JSONata - am I missing something?

General
1

I have a simple JSON

{
  "customers": [
    {
      "id": "XYZ-ABC-DEF",
      "data": {
        "locations": [
          {
            "location_id": 1,
            "location_name": "Main",
            "location_attr": "blue"
          },
          {
            "location_id": 2,
            "location_name": "Secondary"
          }
        ]
      }
    }
  ]
}

I do a simple op:

customers[][data.locations.location_attr = "blue"]

And it works, returns the entire 'customer' object. But only as long as only one record has the "location_attr" field. As soon as I add "location_attr" to a second record - it stops working.

But a different op is still possible:

customers.data.locations[location_attr = "blue"]

However, this defeats the purpose - I need the customer id.

It's driving me nuts :slight_smile: Please point me to the explanation/solution!

2

What about this:

customers.data.locations[location_attr = "blue"].$$

And possibly:

customers.data.locations[location_attr = "blue"].$$.customers.id

To only get the id.

Please, don't ask for an explanation... but $$ and $ are "magical" (at least for me) in Jsonata.

3

Thanks! Your suggestion looked promising, but matches the whole structure :frowning: I need only the customer with the attribute.

4

So it looks like a limitation :frowning:

I had to get rid of embedded structures and add redundancy to the data.

5

Using your json, what exact output do you need?

6

Just stumbled upon your jsonata riddle.

Here is the solution I think:

customers[][data.locations.location_attr[$="blue"]!=[]]

you can check it with following url: https://try.jsonata.org/LimnK1l7D

Explanation:
It returns all customers matching function :

data.locations.location_attr[$="blue"] != []
  • The left part (data.locations.location_attr[$="blue"]) of the function creates an array of all location_attr values belonging to that customer that have value "blue".
  • The right part ([]) is an empty array.
2 Likes
7

Thank you so much! So it looks like this '$' is indeed the magic bullet.

1 Like
8

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