div data-test='b'></div>
    <div data-test='c'>
        <div data-test='d'></div>
    </div>
</div>

to

{
    a: {
        b: null,
        c: {
             d: null
        }
    }
}

The language in question is https://github.com/phoenixframework/tree-sitter-heex and I'm using this query:

const query = new Parser.Query(PhoenixHeexParser, `(
    (attribute
        (attribute_name) @attribute-name
        (quoted_attribute_value
            (attribute_value) @attribute-value
        )
    )
    (#eq? @attribute-name "data-test")
)`);
const captures = query.captures(tree.rootNode)
  .filter((match) => match.name === 'attribute-value');
console.log(captures.map(c => tree.getText(c.node)));

This correctly outputs the ['a', 'b', 'c', 'd'], but there doesn't appear to be any obvious way to gather their relative positions to each other.

What I've been trying to do is look at each matching capture's node and repeatedly use .parent and attempt to find a parent of one match that is the node from another match. But I can't figure out how to verify I've found the right node. And I suspect there's a better way... but documentation on tree-sitter is very sparse, especially on this node implementation. Comparing the text doesn't work, since the node text will have extraneous information (and it isn't guaranteed to be unique)

It might even be possible to do this with the original query itself, but I didn't see anything in the documentation that suggested this would be possible.

[sogaiu]

Regarding:

whether two SyntaxNodes are the same

Just some ideas:

1. At least one of the test files for this repository appears to directly compare nodes for equality. Not sure if that can be relied on.

2. Does comparing the pair of startPosition and endPosition of a node with those of another node work?

   Assuming the root node is not under consideration, is it possible for two nodes to have the same startPosition and endPosition pair?

   node-tree-sitter/tree-sitter.d.ts

   Lines 54 to 55 in 3224eef

   	startPosition: Point;
   	endPosition: Point;

---

On a side note, I think the underlying C library has a way to directly compare two nodes for equality, but I didn't find any mention of ts_node_eq in the node-tree-sitter codebase.

[FFdhorkin]

Thanks, @sogaiu ! I switched to web-tree-splitter, which apparently is not ideal for Node, but it exposes the .equals method you mentioned, plus some other stuff missing from node-tree-sitter.

[RedCMD]

you can specify start and end points to captures(), to gather all nodes that intersect the range

query.captures(tree.rootNode, startPosition, endPosition);

I'm interested in what you came up with
did you manage to find a clean design?

[sogaiu]

@liaodalin19903 AFAIU, ordinary participants (me included) are not given the ability to label an issue (except when we create the issue, I think some labels may apply automatically).

P.S. ahlinc no longer participates in this repository, may be better not to ping / tag him.