Hey there,
I really like this library and it's really easy to use. Sadly somehow something
doesn't want to work and I don't understand why.
I have a class called 'Message', Message contains two private QVariantMaps, one
called Header, the other Data.
I exported both QVariantMaps as a Q_PROPERTY. If I serialize the class, I get a
valid JSon string:
{
"Data" : {
"Code" : 0,
"Message" : "Hello, OK"
},
"Header" : {
"SubType" : "Error",
"Type" : "Message"
},
"ID" : ""
}
When I try to parse it back I get the following error:
Number is out of range: 0
json_parser - syntax error found, forcing abort, Line 1 Column 1
I don't understand what's wrong here. Code itself is typed as int. But when I
type it as an enum I get the same error. So I guess the Json parser has
problems with integers itself, because enums get serialized as integers too.
I serialize/parse the class like so:
inline static Message * ConvertFromString(QString JSon, bool *
parsed = 0)
{
QJson::Parser parser;
QVariant Data = parser.parse(JSon.toUtf8(), parsed);
if(*parsed) // I know this can lead to an error
{
Message * pRequest = new Message();
QJson::QObjectHelper::qvariant2qobject( Data.toMap(),
pRequest );
return pRequest;
}
return new pimote::Message::Message();
}
QString ConvertToString()
{
QJson::Serializer serializer;
serializer.setDoublePrecision(2);
serializer.setIndentMode( QJson::IndentFull );
QVariantMap Data = QJson::QObjectHelper::qobject2qvariant(this);
QString Converted = serializer.serialize(Data);
qDebug() << serializer.errorMessage();
return Converted;
}
I see no mistake here. If I replace the int with a QString it works. But well
.. I want it typed correctly.
Thanks in advance, have a great weekend!
Greetings from Germany
Lukas Möller
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