$FindJSON

Finds data within a JSON record using selection criteria.

Syntax

$findjson(source, argument-1, argument-2, ..., argument-n)

Parameters

source	a field, a variable, a form field or a constant containing a JSON structure
argument-1	selection criteria to be searched within the specified source
argument-2	secondary selection criteria to be searched within the result just found by the previous selection criteria
argument-n	and so on

Return Value

The character string resulting from all arguments searched in succession. If there are no matched criteria or if the source does not contain a well-formed JSON data structure, a $NULL value is returned.

Comments

The source must contain well-formed JSON data according to the JSON protocol; otherwise, a $NULL value is returned by the function.

Example

Given the following JSON file:

out $findjson('{"Example": [123,"XYZ"]}',"Example")

Will produce the output:

[123,"XYZ"]

because the selection criteria was the entire “Example” index. However:

out $findjson('{"Example": [123,"XYZ"]}',"Example[2]")

returns:

"XYZ"

The above example is equivalent to:

out $findjson('{"Example": [123,"XYZ"]}',"Example", "Example[2]")