大家好,我是骡子的新人,所以请放轻松我。首先,我将向您展示SOAP UI上的示例:
这是WSDL文件:wsdl file
非常简单。我想制作相同的mule流(没有输入数据等 - 代码中设置的有效负载)。问题是我很简单,甚至无法启动。我读了教程: http://www.mulesoft.org/documentation/display/current/XML-only+SOAP+Web+Service+Example
但我仍然不能像这里一样做dataMappings。我知道我做错了什么我的整个流程并不大......现在看起来像这样:
<flow doc:name="PostRequest" name="PostRequest">
<http:inbound-endpoint exchange-pattern="request-response" host="localhost" port="8081" path="getPostRequest" doc:name="HTTP" />
<cxf:jaxws-client operation="PostRequest" doc:name="SOAP"
enableMuleSoapHeaders="true" clientClass="pl.execon.integration.axpppk.ws.client.XISGateway"
port="XISGatewaySoap" />
<http:outbound-endpoint address="http://localhost:8889/Service/XISGateway.asmx" />
<object-to-string-transformer doc:name="Object to String" />
</flow>
我想使用PostRequest ...任何建议?教程可以帮忙吗?问题是我需要修改这个信封:
<soap:Body>
<m:PostRequest>
<m:_requestCode>Test</m:_requestCode>
<m:GetGasCustTable>
<m:XMLDocumentTime>2014-07-21T12:24:50</m:XMLDocumentTime>
<m:CustAccount>00043280</m:CustAccount>
</m:GetGasCustTable>
</m:PostRequest>
</soap:Body>
我只是不知道如何
答案 0 :(得分:0)
如果想使用post并映射web服务,只需使用pattern,这是post方法的工作代码:
<pattern:web-service-proxy name="webserviseName">
<inbound-endpoint address="http://localhost:8081/localUWant" exchange-pattern="request-response"/>
<outbound-endpoint address="http://localhost:8889/Service/XISGateway.asmx" exchange-pattern="request-response"/>
</pattern:web-service-proxy>
就这么简单。
这是文档:http://www.mulesoft.org/documentation/display/current/Using+Mule+Configuration+Patterns