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walkerline117's avatar
walkerline117
Contributor
8 years ago
Solved

Show only partial body of a json

Hi If I have below json body, How would I hide certain fields inside the json body and only show partial of it. Input: { "apiVersion": "1.0", "noOfRecords": 41607, "params": { ...
  • tstack's avatar
    tstack
    8 years ago

    There’s a filter() method on objects that you can use to filter properties in objects. For example, to retain only the ‘q’ and ‘qt’ properties in the $params object, you can use the following expression:

    $params.filter((value, key) => ['q', 'qt'].indexOf(key) != -1)

    Since the property names are user-supplied, you’d want to replace the ['q', 'qt'] array literal with array from the user. For example, if the names were in a JSON-encoded array in the “field” pipeline parameter, you would use:

    $params.filter((value, key) => JSON.parse(_fields).indexOf(key) != -1)
walkerline117's avatar
walkerline117
Contributor
8 years ago

Just to add, I know mapper can do this, but we want to have a dynamic filter on this. user will have the ability to indicate which fields they want to see in the response. So I guess somthing in the mapper would make more sense.

tstack's avatar
tstack
Former Employee
8 years ago

There’s a filter() method on objects that you can use to filter properties in objects. For example, to retain only the ‘q’ and ‘qt’ properties in the $params object, you can use the following expression:

$params.filter((value, key) => ['q', 'qt'].indexOf(key) != -1)

Since the property names are user-supplied, you’d want to replace the ['q', 'qt'] array literal with array from the user. For example, if the names were in a JSON-encoded array in the “field” pipeline parameter, you would use:

$params.filter((value, key) => JSON.parse(_fields).indexOf(key) != -1)
  • walkerline117's avatar
    walkerline117
    Contributor
    8 years ago

    Thanks for your solution