API

Name	Author	Last updated by	Updated on
Name	Author	Last updated by	Updated on
LibJSON-1.0	ckknight	ckknight	05 Apr 2009
LibJSON.Deserialize(value) Deserialize a JSON object into a lua object This will error if the JSON object is malformed. Parameters value a string, number, boolean, or nil (or LibJSON.Null()), or a table consisting of those Return value a lua object, either a table, string, number, boolean, or nil Usage LibJSON.Deserialize("1234") == 1234 LibJSON.Deserialize("1.234e+53") == 1234e50 LibJSON.Deserialize('"Hello, friend"') == "Hello, friend" LibJSON.Deserialize('[1, 2, 3]') => { 1, 2, 3 }...