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Fixed wrong example in JSONParseResult.

Michael Alexsander Silva Dias 7 years ago
parent
commit
24e87f97c6
1 changed files with 4 additions and 4 deletions
  1. 4 4
      doc/classes/JSONParseResult.xml

+ 4 - 4
doc/classes/JSONParseResult.xml

@@ -4,7 +4,7 @@
 		Data class wrapper for decoded JSON.
 	</brief_description>
 	<description>
-		Returned by [method JSON.parse], [code]JSONParseResult[/code] contains decoded JSON or error information if JSON source not successfully parsed. You can check if JSON source was successfully parsed with [code]if json_result.error == 0[/code].
+		Returned by [method JSON.parse], [code]JSONParseResult[/code] contains decoded JSON or error information if JSON source not successfully parsed. You can check if JSON source was successfully parsed with [code]if json_result.error == OK[/code].
 	</description>
 	<tutorials>
 	</tutorials>
@@ -26,9 +26,9 @@
 			A [Variant] containing the parsed JSON. Use typeof() to check if it is what you expect. For example, if JSON source starts with curly braces ([code]{}[/code]) a [Dictionary] will be returned, if JSON source starts with braces ([code][][/code]) an [Array] will be returned.
 			[i]Be aware that the JSON specification does not define integer or float types, but only a number type. Therefore, parsing a JSON text will convert all numerical values to float types.[/i]
 			[codeblock]
-			p = JSON.parse('["hello", "world", "!"]')
-			if typeof(p) == TYPE_ARRAY:
-			    print(p[0]) # prints 'hello'
+			var p = JSON.parse('["hello", "world", "!"]')
+			if typeof(p.result) == TYPE_ARRAY:
+			    print(p.result[0]) # prints 'hello'
 			else:
 			    print("unexpected results")
 			[/codeblock]