Fetching xml node nth value using Xpath

HI Experts ,
with reference to below question, i have one more doubt.
https://social.msdn.microsoft.com/Forums/en-US/f894e5e2-8926-4604-9171-616da3f00cd7/xpath-to-fetch-value-in-flatfile-schema?forum=biztalkgeneral
I used below xpath to fetch nth value of full repeating message nodes
indexStr = System.Convert.ToString(index);DATE=xpath(In_FF, "string(//*[local-name() = 'Details']["+indexStr+"])/*[local-name()='DATE'])");index=index + 1;
and i got below error, can any help me in this?
Exception thrown from: segment 2, progress 43
Inner exception: 'string(//*[local-name() = 'Details'][1])/*[local-name()='DATE'])' has an invalid token.
Exception type: XPathException
Source: System.Xml
Target Site: MS.Internal.Xml.XPath.AstNode ParseXPathExpresion(System.String)
The following is a stack trace that identifies the location where the exception occured

your XPATH expression is WRONG.
it should be
DATE=xpath(In_FF,
"string(//*[local-name() = 'Details']["+indexStr+"]/*[local-name()='DATE'])");
There is NO ")" after ["+indexStr+"].
The more reliable way is to use a string variable and use the System.String.Format() call as below
xPathExpr = System.String.Format("string(//*[local-name() = 'Details'][{0}]/*[local-name()='DATE'])", index);
DATE=xpath(In_FF, xPathExpr);
The "invalid token" is because of the mismatched ")" in your XPATH Expression.
Regards.

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The following is a stack trace that identifies the location where the exception occured
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this is my schema:
<?xml version="1.0" encoding="utf-16"
?>
<xs:schema xmlns="http://BizTalk_Server_ProjectRNd2.Schema3" xmlns:b="http://schemas.microsoft.com/BizTalk/2003"
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<xs:element name="Root">
<xs:annotation>
<xs:appinfo>
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<xs:complexType>
<xs:sequence>
<xs:element
name="no" type="xs:string" />
<xs:element
name="name" type="xs:string" />
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:schema>

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<?nameStyle 0?>
<Person ID="5" />
849-555-0139<Person><Name><First>Gail</First><Middle>A</Middle><Last>Erickson</Last></Name><Email>[email protected]</Email></Person></row>
<row>
<?nameStyle 0?>
<Person ID="6" />
122-555-0189<Person><Name><First>Jossef</First><Middle>H</Middle><Last>Goldberg</Last></Name><Email>[email protected]</Email></Person></row>
<row>
<?nameStyle 0?>
<Person ID="7" />
181-555-0156<Person><Name><First>Dylan</First><Middle>A</Middle><Last>Miller</Last></Name><Email>[email protected]</Email></Person></row>
<row>
<?nameStyle 0?>
<Person ID="8" />
815-555-0138<Person><Name><First>Diane</First><Middle>L</Middle><Last>Margheim</Last></Name><Email>[email protected]</Email></Person></row>
<row>
<?nameStyle 0?>
<Person ID="9" />
185-555-0186<Person><Name><First>Gigi</First><Middle>N</Middle><Last>Matthew</Last></Name><Email>[email protected]</Email></Person></row>
<row>
<?nameStyle 0?>
<Person ID="10" />
330-555-2568<Person><Name><First>Michael</First><Last>Raheem</Last></Name><Email>[email protected]</Email></Person></row>
<row>
<?nameStyle 0?>
<Person ID="11" />
719-555-0181<Person><Name><First>Ovidiu</First><Middle>V</Middle><Last>Cracium</Last></Name><Email>[email protected]</Email></Person></row>
<row>
I feel this xml output is not like the regular xml output. Do you know why it is diffrent from the regular xml xml output? Please comment on this matter.
Thanks,
Scott Chang
What do you mean by regular xml document? Are you referring to fact that its missing a root element? if yes it can be added as below
USE AdventureWorks;
GO
SELECT
p.NameStyle AS "processing-instruction(nameStyle)",
p.BusinessEntityID AS "Person/@ID",
p.ModifiedDate AS "comment()",
pp.PhoneNumber AS "text()",
FirstName AS "Person/Name/First",
MiddleName AS "Person/Name/Middle",
LastName AS "Person/Name/Last",
EmailAddress AS "Person/Email"
FROM Person.Person p
INNER JOIN Person.EmailAddress e
ON p.BusinessEntityID = e.BusinessEntityID
INNER JOIN Person.PersonPhone pp
ON p.BusinessEntityID = pp.BusinessEntityID
FOR XML PATH('ElementName'),ROOT('RootName');
replace ElementName and RootName with whatever name you need to set for element as well as the root element
Please Mark This As Answer if it solved your issue
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Visakh
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Failing to extract xml value using Jdom & Xpath

I have a method (getSingleNodeValue()) which when passed an xpatch expression will extract the value of the specified element in the xml document refered to in 'doc'. Assume doc at this point has been initialised as shown below and xmlInput is the buffer containing the xml content.
 SAXBuilder builder = null;
 Document doc = null;
 XPath xpathInstance = null;
 doc = builder.build(new StringReader(xmlInput));When i call the method, i pass the following xpath xpression
 /TOP4A/PERLODSUMDEC/TINPLD1/text()Here is the method. It basically just takes an xml buffer and uses xpath to extract the value:
 public static String getSingleNodeValue(String xpathExpr) throws Exception{
 Text list = null;
 try {
 xpathInstance = XPath.newInstance(xpathExpr);
 list = (Text) xpathInstance.selectSingleNode(doc);
 } catch (JDOMException e) {
 throw new Exception(e);
 }catch (Exception e){
 throw new Exception(e);
 return list==null ? "?" : list.getText();
 }The above method always returns "?" i.e. nothing is found so 'list' is null.
The xml document it looks at is
 <TOP4A xmlns="http://www.testurl.co.uk/enment/gqr/3232/1">
 <HEAD>
 <Doc>ABCDUK1234</Doc>
 </HEAD>
 <PERLODSUMDEC>
 <TINPLD1>10109000000000000</TINPLD1>
 </PERLODSUMDEC>
 </TOP4A>The same method works with other xml documents so i am not sure what is special about this one. There is no exception so the xml is valid xml. Its just that the method always sets 'list' to null. Any ideas?
Edit
Here is a running program testing the above:
 import org.jdom.*;
 import org.jdom.input.*;
 import org.jdom.xpath.*;
 import java.io.IOException;
 import java.io.StringReader;
 public class XpathTest {
 public static String getSingleNodeValue(String xpathExpr, String xmlInput) throws Exception{
 Text list = null;
 SAXBuilder builder = null;
 Document doc = null;
 XPath xpathInstance = null;
 try {
 builder = new SAXBuilder();
 doc = builder.build(new StringReader(xmlInput));
 xpathInstance = XPath.newInstance(xpathExpr);
 list = (Text) xpathInstance.selectSingleNode(doc);
 } catch (JDOMException e) {
 throw new Exception(e);
 }catch (Exception e){
 throw new Exception(e);
 return list==null ? "Nothing Found" : list.getText();
 public static void main(String[] args){
 String xmlInput1 = "<TOP4A xmlns=\"http://www.testurl.co.uk/enment/gqr/3232/1\"><HEAD><Doc>ABCDUK1234</Doc></HEAD><PERLODSUMDEC><TINPLD1>10109000000000000</TINPLD1></PERLODSUMDEC></TOP4A>";
 String xpathExpr = "/TOP4A/PERLODSUMDEC/TINPLD1/text()";
 XpathTest xp = new XpathTest();
 try {
 System.out.println(xp.getSingleNodeValue(xpathExpr, xmlInput1));
 } catch (Exception e) {
 // TODO Auto-generated catch block
 e.printStackTrace();
 }When i run the above, the output is
 Nothing foundEdit
I have run some further testing and it appears that if i remove the namespace url it does work. Not sure why yet. Is there any way i can tell it to ignore the namespace?
Edited by: ziggy on Sep 3, 2011 4:57 PM

ziggy wrote:
<TOP4A xmlns="http://www.testurl.co.uk/enment/gqr/3232/1">
<HEAD>
<Doc>ABCDUK1234</Doc>
</HEAD>
<PERLODSUMDEC>
<TINPLD1>10109000000000000</TINPLD1>
</PERLODSUMDEC>
</TOP4A>
It works fine, the problem is not with namespace, it is with url have given.
Editing:
Found a way to say the program to ignore namespace.
You have to use Xpath.addNamespace(prefix,uri), and pass the prefix to your pattern string.
If not clear refer the below code:
import org.jdom.*;
 import org.jdom.input.*;
 import org.jdom.xpath.*;
 import java.io.IOException;
 import java.io.StringReader;
 public class XpathTest {
 public static String getSingleNodeValue(String xpathExpr, String xmlInput) throws Exception{
 Text list = null;
 SAXBuilder builder = null;
 Document doc = null;
 XPath xpathInstance = null;
 try {
 builder = new SAXBuilder();
 doc = builder.build(new StringReader(xmlInput));
 xpathInstance = XPath.newInstance(xpathExpr);
 xpathInstance.addNamespace("ns","http://www.testurl.co.uk/enment/gqr/3232/1");
 list = (Text) xpathInstance.selectSingleNode(doc);
 } catch (JDOMException e) {
 throw new Exception(e);
 }catch (Exception e){
 throw new Exception(e);
 return list==null ? "Nothing Found" : list.getText();
 public static void main(String[] args){
 String xmlInput1 = "<TOP4A xmlns=\"http://www.testurl.co.uk/enment/gqr/3232/1\"><HEAD><Doc>ABCDUK1234</Doc></HEAD><PERLODSUMDEC><TINPLD1>10109000000000000</TINPLD1></PERLODSUMDEC></TOP4A>";
 String xpathExpr = "/ns:TOP4A/ns:PERLODSUMDEC/ns:TINPLD1/text()";
 XpathTest xp = new XpathTest();
 try {
 System.out.println(xp.getSingleNodeValue(xpathExpr, xmlInput1));
 } catch (Exception e) {
 // TODO Auto-generated catch block
 e.printStackTrace();
 }Edited by: 833545 on Sep 8, 2011 11:35 PM

Using XSLT to extract value of a XML node with namespace

I have a XML source code here.
<?xml version="1.0" encoding="utf-8" ?>
<rss version="2.0" xmlns:job="http://www.pageuppeople.com">
<channel>
 <title>SMH Jobs</title>
 <link>internalrecruitment.smhgroup.com.au/jobsrss.ashx?stp=di</link>
 <description>A listing of jobs available here</description>
 <item>
 <title>eCommerce Optimisation Advisor</title>
 <description>A new and exciting opportunity exists for an experienced eCommerce Advisor to join</description>
 <job:location PUReferenceID="3711">Sydney - Inner Suburbs & CBD</job:location>
 </item>
</channel>
</rss>
I want to use XSLT to extract value of a XML node with namespace <job:location>, and the returned value should be string 'Sydney - Inner Suburbs & CBD'. I tried a few XSL code below, but failed with error or nothing was returned.
<xsl:value-of select="job:location" disable-output-escaping="yes"/>
<xsl:value-of select="job/location" disable-output-escaping="yes"/>
<xsl:value-of select="job\location" disable-output-escaping="yes"/>
<xsl:value-of select="location" disable-output-escaping="yes"/>
This might be an easy question for you, but I would appreciate if anyone can help.

Hi Suncorp IT Learner,
We need to tell the XSLT that some elements are in another namespace. Copy the xmls declarations for the prefixes you need to use. Then use the xsl format as:
<xsl: value-of select=”job:location/@PUReferenceID”/>
In following issue, Chriztian has a good explanation:
http://our.umbraco.org/forum/developers/xslt/33353-XSLT-reading-XML-attribute-value
Thanks,
Qiao Wei
TechNet Community Support

Fetching xml node nth value using Xpath

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