@XMLJavaTypeAdapter不工作

Question

@XmlRootElement(name="info")
@XmlAccessorType(XmlAccessType.NONE)
public class InfoVO  { 
    private int infoId;
    @XmlElement
    @XmlJavaTypeAdapter(DateAdapter.class)
    private Date createTime;
//...get/set

}

import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;

import javax.ws.rs.WebApplicationException;
import javax.xml.bind.annotation.adapters.XmlAdapter;

public class DateAdapter extends XmlAdapter<String, Date> {

    private static final SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");

    @Override
    public String marshal(Date v) {
        return dateFormat.format(v);
    }

    @Override
    public Date unmarshal(String v) {
        try {
            return dateFormat.parse(v);
        } catch (ParseException e) {
            throw new WebApplicationException();
        }
    }
}


    <dependency>
        <groupId>javax.annotation</groupId>
        <artifactId>jsr250-api</artifactId>
        <version>1.0</version>
    </dependency>

2013-06-12 11:11:13.363:warn::/xa/info/save/12121:org.codehaus.jackson.map.JSONMappingException：无法从字符串值“2013-06-08 08:00:00”构造java.util.date实例：无效表示（错误：无法解析日期“2013-06-08 08:00:00”：与任何标准形式（“yyyy-mm-dd't'hh:mm:ss.sssz”、“yyyy-mm-dd't'hh:mm:ss.sss'z'”、“eee,dd MMM”yyyy hh：mm：ssszz“、”yyyy-mm“）不兼容-dd”））在[源:java.io.StringReader@b0ff5e1；行：8,列：23]（通过引用链:com.xchange.me.vo.infovo[“CreateTime”])

Answer 1

当我添加了一个定制的MessageBodyReader时，我找到了根本原因，

  @Provider
    @Consumes("application/json")
    public class CustomJsonReader<T> implements MessageBodyReader<T> {
        @Override
        public boolean isReadable(Class<?> type, Type genericType, Annotation[] annotations, MediaType mediaType) {
            return true;
        }

    @Override
    public T readFrom(Class<T> type, Type genericType, Annotation[] annotations, MediaType mediaType, MultivaluedMap<String, String> httpHeaders,
            InputStream entityStream) throws IOException, WebApplicationException {

        /*
         * Copy the input stream to String. Do this however you like. Here I use
         * Commons IOUtils.
         */
        StringWriter writer = new StringWriter();
        IOUtils.copy(entityStream, writer, "UTF-8");
        String json = writer.toString();

        /*
         * if the input stream is expected to be deserialized into a String,
         * then just cast it
         */
        if (String.class == genericType)
            return type.cast(json);

        /*
         * Otherwise, deserialize the JSON into a POJO type. You can use
         * whatever JSON library you want, here's a simply example using GSON.
         */
        ObjectMapper objectMapper = new ObjectMapper();
        return objectMapper.readValue(json, type);
    }
    }

因此，当收到json数据时，总是进入readFrom方法，然后在返回ObjectMapper.ReadValue（json,type）行中抛出异常；

所以我认为根本原因是ObjectMapper.ReadValue忽略了注释@XMLJavaTypeAdapter