java – 如何在返回多种类型的XML的URL上使用Spring RestTemplat

我有如下代码：

Job job = getRestTemplate().postForObject(url,postData,Job.class);

我的applicationContext.xml看起来像：

<bean id="restTemplate" class="org.springframework.web.client.RestTemplate">
    <constructor-arg ref="httpClientFactory"/>

    <property name="messageConverters">
        <list>
            <bean class="org.springframework.http.converter.xml.MarshallingHttpMessageConverter">
                <property name="marshaller" ref="jaxbMarshaller"/>
                <property name="unmarshaller" ref="jaxbMarshaller"/>
            </bean>
            <bean class="org.springframework.http.converter.FormHttpMessageConverter"/>
            <bean class="org.springframework.http.converter.StringHttpMessageConverter"/>
        </list>
    </property>
</bean>

<bean id="jaxbMarshaller" class="org.springframework.oxm.jaxb.Jaxb2Marshaller">
    <property name="classesToBeBound">
        <list>
            <value>domain.fullspec.Job</value>
            <value>domain.fullspec.Exception</value>
        </list>
    </property>
</bean>

当我尝试使这个电话和服务失败,我得到：

Failed to convert value of type 'domain.fullspec.Exception' to required type 'domain.fullspec.Job'

在postForObject()调用中,我要求一个Job.class,而不是得到一个,它正在变得不安.

我想我需要能够做的事情是：

Object o = getRestTemplate().postForObject(url,Object.class);
if (o instanceof Job.class) {
   ...
else if (o instanceof Exception.class) {
}

但是这不行,因为JAXB抱怨说它不知道如何组织到Object.class – 并不奇怪.

我试图创建MarshallingHttpMessageConverter的子类并重写readFromSource()

protected Object readFromSource(类clazz,HttpHeaders头,源代码){

Object o = null;
    try {
        o = super.readFromSource(clazz,headers,source);
    } catch (Exception e) {
        try {
            o = super.readFromSource(MyCustomException.class,source);
        } catch (IOException e1) {
            log.info("Failed readFromSource "+e);
        }
    }

    return o;
}

不幸的是,这不工作,因为在我重试之前,源代码中的底层输入流已被关闭.

有意见的建议,

汤姆

更新：我已经通过获取一个inputStream的副本来工作

protected Object readFromSource(Class<?> clazz,HttpHeaders headers,Source source) {
    InputStream is = ((StreamSource) source).getInputStream();

    // Take a copy of the input stream so we can use it for initial JAXB conversion
    // and if that fails,we can try to convert to Exception
    CopyInputStream copyInputStream = new CopyInputStream(is);

    // input stream in source is empty now,so reset using copy
    ((StreamSource) source).setInputStream(copyInputStream.getCopy());

    Object o = null;
    try {
        o = super.readFromSource(clazz,source);
      // we have failed to unmarshal to 'clazz' - assume it is <exception> and unmarshal to MyCustomException

    } catch (Exception e) {
        try {

            // reset input stream using copy
            ((StreamSource) source).setInputStream(copyInputStream.getCopy());
            o = super.readFromSource(MyCustomException.class,source);

        } catch (IOException e1) {
            e1.printStackTrace();  
        }
        e.printStackTrace();
    }
    return o;

}

CopyInputStream取自http://www.velocityreviews.com/forums/t143479-how-to-make-a-copy-of-inputstream-object.html,我将其粘贴到这里.

import java.io.ByteArrayInputStream;
import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.io.InputStream;

public class CopyInputStream
{
private InputStream _is;
private ByteArrayOutputStream _copy = new ByteArrayOutputStream();

/**
 * 
 */
public CopyInputStream(InputStream is)
{
    _is = is;

    try
    {
        copy();
    }
    catch(IOException ex)
    {
        // do nothing
    }
}

private int copy() throws IOException
{
    int read = 0;
    int chunk = 0;
    byte[] data = new byte[256];

    while(-1 != (chunk = _is.read(data)))
    {
        read += data.length;
        _copy.write(data,chunk);
    }

    return read;
}

public InputStream getCopy()
{
    return (InputStream)new ByteArrayInputStream(_copy.toByteArray());
}
}