我正在尝试解组XML文件,但一直收到此错误:
javax.xml.bind.UnmarshalException: unexpected element (uri:"MyProtocol.xsd", local:"MyFrame"). Expected elements are (none)
我正在使用Axis2从我的wsdl文件生成带有ADB绑定的类。这是.wsdl:
的根源<?xml version="1.0" encoding="utf-8"?>
<definitions xmlns:xs="http://www.w3.org/2001/XMLSchema/"
xmlns:http="http://schemas.xmlsoap.org/wsdl/http/" xmlns:soap="http://schemas.xmlsoap.org/wsdl/soap/"
xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/" xmlns:tns="MyProtocol.xsd"
xmlns:mime="http://schemas.xmlsoap.org/wsdl/mime/" xmlns:wsi="http://ws-i.org/profiles/basic/1.1/xsd"
xmlns="http://schemas.xmlsoap.org/wsdl/" targetNamespace="MyProtocol.xsd">
<types>
<xs:schema targetNamespace="MyProtocol.xsd"
xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns="MyProtocol.xsd"
elementFormDefault="qualified">
<xs:import namespace="http://ws-i.org/profiles/basic/1.1/xsd"
schemaLocation="http://ws-i.org/profiles/basic/1.1/swaref.xsd" />
<xs:element name="MyFrame">
<xs:complexType>
<xs:choice maxOccurs="unbounded">
...
这是我必须解组的代码,我假设它不适用于Axis2和ADB绑定。
public class Foobar<T>
{
private T obj;
private Class<T> type;
public Foobar(Class<T> type) {
this.type = type;
}
public void unmarshalXML(String xml) {
JAXBContext jaxbContext;
Unmarshaller unmarshaller;
StringReader reader;
try {
jaxbContext = JAXBContext.newInstance(type);
unmarshaller = jaxbContext.createUnmarshaller();
reader = new StringReader(xml);
obj = (T) unmarshaller.unmarshal(reader);
} catch(JAXBException e) {
e.printStackTrace();
}
}
}
Axis2生成一个存根类,它包含我的模式中选项的所有getter,setter和类。我使用该类并调用unmarshal
方法,如下所示:
Foobar<MyStub.MyFrame> foobar = new Foobar<MyStub.MyFrame>(MyStub.MyFrame.class);
// Unmarshal
String xml = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>"
+ "<MyFrame xmlns=\"MyProtocol.xsd\">"
+ "<YC>"
+ "<SenderId>172</SenderId>"
+ "<RequestId>123saA</RequestId>"
+ "<SubNumber>5558879876</SubNumber>"
+ "</YC>"
+ "</MyFrame>";
foobar.unmarshalXML(xml);
然后我收到上面发布的错误。为什么会这样?我的代码错了吗?
答案 0 :(得分:0)
没关系,我明白了。如果你有这个XML:
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xml>
<MyFrame>
<Data>
<ID>172</ID>
<Name>101</Name>
<Date>11241987</Date>
</Data>
</MyFrame>
您可以使用以下代码解组它。 OP中的代码仅在使用JAXB时才有效。但是,在我的方案中,我使用的是ADB。
<强>解组强>
XMLStreamReader reader = XMLInputFactory.newInstance()
.createXMLStreamReader(new ByteArrayInputStream(someXMLString.getBytes()));
SomeClass myClass = SomeClass.Factory.parse(reader);
Marshal (如果您想从类中获取上述XML):
OMElement omElement = myClass.getOMElement
(SomeClass.MY_QNAME, OMAbstractFactory.getSOAP12Factory());
String someXMLString = omElement.toStringWithConsume();