小编Mar*_*man的帖子

我有一个看起来像的模式片段

<xs:element name="dataValue">
        <xs:complexType>
            <xs:sequence>           
                <xs:element name="value" type="xs:anyType"\>
            </xs:sequence>
        </xs:complexType>
</xs:element>

hyperjaxb3生成的类包含以下片段:

@XmlElement(required = true)
protected Object value;

@Transient
public Object getValue() {
    return value;
}

public void setValue(Object value) {
    this.value = value;
}

@Basic
@Column(name = "VALUEOBJECT")
public String getValueObject() {
    if (JAXBContextUtils.
       isMarshallable("org.lcogt.schema", this.getValue())) {
        return JAXBContextUtils.unmarshall("org.lcogt.schema", this.getValue());
    } else {
        return null;
    }
}

据我所知,hibernate将很难持久化纯对象,因此hyperjaxb假设该对象可以被解组为XML字符串并且生成的String是持久的.在我的情况下,这不是真的,但我可以保证toString()方法将返回一些有用的东西.我希望生成的代码看起来更像:

@XmlElement(required = true)
protected Object value;

@Transient
public Object getValue() {
    return value;
}

public void setValue(Object value) {
    this.value = …

某些设备(例如webrelay)返回原始XML以响应HTTPGet请求.也就是说,回复不包含有效的HTTP标头.多年来,我使用以下代码从这些设备中检索信息:

private InputStream doRawGET(String url) throws MalformedURLException, IOException
{
    try
    {
        URL url = new URL(url);
        HttpURLConnection con = (HttpURLConnection) url.openConnection();
        con.setConnectTimeout(5000);
        con.setReadTimeout(5000);
        return con.getInputStream();
    }
    catch (SocketTimeoutException ex)
    {
        throw new IOException("Timeout attempting to contact Web Relay at " + url);
    }
}

在openJdk 7中,sun.net.www.protocol.http.HttpURLConnection添加了以下行,这意味着任何带有无效标头的HTTP响应都会生成IOException:

1325                respCode = getResponseCode();
1326                if (respCode == -1) {
1327                    disconnectInternal();
1328                    throw new IOException ("Invalid Http response");
1329                }

如何从Java 7新世界中需要HTTPGet请求的服务器获取"无头"XML？