jaxb unmarshal时间戳

Question

我无法让JAXB在Resteasy JAX-RS服务器应用程序中解组时间戳.

我的班级看起来像这样:

@XmlAccessorType(XmlAccessType.NONE)
@XmlRootElement(name = "foo")
public final class Foo {
    // Other fields omitted

    @XmlElement(name = "timestamp", required = true)
    protected Date timestamp;

    public Foo() {}

    public Date getTimestamp() {
        return timestamp;
    }

    public String getTimestampAsString() {
        return (timestamp != null) ? new SimpleDateFormat("yyyy-MM-dd HH:mm:ss").format(timestamp) : null;
    }

    public void setTimestamp(final Date timestamp) {
        this.timestamp = timestamp;
    }

    public void setTimestamp(final String timestampAsString) {
        try {
            this.timestamp = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss").parse(timestampAsString);
        } catch (ParseException ex) {
            this.timestamp = null;
        }
    }
}

有任何想法吗？

谢谢.

Answer 1

JAXB可以处理java.util.Date类.但是它期望格式:

"yyyy-MM-dd'T'HH:mm:ss"而不是"yyyy-MM-dd HH:mm:ss"

如果您想使用该日期格式,我建议使用XmlAdapter,它看起来如下所示:

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

import javax.xml.bind.annotation.adapters.XmlAdapter;

public class DateAdapter extends XmlAdapter<String, Date> {

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

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

    @Override
    public Date unmarshal(String v) throws Exception {
        return dateFormat.parse(v);
    }

}

然后,您可以在timestamp属性上指定此适配器:

import java.util.Date;

import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.adapters.XmlJavaTypeAdapter;

@XmlAccessorType(XmlAccessType.NONE) 
@XmlRootElement(name = "foo") 
public final class Foo { 
    // Other fields omitted 

    @XmlElement(name = "timestamp", required = true) 
    @XmlJavaTypeAdapter(DateAdapter.class)
    protected Date timestamp; 

    public Foo() {} 

    public Date getTimestamp() { 
        return timestamp; 
    } 

    public void setTimestamp(final Date timestamp) { 
        this.timestamp = timestamp; 
    } 

}

Answer 2

JAXB不能Date直接编组对象,因为它们没有足够的信息来明确.JAXB XmlGregorianCalendar为此目的引入了类,但直接使用它是非常不愉快的.

我建议将您的timestamp字段更改为a XmlGregorianCalendar,并在可能的情况下更改您的各种方法以更新此字段,同时保留您已有的公共接口.

如果你想保留这个Date字段,那么你需要实现自己的XmlAdapter类来告诉JAXB如何转换你的DateXML.