[英]Apache XML-RPC Exception Handling
从Apache的XML-RPC实现返回的异常中提取原始异常的最简单方法是什么?
It turns out that getting the cause exception from the Apache exception is the right one. 事实证明,从Apache异常中获取原因异常是正确的。
} catch (XmlRpcException rpce) {
Throwable cause = rpce.getCause();
if(cause != null) {
if(cause instanceof ExceptionYouCanHandleException) {
handler(cause);
}
else { throw(cause); }
}
else { throw(rpce); }
}
According to the XML-RPC Spec it returns the "fault" in the xml. 根据XML-RPC Spec,它返回xml中的“fault”。
Is this the "Exception" you are referring to or are you refering to a Java Exception generated while making the XML-RPC call? 这是您所指的“异常”,还是指的是在进行XML-RPC调用时生成的Java异常?
Fault example 故障示例
HTTP/1.1 200 OK
Connection: close
Content-Length: 426
Content-Type: text/xml
Date: Fri, 17 Jul 1998 19:55:02 GMT
Server: UserLand Frontier/5.1.2-WinNT
<?xml version="1.0"?>
<methodResponse>
<fault>
<value>
<struct>
<member>
<name>faultCode</name>
<value><int>4</int></value>
</member>
<member>
<name>faultString</name>
<value>
<string>Too many parameters.</string>
</value>
</member>
</struct>
</value>
</fault>
</methodResponse>
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