Question

@XmlRootElement(name = "InitiatePhoneCall", namespace = "namespace")
@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "InitiatePhoneCall",
        namespace = "namespace",
        propOrder = {"messageParams"})

public class InitiatePhoneCall implements IRequest {

   // variables

    @XmlElement(name = "MessageParams", namespace = "namespace")
    private HashMap<String, Object> messageParams;

    public HashMap<String, Object> getMessageParams() {
        return messageParams;
    }

    public void setMessageParams(HashMap<String, Object> messageParams) {
        this.messageParams = messageParams;
    }

    //getter setters

}

我在我的应用程序中有上面的代码块。我正在使用带有JDK 1.7的Weblogic 12c服务器。我在Web服务部署中得到了这个：

javax.xml.ws.WebServiceException：class package.InitiatePhoneCall在com.sun.xml.ws.server.sei.EndpointArgumentsBuilder $ DocLit中没有名称{namespace} MessageParams的属性。（EndpointArgumen tsBuilder.java ：610）com.sun.xml.ws.server.sei.TieHandler.createArgumentsBuilder（TieHandler.java:143）at com.sun.xml.ws.server.sei.TieHandler。（TieHandler.java:115）at com .sun.xml.ws.db.DatabindingImpl。（DatabindingImpl.java:116）

Answer 1

1＃解决方案

使用此InitiatePhoneCall类

@XmlRootElement(name = "InitiatePhoneCall", namespace = "namespace")
@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "InitiatePhoneCall",
        namespace = "namespace",
        propOrder = {"messageParams"})
public class InitiatePhoneCall implements IRequest {

    @XmlElementWrapper(name="MessageParams",namespace="namespace")
    private HashMap<String, Object> messageParams = new HashMap<String, Object>();

    public HashMap<String, Object> getMessageParams() {
        return messageParams;
    }

    public void setMessageParams(HashMap<String, Object> messageParams) {
        this.messageParams = messageParams;
    }
}

2＃解决方案

您还可以使用@XmlAdapter

import java.util.*;
import javax.xml.bind.annotation.adapters.XmlAdapter;

public class MapAdapter extends XmlAdapter<MapAdapter.AdaptedMap, Map<String, Object>> {

    public static class AdaptedMap {
        public List<Entry> entry = new ArrayList<Entry>();
    }

    public static class Entry {
        public String key;
        public Object value;
    }

    @Override
    public Map<String, Object> unmarshal(AdaptedMap adaptedMap) throws Exception {
        Map<String, Object> map = new HashMap<String, Object>();
        for(Entry entry : adaptedMap.entry) {
            map.put(entry.key, entry.value);
        }
        return map;
    }

    @Override
    public AdaptedMap marshal(Map<String, Object> map) throws Exception {
        AdaptedMap adaptedMap = new AdaptedMap();
        for(Map.Entry<String, Object> mapEntry : map.entrySet()) {
            Entry entry = new Entry();
            entry.key = mapEntry.getKey();
            entry.value = mapEntry.getValue();
            adaptedMap.entry.add(entry);
        }
        return adaptedMap;
    }
}

InitiatePhoneCall上课

@XmlRootElement(name = "InitiatePhoneCall", namespace = "namespace")
@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "InitiatePhoneCall",
        namespace = "namespace",
        propOrder = {"messageParams"})
public class InitiatePhoneCall implements IRequest {

    @XmlElement(name="MessageParams",namespace="namespace")
    @XmlJavaTypeAdapter(MapAdapter.class)
    private HashMap<String, Object> messageParams = new HashMap<String, Object>();

    public HashMap<String, Object> getMessageParams() {
        return messageParams;
    }

    public void setMessageParams(HashMap<String, Object> messageParams) {
        this.messageParams = messageParams;
    }
}

HashMap不是JAXB类中的有效属性

1 个答案: