仅查找并选择具有给定属性且其值包含唯一数据的元素

Question

Following is the XMl file

enter code here <items> <item itemcode= "ABC10145" code= "74582 >10</item><item itemcode = "CBD748   Code = "9636">20</item> </items>

I want to write a XSLT code that only prints an element that contains an attribute with value = "ABC10145" and also it's item value = "10" . The rest of data should be eliminated.

Answer 1

I have made couple of changes in the input xml

<?xml version="1.0" encoding="UTF-16"?>
<items> 
<item itemcode= "ABC10145" code= "74582" >10</item>
<item itemcode = "CBD748"   code = "9636">20</item> 
</items>

You can use below template to print only attribute with value = "ABC10145" and also it's item value = "10"

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format">

 <xsl:template match="@*|node()">
  <xsl:copy>
    <xsl:apply-templates select="@*|node()"/>
  </xsl:copy>
</xsl:template>
<xsl:template match="/items/item[@itemcode!='ABC10145' and .!=10]"/>

</xsl:stylesheet>

Answer 2

The expected result can achieved quite simply by:

XSLT 1.0

<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>

<xsl:param name="itemcode" select="'ABC10145'"/>
<xsl:param name="value" select="'10'"/>

<xsl:template match="/items">
    <xsl:copy>
        <xsl:for-each select="item[@itemcode=$itemcode and .=$value]">
            <item itemcode="{@itemcode}">
                <xsl:value-of select="."/>
            </item>
        </xsl:for-each>
    </xsl:copy>
</xsl:template>

</xsl:stylesheet>

I am assuming you would want to parametrize the search values.