根据条件将节点分组在一起并使用 Schematron 进行验证

Question

我正在使用 Schematron 编写规则来验证下面的数据。要求是验证患者在过去 12 个月内是否至少有过一次接触。如果每位患者有多次遭遇，请使用最后一次遭遇。

<root>
   <entry>
    <resource>
      <resourceType>Encounter</resourceType>
      <subject>
        <id>Patient/12345</id>
      </subject>
      <encounterDate>2018-04-10T10:00:00</encounterDate>
    </resource>
  </entry>
   <entry>
    <resource>
      <resourceType>Encounter</resourceType>
      <subject>
        <id>Patient/abcde</id>
      </subject>
      <encounterDate>2020-04-10T10:00:00</encounterDate>
    </resource>
  </entry>
  <entry>
   <resource>
      <resourceType>Encounter</resourceType>
      <subject>
        <id>Patient/abcde</id>
      </subject>
      <encounterDate>2019-05-10T10:00:00</encounterDate>
    </resource>
  </entry>
</root>

以上数据应该可以通过验证，因为最近一次相遇还不到一年前。我想知道的是，如果我编写一个模板，按患者 ID 将遇到的情况分组在一起，是否有办法将该模板传递给规则上下文？如果没有，还有其他方法吗？我对 xslt 和 Schematron 都是全新的，这是我目前所拥有的：

<schema xmlns="http://purl.oclc.org/dsdl/schematron" >
   <pattern>
   <key name="patientId" match="entry" use="/resouce/subject/id/text()"/>
   <template name="dateByPatient" match="entry">
   <root>
     <for-each select="resource/subject/id">
       <patient >
         <for-each select="key('patientId',text())">
           <effectiveDateTime><value-of select="./resource/encounterDate"/></effectiveDateTime>
          </for-each>
       </patient>
     </for-each>
     </root>
   </template>
   <let name="template">
    <dateByPatient/>
   </let>
   <let name="latest">
   <root>
   <for-each select="$template/root/patient">
   <patient >
    <sort select="effectiveDateTime" order="descending" />
       <if test="position() = 1">
       <effectiveDateTime><value-of select="effectiveDateTime" /></effectiveDateTime>
       </if>
    </patient>
   </for-each>
     </root>
   </let>
     <rule context="$latest/root/patient/effectiveDateTime">
     <let name="days" value="days-from-duration(fn:current-dateTime() - xs:dateTime(text()))" />
     <assert test="days-from-duration(fn:current-dateTime() - xs:dateTime(text())) &lt; 365">
       Encounter date more than a year : <value-of select="$days" /> days 
     </assert>
   </rule>
 </pattern>
</schema>

Answer 1

在 XSLT 3 的基础上，您可以使用

<?xml version="1.0" encoding="UTF-8"?>
<sch:schema xmlns:sch="http://purl.oclc.org/dsdl/schematron" queryBinding="xslt3"
    xmlns:sqf="http://www.schematron-quickfix.com/validator/process">
    <sch:ns prefix="map" uri="http://www.w3.org/2005/xpath-functions/map"/>
    <sch:pattern>
        <sch:rule context="root">
            <sch:let name="groups"
                value="let $encounter-resources := entry/resource[resourceType = 'Encounter']
                       return map:merge( 
                         $encounter-resources
                         ! 
                         map { 
                           data(subject/id) : xs:dateTime(encounterDate) 
                         }, 
                         map { 'duplicates' : 'combine' }
                       )"/>
            <sch:assert 
                test="every $patient in map:keys($groups) 
                      satisfies 
                      (current-dateTime() - max($groups($patient))) 
                      lt xs:dayTimeDuration('P365D')">At least one patient with latest encounter more than a year ago.</sch:assert>
        </sch:rule>
    </sch:pattern>
</sch:schema>

或者输出更详细的信息，只处理Encounter类型的资源：

<?xml version="1.0" encoding="UTF-8"?>
<sch:schema xmlns:sch="http://purl.oclc.org/dsdl/schematron" queryBinding="xslt3"
    xmlns:sqf="http://www.schematron-quickfix.com/validator/process">
    <sch:ns prefix="map" uri="http://www.w3.org/2005/xpath-functions/map"/>
    <sch:pattern>
        <sch:rule context="root">
            <sch:let name="groups"
                value="let $encounter-resources := entry/resource[resourceType = 'Encounter']
                       return map:merge( 
                         $encounter-resources 
                         ! 
                         map { 
                           data(subject/id) : xs:dateTime(encounterDate) 
                         }, 
                         map { 'duplicates' : 'combine' }
                       )"/>
            <sch:let name="failing-patients"
                value="map:keys($groups)[(current-dateTime() - max($groups(.))) gt xs:dayTimeDuration('P365D')]"/>
            <sch:report 
                test="exists($failing-patients)">Patients <sch:value-of select="$failing-patients"/> with latest encounter more than a year ago.</sch:report>
        </sch:rule>
    </sch:pattern>
</sch:schema>

我认为您不能像您的代码那样自由混合使用 Schematron 和 XSLT，您需要设置 XProc 管道以使用 p:xslt 对原始输入进行分组，然后进行验证步骤使用 Schematron 验证。

至于运行第二个示例 node-schematron 的问题，它使用的 XPath 实现似乎不支持 XPath 3.1 sort 函数，node-schematron 也无法将映射作为 Schematron 变量的中间结果来处理，因此似乎只能将所有内容填充到一个变量表达式中；两个例子有效：

<sch:schema xmlns:sch="http://purl.oclc.org/dsdl/schematron" queryBinding="xslt3"
    xmlns:sqf="http://www.schematron-quickfix.com/validator/process">
    <sch:ns prefix="map" uri="http://www.w3.org/2005/xpath-functions/map"/>
    <sch:pattern>
        <sch:rule context="root">
            <sch:let name="failing-patients"
                value="let $encounter-resources := entry/resource[resourceType = 'Encounter'],
                         $groups := map:merge( 
                         $encounter-resources
                         ! 
                         map { 
                           data(subject/id) : xs:dateTime(encounterDate) 
                         }, 
                         map { 'duplicates' : 'combine' }
                       )
                       return map:keys($groups)[(current-dateTime() - max($groups(.))) gt xs:dayTimeDuration('P365D')]"/>

            <sch:report 
                test="exists($failing-patients)">Patients <sch:value-of select="$failing-patients"/> with latest encounter more than a year ago.</sch:report>
        </sch:rule>
    </sch:pattern>
</sch:schema>

或

<sch:schema xmlns:sch="http://purl.oclc.org/dsdl/schematron" queryBinding="xslt3"
    xmlns:sqf="http://www.schematron-quickfix.com/validator/process">
    <sch:ns prefix="map" uri="http://www.w3.org/2005/xpath-functions/map"/>
    <sch:pattern>
        <sch:rule context="root">
            <sch:let name="failing-patients"
                value="let 
                         $encounter-resources := entry/resource[resourceType = 'Encounter'],
                         $groups := fold-left(
                            $encounter-resources, 
                            map{}, 
                            function($m, $e) { 
                                map:put(
                                    $m, 
                                    data($e/subject/id),
                                    max((xs:dateTime($e/encounterDate), map:get($m, data($e/subject/id))))
                                )
                            })
                      return map:keys($groups)[(current-dateTime() - $groups(.)) gt xs:dayTimeDuration('P365D')]"/>
            <sch:report test="exists($failing-patients)">Patients <sch:value-of
                    select="$failing-patients"/> with latest encounter more than a year
                ago.</sch:report>
        </sch:rule>
    </sch:pattern>
</sch:schema>

如果您需要一个失败的断言，请将 sch:report 替换为

<sch:assert 
            test="empty($failing-patients)">Patients <sch:value-of select="$failing-patients"/> with latest encounter more than a year ago.</sch:assert>

根据条件将节点分组在一起并使用 Schematron 进行验证

Group nodes together based on condition and validate using Schematron

xslt

schematron