Validation 骆驼验证错误消息

我有下面的骆驼路线

<route>
    <from uri="target/in"/>
    <doTry> 
        <to uri="validator:schema.xsd"/>
        <to uri="file:target/messages/validation/valid?fileName=a.xml"/>
        <doCatch> 
            <exception>org.apache.camel.ValidationException</exception>             
            <to uri="file:target/messages/validation/invalid?fileName=a.xml"/>
       </doCatch>
    </doTry> 
</route>

org.apache.camel.ValidationException

我想在XML文件未通过本文中所述的验证时获得错误消息

---

但是，在Spring DSL中，我该如何做到这一点呢？

就像克劳斯在引用的线程中所说的那样：

异常原因=exchange.getProperty（exchange.Exception\u被捕获， 例外情况（类别）

因此，此路由应保存您的异常：

<route>
    <from uri="target/in"/>
    <doTry> 
        <to uri="validator:schema.xsd"/>
        <to uri="file:target/messages/validation/valid?fileName=a.xml"/>
        <doCatch> 
            <exception>org.apache.camel.ValidationException</exception>     
            <transform>
               <simple>${property.CamelExceptionCaught}</simple>
            </transform 
            <to uri="file:target/messages/validation/invalid?fileName=a.xml"/>
       </doCatch>
    </doTry> 
</route>

org.apache.camel.ValidationException
${property.CamelExceptionCaught}
与引用线程中的克劳斯状态类似：

异常原因=exchange.getProperty（exchange.Exception\u被捕获，
例外情况（类别）

因此，此路由应保存您的异常：

<route>
    <from uri="target/in"/>
    <doTry> 
        <to uri="validator:schema.xsd"/>
        <to uri="file:target/messages/validation/valid?fileName=a.xml"/>
        <doCatch> 
            <exception>org.apache.camel.ValidationException</exception>     
            <transform>
               <simple>${property.CamelExceptionCaught}</simple>
            </transform 
            <to uri="file:target/messages/validation/invalid?fileName=a.xml"/>
       </doCatch>
    </doTry> 
</route>


org.apache.camel.ValidationException
${property.CamelExceptionCaught}
如果希望将异常信息存储在文件中，如示例代码中所示，则需要将其转换为字符串。否则，您将得到另一个关于无法存储文件的异常，因为Camel无法将架构验证异常转换为java.io.inputStream

<route>
    <from uri="target/in"/>
    <doTry> 
        <to uri="validator:schema.xsd"/>
        <to uri="file:target/messages/validation/valid?fileName=a.xml"/>
        <doCatch> 
            <exception>org.apache.camel.ValidationException</exception>     
            <transform>
               <simple>${property.CamelExceptionCaught}</simple>
            </transform>
            <transform>
               <simple>${bodyAs(String)}</simple>
            </transform>
            <to uri="file:target/messages/validation/invalid?fileName=a.xml"/>
       </doCatch>
    </doTry> 
</route>


org.apache.camel.ValidationException
${property.CamelExceptionCaught}
${bodyAs（字符串）}
如果要将异常信息存储在示例代码中的文件中，则需要将其转换为字符串。否则，您将得到另一个关于无法存储文件的异常，因为Camel无法将架构验证异常转换为java.io.inputStream

<route>
    <from uri="target/in"/>
    <doTry> 
        <to uri="validator:schema.xsd"/>
        <to uri="file:target/messages/validation/valid?fileName=a.xml"/>
        <doCatch> 
            <exception>org.apache.camel.ValidationException</exception>     
            <transform>
               <simple>${property.CamelExceptionCaught}</simple>
            </transform>
            <transform>
               <simple>${bodyAs(String)}</simple>
            </transform>
            <to uri="file:target/messages/validation/invalid?fileName=a.xml"/>
       </doCatch>
    </doTry> 
</route>


org.apache.camel.ValidationException
${property.CamelExceptionCaught}
${bodyAs（字符串）}
您可以在一个文件中发送异常，在另一个文件中发送错误的xml

<route>
    <from uri="target/in"/>
    <doTry> 
        <to uri="validator:schema.xsd"/>
        <to uri="file:target/messages/validation/valid?fileName=a.xml"/>
        <doCatch> 
            <exception>org.apache.camel.ValidationException</exception>
            <setHeader headerName="CamelOverruleFileName">
                <simple>${file:onlyname.noext}.${date:now:yyyyMMdd_HHmmssSSS}.xml</simple>
            </setHeader>
            <to uri="file:target/messages/validation/invalid/"/>
            <setBody>
                <simple>Got "${exception.message}" with this stack\n${exception.stacktrace}\n${body}</simple>
            </setBody>
            <setHeader headerName="CamelOverruleFileName">
                <simple>${file:onlyname.noext}.${date:now:yyyyMMdd_HHmmssSSS}.xml.error</simple>
            </setHeader>
            <to uri="file:target/messages/validation/invalid/"/>
        </doCatch>
    </doTry> 
</route>


org.apache.camel.ValidationException
${file:onlyname.noext}.${date:now:yyyyymmdd_hhmmssss}.xml
已使用此堆栈获取“${exception.message}”\n${exception.stacktrace}\n${body}
${file:onlyname.noext}.${date:now:yyyyymmdd_hhmmssss}.xml.error
您可以在一个文件中发送异常，在另一个文件中发送错误的xml

<route>
    <from uri="target/in"/>
    <doTry> 
        <to uri="validator:schema.xsd"/>
        <to uri="file:target/messages/validation/valid?fileName=a.xml"/>
        <doCatch> 
            <exception>org.apache.camel.ValidationException</exception>
            <setHeader headerName="CamelOverruleFileName">
                <simple>${file:onlyname.noext}.${date:now:yyyyMMdd_HHmmssSSS}.xml</simple>
            </setHeader>
            <to uri="file:target/messages/validation/invalid/"/>
            <setBody>
                <simple>Got "${exception.message}" with this stack\n${exception.stacktrace}\n${body}</simple>
            </setBody>
            <setHeader headerName="CamelOverruleFileName">
                <simple>${file:onlyname.noext}.${date:now:yyyyMMdd_HHmmssSSS}.xml.error</simple>
            </setHeader>
            <to uri="file:target/messages/validation/invalid/"/>
        </doCatch>
    </doTry> 
</route>


org.apache.camel.ValidationException
${file:onlyname.noext}.${date:now:yyyyymmdd_hhmmssss}.xml
已使用此堆栈获取“${exception.message}”\n${exception.stacktrace}\n${body}
${file:onlyname.noext}.${date:now:yyyyymmdd_hhmmssss}.xml.error
谢谢，这很有效。它会发回一条消息，但这不是我所期望的消息。我正在获取soap:服务器无法存储文件：target\messages\validation\invalid\a.xml引发异常后，将xml文件放在另一个文件夹中是否有问题？谢谢，可以。它会发回一条消息，但这不是我所期望的消息。我正在获取soap:服务器无法存储文件：target\messages\validation\invalid\a.xml引发异常后，将xml文件放在另一个文件夹中是否有问题？