xstream CannotResolveClassException

Ich versuche, Xstream 1.4.2 zu verwenden, um XML in Objekt zu konvertieren. Es funktioniert einwandfrei für mich, bis ich die Klassendatei des Objekts in einem separaten Paket abgelegt habe, als wo der Hauptcode ausgeführt wird. Dann bekomme ich eine CannotResolveClassException. Ich habe versucht, die von anderen empfohlene setClassLoader-Methode zu verwenden, aber das hilft nicht.

<code>Exception in thread "main" com.thoughtworks.xstream.mapper.CannotResolveClassException: result
at com.thoughtworks.xstream.mapper.DefaultMapper.realClass(DefaultMapper.java:56)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.DynamicProxyMapper.realClass(DynamicProxyMapper.java:55)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.PackageAliasingMapper.realClass(PackageAliasingMapper.java:88)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.ClassAliasingMapper.realClass(ClassAliasingMapper.java:79)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.ArrayMapper.realClass(ArrayMapper.java:74)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30)
at com.thoughtworks.xstream.mapper.CachingMapper.realClass(CachingMapper.java:45)
at com.thoughtworks.xstream.core.util.HierarchicalStreams.readClassType(HierarchicalStreams.java:29)
at com.thoughtworks.xstream.core.TreeUnmarshaller.start(TreeUnmarshaller.java:133)
at com.thoughtworks.xstream.core.AbstractTreeMarshallingStrategy.unmarshal(AbstractTreeMarshallingStrategy.java:32)
at com.thoughtworks.xstream.XStream.unmarshal(XStream.java:1052)
at com.thoughtworks.xstream.XStream.unmarshal(XStream.java:1036)
at com.thoughtworks.xstream.XStream.fromXML(XStream.java:912)
at com.thoughtworks.xstream.XStream.fromXML(XStream.java:903)
at main.readClass(main.java:48)
at main.main(main.java:28)
</code>

Antworten: xstream erwartet, dass die XML-Struktur relativ zu dem Paket ist, aus dem sie (das Objekt) stammt. Daher muss xstream.alias verwendet werden, um die xml-Struktur zu aliasen.

<code>xstream.alias("something", Something.class);
</code>

Andernfalls erwartet xstream, dass "Something" im Standardpaket und nicht in dem Paket enthalten ist, in dem es Mitglied ist.

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