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Concatenating multiple occurrences of the same field into one string

Posted on 2014-10-06
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Last Modified: 2014-10-13
Hi

I have to concatenate the multiple occurrences of the same field as follows, FYI, I am using xslt 1.0

Input code fragment
 <E1EDKT1 SEGMENT="1">
         <TDID>ZTDT</TDID>
         <E1EDKT2 SEGMENT="1"/>
            <TDLINE>ABC</TDLINE>
         </E1EDKT2>
        <E1EDKT2 SEGMENT="1"/>
            <TDLINE>DEF</TDLINE>
         </E1EDKT2>
     </E1EDKT1>

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Output desired

<Reference> ABCDEF </Reference>

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Code  that I am trying to apply by using string-join function
<Reference>
			    <xsl:when test="normalize-space(E1EDKT1/TDID) = 'ZTDT'">
				  <xsl:value-of select="fn:string-join(/E1EDKT2/TDLINE, ' ')" />
				</xsl:when>
  </Reference>

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0
Comment
Question by:kalyangkm
  • 3
  • 3
8 Comments
 

Author Comment

by:kalyangkm
ID: 40365193
As String-join is not supported in XSLT 1.0 I am trying in a different way but having trouble achieving the concatenation. Please advice.

<xsl:for-each select="E1EDKT1">
			      <xsl:choose>
			         <xsl:when test="normalize-space(TDID) = 'F01'">
				   <xsl:for-each select="E1EDKT2/TDLINE">
					<Reference>
				            <xsl:value-of select="." />              
                                          </Reference>
				    </xsl:for-each>
				   </xsl:when>
				  </xsl:choose>
				 
		 </xsl:for-each> 

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0
 

Assisted Solution

by:kalyangkm
kalyangkm earned 0 total points
ID: 40365234
I got the target xml with slight modification

<xsl:for-each select="E1EDKT1">
								<xsl:choose>
									<xsl:when test="normalize-space(TDID) = 'F01'">
										<Reference>
											<xsl:for-each select="E1EDKT2/TDLINE">
												<xsl:value-of select="."/>
											</xsl:for-each>
										</Reference>
									</xsl:when>
								</xsl:choose>
							</xsl:for-each>

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LVL 60

Accepted Solution

by:
Geert Bormans earned 400 total points
ID: 40365695
<Reference>
  <xsl:for-each select="E1EDKT2/TDLINE">
    <xsl:value-of select="."/>
  </xsl:for-each>
</Reference>

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Is better done like this

<Reference>
   <xsl:value-of select="E1EDKT2/TDLINE"/>
</Reference>

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LVL 60

Expert Comment

by:Geert Bormans
ID: 40365699
value-of does concatenate all the string nodes found

you can also use concat()
<Reference>
   <xsl:value-of select="concat(E1EDKT2[1]/TDLINE, ' ', E1EDKT2[2]/TDLINE)"/>
</Reference>

If you want a space in between (and you know there is only two)

Depending on the exact requirements there is a lot more you can do
0
 
LVL 28

Assisted Solution

by:Pravin Asar
Pravin Asar earned 100 total points
ID: 40367599
This should work for you.

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:fn="http://www.w3.org/2005/xpath-functions">
<xsl:output method="text"/>
<xsl:template match="/E1EDKT1">
<xsl:for-each select="E1EDKT2/TDLINE">
<xsl:value-of select="concat(., '=')"/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>

To test, I used following XML data

<?xml version="1.0" encoding="UTF-8"?>
<E1EDKT1 SEGMENT="1">
      <TDID>ZTDT</TDID>
      <E1EDKT2 SEGMENT="1">
            <TDLINE>ABC</TDLINE>
      </E1EDKT2>
      <E1EDKT2 SEGMENT="1">
            <TDLINE>DEF</TDLINE>
      </E1EDKT2>
</E1EDKT1>
0
 
LVL 60

Assisted Solution

by:Geert Bormans
Geert Bormans earned 400 total points
ID: 40367728
some comments with @pravinasar's code

I understand in the real XML, E1EDKT1 is not the root,
so this template <xsl:template match="/E1EDKT1"> will never fire
you should remove the '/' to make it work

this code is adding a '=' between all te nodes
if that would be the requirement, here is how to avoid the trailing extra '='
<xsl:for-each select="E1EDKT2/TDLINE">
   <xsl:if test="not(position() = 1)">
      <xsl:text>=</xsl:text>
</xsl:if>
    <xsl:value-of select="."/>
 </xsl:for-each>

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instead of
<xsl:for-each select="E1EDKT2/TDLINE">
 <xsl:value-of select="concat(., '=')"/>
 </xsl:for-each>

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Author Closing Comment

by:kalyangkm
ID: 40376639
Because it is the correct solution
0

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