Fixed wrong example in JSONParseResult.

This commit is contained in:
Michael Alexsander Silva Dias 2018-02-04 19:52:27 -02:00
parent ea99b90a77
commit 24e87f97c6
1 changed files with 4 additions and 4 deletions

View File

@ -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]