$FindJSON
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$FindJSON
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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]")See Also
Updated
ByCelso Bressan
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