1

I have a document with a tree structure:

{
    "_id": "62e1f19f094a5696fd18f4e9",
    "parent": null,
    "children": [
          {
              "_id": "44e1f19f094a5696fd18f4o7",
              "parent": "62e1f19f094a5696fd18f4e9",
              "children": [
                    {
                        "_id": "62e1f19f094a5696fd18f4e9",
                        "parent": "44e1f19f094a5696fd18f4o7",
                        "children": []
                    }
               ] 
          }
    ] 
}

I want to add a new field "id" (without the underscore) which refers to "_id" to each document even if it is a child or parent I tried to do something like this:

$addFields: {id: $_id, children.id: $children._id} but in doesn't work

so, the final result I want to get

{
        "_id": "62e1f19f094a5696fd18f4e9",
        "id": "62e1f19f094a5696fd18f4e9",
        "parent": null,
        "children": [
              {
                  "_id": "44e1f19f094a5696fd18f4o7",
                  "id": "44e1f19f094a5696fd18f4o7",
                  "parent": "62e1f19f094a5696fd18f4e9",
                  "children": [
                        {
                            "_id": "62e1f19f094a5696fd18f4e9",
                            "id": "62e1f19f094a5696fd18f4e9",
                            "parent": "44e1f19f094a5696fd18f4o7",
                            "children": []
                        }
                   ] 
              }
        ] 
    }
uchiha itachi
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  • do you want to add this permanently in a database or just need to show it in the result? and how many nested level will be there? – turivishal Jul 28 '22 at 08:01
  • I just need to show in the result. There can be many nested levels – uchiha itachi Jul 28 '22 at 08:05
  • There is no straight way to do this, and also this is not a good idea to do this operation every time when you fetch data if you really want this field i would suggest you to store it when you insert a child element. – turivishal Jul 28 '22 at 08:42

1 Answers1

2

Based on a question by @rickhg12hs:

It can be done, but I agree with @turivishal that if you are doing it, it is reasonable to store it once (a minor change to the query):

db.collection.aggregate([
  {
    $replaceRoot: {
      newRoot: {
        $function: {
          body: "function drill(r) {r.id = r._id; if (r.children.length > 0) { for (let elem of r.children) { drill(elem)}} return r};",
          args: [
            "$$ROOT"
          ],
          lang: "js"
        }
      }
    }
  }
])

See how it works on the playground example

For storing on the db, instead of aggregate use an update with pipeline, like this

nimrod serok
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