XPath: How to find node contaning substring of value?

Hi!
I am new to XML and I would appreciate any help. I need to find all XML nodes which have value which includes given substring. For example:
<name> abcd </name>
<name> cdfg </name>
<name> cbcs </name>
I will need to get list of nodes contaning substring "bc" (first and third nodes). I know how to search xml to match entire value string. But what should I do to match substring of value string?
LVL 8
gelbertAsked:
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dualsoulCommented:
i' assuming you have xml like this, for instance:

<?xml version="1.0" encoding="UTF-8"?>
<root>
      <name> abcd </name>
      <name> cdfg </name>
      <name> cbcs </name>
</root>

so, xslt to get  what you want:
....................................
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<output>
    <xsl:for-each select="/root/name[contains(.,'bc')]">
                <xsl:copy-of select="." />
    </xsl:for-each>
</output>
</xsl:template>
</xsl:stylesheet>
.....................................

will produce:
<output>
<name> abcd </name>
<name> cbcs </name>
</output>
, or if you don't use xslt, just use following XPath:

/root/name[contains(.,'bc')]

to get nodes with substring 'bc'
0
gelbertAuthor Commented:
Thanks  dualsoul, but I forgot to mention that I need XPath, not XSLT
0
dualsoulCommented:
> Thanks  dualsoul, but I forgot to mention that I need XPath, not XSLT
you read inattentive :)
i wrote:
  if you don't use xslt, just use following XPath:  /root/name[contains(.,'bc')]  
0
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rdcproCommented:
But, if you're using MSXML, first make sure you tell the DOM document object that you're using XPath (it defaults to XSL Patterns):

var xmlDoc = new ActiveXObject("Msxml2.DomDocument");
xmlDoc.setProperty("SelectionLanguage", "XPath");
xmlDoc.load("path/to/xml.xml")

var oNodeList = xmlDoc.selectNodes("/root/name[contains(.,'bc')]")

Regards,
Mike Sharp
0
gelbertAuthor Commented:
Thanks guys
0
dualsoulCommented:
ok, if you haven't more questions, please close the question.
0

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