BizTalk Map Group Person by Key

Question

My existing map, transforms a person from the source schema to a person from the destination schema using the functoids within the map based on type and a number of other logica. With a recent change from the destination schema I need to amend the transformation to use a key (guid) node so that we can group the data.

I would like to implement an inline XSLT script to for something like //Person*/[Key=same] but not sure on the XSLT for this.

Example input:

<Person>
    <PersonType>PartnerParentsFarther</PersonType>
    <Name>
        <title>MR</title>
        <Gender>MALE</Gender>
        ...
    </Name>
    <Address/>
    <PersonKey>
        <Key>a1093c4a-7c53-4de5-ad30-cb4140aec58</Key>
        <ParentKey/>
    </PersonKey>
</Person>
<Person>
    <PersonType>PartnerParentsMother</PersonType>
    <Name>
        <title>MRS</title>
        <Gender>FEMALE</Gender>
        ...
    </Name>
    <Address/>
    <PersonKey>
        <Key>a1093c4a-7c53-4de5-ad30-cb4140aec58</Key>
        <ParentKey/>
    </PersonKey>
</Person>
<Person>
    <PersonType>PartnerParentsFarther</PersonType>
    <Name>
        <title>MR</title>
        <Gender>MALE</Gender>
        ...
    </Name>
    <Address/>
    <PersonKey>
        <Key>a1093c4a-7c53-4de5-ad30-cb4140aec59</Key>
        <ParentKey/>
    </PersonKey>
</Person>
<Person>
    <PersonType>PartnerParentsMother</PersonType>
    <Name>
        <title>MRS</title>
        <Gender>FEMALE</Gender>
        ...
    </Name>
    <Address/>
    <PersonKey>
        <Key>a1093c4a-7c53-4de5-ad30-cb4140aec59</Key>
        <ParentKey/>
    </PersonKey>
</Person>

Example output currently:

<PartnerParents>
    <Father>
        <Title>MR</Title>
        <Gender>MALE</Gender>
    </Father>
    <Father>
        <Title>MR</Title>
        <Gender>MALE</Gender>
    </Father>
    <Mother>
        <Title>MRS</Title>
        <Gender>FEMALE</Gender>
    </Mother>
    <Mother>
        <Title>MRS</Title>
        <Gender>FEMALE</Gender>
    </Mother>
</PartnerParents>

Example required output, based on same key values:

<PartnerParents>
    <Father>
        <Title>MR</Title>
        <Gender>MALE</Gender>
    </Father>
    <Mother>
        <Title>MRS</Title>
        <Gender>FEMALE</Gender>
    </Mother>
</PartnerParents>
<PartnerParents>
    <Father>
        <Title>MR</Title>
        <Gender>MALE</Gender>
    </Father>
        <Mother>
        <Title>MRS</Title>
        <Gender>FEMALE</Gender>
    </Mother>
</PartnerParents>

This appears to be a grouping question. See: http://www.jenitennison.com/xslt/grouping/muenchian.html — michael.hor257k, Mar 22 '16 at 08:18

user1104946 · Accepted Answer · 2016-03-24T14:57:35.373

When this XSLT 1.0 solution:

<xsl:key name="groups" match="Test/Person/PersonKey/Key/text()" use="."/>

  <xsl:template match="/">
    <xsl:apply-templates select="/Test" />
  </xsl:template>

<xsl:template match="/Test">
<Test>
  <xsl:for-each select="Person/PersonKey/Key/text()[generate-id()=generate-id(key('groups',.))]">
    <PartnerParents>
      <Father>
        <xsl:for-each select="key('groups',.)">
          <xsl:if test="../../../Name/Gender/text() = 'MALE'">
            <Title>
               <xsl:value-of select="../../../Name/title/text()" />
            </Title>
            <Gender>
               <xsl:value-of select="../../../Name/Gender/text()" />
            </Gender>
          </xsl:if>
        </xsl:for-each>
      </Father>
      <Mother>
        <xsl:for-each select="key('groups',.)">
          <xsl:if test="../../../Name/Gender/text() = 'FEMALE'" >
            <Title>
               <xsl:value-of select="../../../Name/title/text()" />
            </Title>
            <Gender>
               <xsl:value-of select="../../../Name/Gender/text()" />
            </Gender>
          </xsl:if>
        </xsl:for-each>
       </Mother>
    </PartnerParents >
  </xsl:for-each>
</Test>

...is applied to the below XML:

<Test>
  <Person>
    <PersonType>PartnerParentsFarther</PersonType>
    <Name>
      <title>MR</title>
       <Gender>MALE</Gender>
    </Name>
    <Address/>
    <PersonKey>
      <Key>a1093c4a-7c53-4de5-ad30-cb4140aec58</Key>
      <ParentKey/>
    </PersonKey>
 </Person>
 <Person>
   <PersonType>PartnerParentsMother</PersonType>
   <Name>
     <title>MRS</title>
     <Gender>FEMALE</Gender>
   </Name>
   <Address/>
   <PersonKey>
      <Key>a1093c4a-7c53-4de5-ad30-cb4140aec58</Key>
      <ParentKey/>
   </PersonKey>
 </Person>
 <Person>
   <PersonType>PartnerParentsFarther</PersonType>
   <Name>
     <title>MR</title>
     <Gender>MALE</Gender>
   </Name>
   <Address/>
   <PersonKey>
    <Key>a1093c4a-7c53-4de5-ad30-cb4140aec59</Key>
    <ParentKey/>
 </PersonKey>
 </Person>
 <Person>
  <PersonType>PartnerParentsMother</PersonType>
  <Name>
    <title>MRS</title>
    <Gender>FEMALE</Gender>
  </Name>
  <Address/>
   <PersonKey>
     <Key>a1093c4a-7c53-4de5-ad30-cb4140aec59</Key>
     <ParentKey/>
   </PersonKey>
 </Person>
</Test>

...the wanted result is produced:

<Test>
 <PartnerParents>
   <Father>
     <Title>MR</Title>
     <Gender>MALE</Gender>
   </Father>
   <Mother>
    <Title>MRS</Title>
    <Gender>FEMALE</Gender>
   </Mother>
 </PartnerParents>
 <PartnerParents>
   <Father>
     <Title>MR</Title>
     <Gender>MALE</Gender>
   </Father>
   <Mother>
     <Title>MRS</Title>
     <Gender>FEMALE</Gender>
   </Mother>
 </PartnerParents>
</Test>

It turns out that some of the data will be missing the key and so I need to cater for mapping the data where there is no key mapping also. I current am checking for Parent key ='' but need to work out how to get that added to a template in line with the other parentkey and key pair combo any ideas. — PuffTMD, Jul 17 '16 at 18:30

BizTalk Map Group Person by Key

1 Answers1