【发布时间】:2019-04-04 07:34:44
【问题描述】:
我正在尝试获取具有相同名称数据的内部递归节点。 下面是我的示例 XML,我需要以格式输出
**
RegistrationIncludedProduct-43756.regPackagingHierarchyList-43767.regChildPackagingHierarchyList-43765.regChildPackagingHierarchyList-43763.regChildPackagingHierarchyList-43760
**
<agConnectXML>
<SourceData>
<SKUIDOut noNamespaceSchemaLocation="file:///c:/Users/BODDUAV1/OneDrive%20-%20Novartis%20Pharma%20AG/Avanthi/NovaRIM/Documents/SHAPE/stockKeepingUnit.xsd" schemaVersion="1.0">
<SystemMessageHeader>
<CreationDateTime>2002-10-10T12:00:00-05:00</CreationDateTime>
<SenderID>sandoz</SenderID>
<BusinessSystemID>SHAPE-P34-SKU</BusinessSystemID>
<MessageID>678678-2389789-4893947-473946</MessageID>
</SystemMessageHeader>
<stockKeepingUnit>30</stockKeepingUnit>
<stockKeepingUnitStatus>Approved</stockKeepingUnitStatus>
<nationalTradeItemNumber>098098</nationalTradeItemNumber>
<registrationId>REG-00000023</registrationId>
<finishedDosageFormId>FDF-002</finishedDosageFormId>
<activeSusbstanceId>6437</activeSusbstanceId>
<tenant>sandoz</tenant>
</SKUIDOut>
</SourceData>
<LSRIMSData>
<agl_result>
<agl_service_headers>
<serviceId>CustgetRegPackDetails</serviceId>
<messageProducer>agidmp</messageProducer>
<internalVersion>12077</internalVersion>
<uuid>a94c1128-b145-402b-a139-3bbe44cb04eb</uuid>
<dateFormat>yyyy-MM-dd H:mm:ss</dateFormat>
<generatedTimeStamp>2019-04-04 10:36:10</generatedTimeStamp>
<user>system</user>
</agl_service_headers>
<agl_pagination_details>
<start>0</start>
<limit>10</limit>
<totalRecordsCount>1</totalRecordsCount>
</agl_pagination_details>
<RegistrationPackaging>
<productPackaging>
<ProductPackaging>
<packagingItemName>FDF-002</packagingItemName>
</ProductPackaging>
</productPackaging>
<regIncludedProduct>
<RegistrationIncludedProduct>
<recordId>43756</recordId>
<registration>
<Registration>
<dataState>C</dataState>
<recordId>43750</recordId>
<registrationUID>REG-00000023</registrationUID>
</Registration>
</registration>
</RegistrationIncludedProduct>
</regIncludedProduct>
<regPackagingHierarchyList>
<RegistrationPackagingHierarchy>
<recordId>43767</recordId>
<regChildPackagingHierarchyList></regChildPackagingHierarchyList>
<regParentPackagingHierarchy>
<RegistrationPackagingHierarchy>
<recordId>43765</recordId>
<regParentPackagingHierarchy>
<RegistrationPackagingHierarchy>
<recordId>43763</recordId>
<regParentPackagingHierarchy>
<RegistrationPackagingHierarchy>
<recordId>43760</recordId>
</RegistrationPackagingHierarchy>
</regParentPackagingHierarchy>
</RegistrationPackagingHierarchy>
</regParentPackagingHierarchy>
</RegistrationPackagingHierarchy>
</regParentPackagingHierarchy>
<regPkgHierarchyDataCarrierList></regPkgHierarchyDataCarrierList>
<regErpCodesList></regErpCodesList>
</RegistrationPackagingHierarchy>
</regPackagingHierarchyList>
</RegistrationPackaging>
</agl_result>
</LSRIMSData>
</agConnectXML>
【问题讨论】:
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到目前为止您尝试过什么?你能分享你所做的努力吗?谢谢。
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是否总是跟在 标签后面?您可能只在 上使用匹配项,在其中应用“following::regPackagingHierarchyList[1]//recordId” -
谢谢@ChristianMosz。我尝试使用模板匹配。但没有任何明确的想法。但问题是这样的。我想从这个 XML 中输出 Level1、Level2、Level3、Level4、Level5 ...... XML 结构。是一个内部递归结构。我对此知之甚少。能否请您帮忙编写详细的代码。
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<regPackagingHierarchyList> <RegistrationPackagingHierarchy> <recordId>level5</recordId> <regParentPackagingHierarchy> <RegistrationPackagingHierarchy> <recordId>level4</recordId> <regParentPackagingHierarchy> <RegistrationPackagingHierarchy> <recordId>level3</recordId> </RegistrationPackagingHierarchy> </regParentPackagingHierarchy> </RegistrationPackagingHierarchy> </regParentPackagingHierarchy> </RegistrationPackagingHierarchy> </regPackagingHierarchyList> -
@ChristianMosz 由于评论的限制,我发送了单独的代码。假设仍然存在多个级别,并且在所有级别中都存在 recordid。需要通过将recordId与逗号连接来输出
标签: xml xslt xslt-1.0 xslt-2.0 wso2esb