how to replace string in xslt

Question

I am new to xslt and trying to convert below xml using a xslt. I am trying to modify 2 fields (one is number and another of string type)using below xslt.

<items>
 <item>
  <item_id>27989498</item_id>
  <service>Test Fee1</service>
  <rate>350</rate>
  <quantity>1</quantity>
  <tax_rate>0</tax_rate>
  <item_date>2010-11-17-05:00</item_date>
 </item>
</items>

XSLT :

<xsl:stylesheet  xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

<xsl:output method="xml" indent="yes"/>
<xsl:template match="node()|@*">
    <xsl:copy>
        <xsl:apply-templates select="node()|@*"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="rate[ . &gt; 0 ]">
    <xsl:copy>100</xsl:copy>
</xsl:template>

<xsl:template match="service[ . &eq; 'Test Fee1' ]">
    <xsl:copy>T1</xsl:copy>
</xsl:template>

Output :

<items>
 <item>
 <item_id>27989498</item_id>
 <service>Test Fee1</service>
 <rate>100</rate>
 <quantity>1</quantity>
 <tax_rate>0</tax_rate>
 <item_date>2010-11-17-05:00</item_date>
 </item>
 </items>

Above XSLT is replacing rate field correctly but for string field "service" it's not replacing the intended value. How can I compare and replace string in xslt as "eq" method is not working.

Is it possible to select the replace value from a list of values instead of hard coding? Example : Test Fee1 = T1 , Test Fee2 = T2 , Test Fee3 =T3 etc.

Update :

In the request xml service tag will not have a value from series instead it will be a random value, which needs to be replaced with some keys like state names of a country should be replaced by state code without taking case sensitivity into consideration.

Example : New York = NY or new yoRk = NY, Arizona = AR etc.

You changed the question in a way that the answers are not longer corresponding. Please change it back and extend your question instead of overwriting it. — friedemann_bach
– friedemann_bach, Commented Dec 8, 2016 at 9:08
I have updated the question with the changes, can you please suggest a method so compare with case insensitivity and also picking value from list? — springenthusiast
– springenthusiast, Commented Dec 9, 2016 at 5:46
@springbatcher It would be better to accept the answer given here and post a new question with the new problem. In any case, we need to know if you're using XSLT 1.0 or XSLT 2.0. Please make it a habit to pick one of these tags in addition to xslt when asking about XSLT. — michael.hor257k
– michael.hor257k, Commented Dec 9, 2016 at 6:47

Martin Honnen · Accepted Answer · 2016-12-06 13:45:03Z

1

Assuming an XSLT 1.0 (or later) processor you simply use match="service[. = 'Test Fee1' ]". With an XSLT 2.0 processor you could also use match="service[. eq 'Test Fee1' ]". Also match="rate[ . > 0 ]" can be simplified to match="rate[ . > 0 ]".

answered Dec 6, 2016 at 13:45

Martin Honnen

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friedemann_bach · Accepted Answer · 2016-12-06 13:51:20Z

1

A very simplistic approach in XSLT 1.0:

<xsl:template match="service[starts-with(., 'Test Fee')
         and number(substring-after(., 'Test Fee')) &gt; 0]">
    <service>
        <xsl:text>T</xsl:text>
        <xsl:value-of select="substring-after(., 'Test Fee')"/>
    </service>
</xsl:template>

This could be further refined to your requirements, depending on which values you actually allow for <service>.

answered Dec 6, 2016 at 13:51

friedemann_bach

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