Ich versuche Xstream 1.4.2 zu verwenden, um XML in Objekt zu konvertieren. Es funktioniert völlig in Ordnung für mich, bis ich die Klassendatei des Objekts in ein separates Paket als wo der Hauptcode ausgeführt wird. Dann bekomme ich eine CannotResolveClassException. Ich habe versucht, die setClassLoader-Methode wie von anderen empfohlen, aber das hilft nicht.xstream CannotResolveClassException
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)
Antwort: xstream erwartet, dass die XML-Struktur an die Verpackung relativ zu sein, in dem es (das Objekt) stammt. Also muss xstream.alias verwendet werden, um die XML-Struktur zu aliasieren.
xstream.alias("something", Something.class);
Sonst wird xstream „Something“ erwartet im Standard-Paket zu sein, anstatt das Paket ist ein Mitglied.
Paket? Was meinst du mit Paket? –
können Sie die xml und die Java-Klasse, die Sie verwenden, posten –
Der voll qualifizierte Name funktioniert gut 'xstream.alias (...)', aber wie funktioniert es mit Annotationen? – lvr123