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Xslt Transform XML 2 XML

Hi Guru's

I have an input xml and transform to another xml format thru Xslt.

Input xml

<?xml version="1.0"?>
<Stockpile xmlns="http://www.qr.com.au/schema/2009/2.1/Stockpile">
  <Header>
    <Reference>
      <Identifier>B47M_2010-01-22_2010-01-22T105926</Identifier>
      <DateTime>2010-01-22T10:59:26.667+10:00</DateTime>
    </Reference>
    <Sender>
      <Identifier>BS.SPT</Identifier>
    </Sender>
    <Receiver>
      <Identifier type="Logon Name">PA</Identifier>
    </Receiver>
  </Header>
  <Body>
    <ServiceId>B47M</ServiceId>
    <ServiceDate>2010-01-22</ServiceDate>
    <Consignments>
      <Consignment>
        <Pit>Pit 3</Pit>
        <Route>E</Route>
        <Location>
          <Code>GLD</Code>
        </Location>
      </Consignment>
    </Consignments>
  </Body>
</Stockpile>

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Output XML file

<?xml version="1.0"?>
<Message>
 <MessageHeader>
  <Reference>
   <Identifier>MessageResponse_2008-05-19T09:18:01.807+10:00_B79M_2008-05-
    19_2008-05-19T091617</Identifier>
   <ReferenceDate>2008-05-19T09:18:01.807+10:00</ReferenceDate>
  </Reference>
  <Sender>
   <Identifier>PA</Identifier>
  </Sender>
  <Receiver>
   <Identifier>Bus Service</Identifier>
  </Receiver>
 </MessageHeader>
 <MessageBody>
  <MessageResponse>
   <OriginalReference>
    <Identifier>B79M_2008-05-19_2008-05-19T091617</Identifier>
    <ReferenceDate>2008-05-19T09:16:17.207+10:00</ReferenceDate>
   </OriginalReference>
  <ResponseCode>accepted</ResponseCode>
  <Comments>
   <Comment type="informational">automatically approved</Comment>
  </Comments>
  </MessageResponse>
 </MessageBody>
</Message>

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Please help me how can i acheive thru XSLT using C#.

Thanks in Advance

Rimvis

Hi Sha1395,
Please refer to this link on how to apply XSLT transformations to XML:

XML transformation using Xslt in C#
http://www.csharpfriends.com/Articles/getArticle.aspx?articleID=63

If you need help to construct your XSLT, you will need to clarify some details:

1)
In you destination XML there are 2 reference dates:
<ReferenceDate>2008-05-19T09:18:01.807+10:00</ReferenceDate>
<ReferenceDate>2008-05-19T09:16:17.207+10:00</ReferenceDate>

In source, there is only one. How dates in destination XML are generated?

2) Are there values constants?
Identifier>Bus Service</Identifier>
<ResponseCode>accepted</ResponseCode>
<Comment type="informational">automatically approved</Comment>

Sha1395

ASKER

Hi Rimvis,

Thanks for your help

1)
In you destination XML there are 2 reference dates: Am generating the date and time here and i don't know how to at this moment.
<ReferenceDate>2008-05-19T09:18:01.807+10:00</ReferenceDate>
<ReferenceDate>2008-05-19T09:16:17.207+10:00</ReferenceDate>

2) Are there values constants? : YES
Identifier>Bus Service</Identifier>
<ResponseCode>accepted</ResponseCode>
<Comment type="informational">automatically approved</Comment>

Sha1395

ASKER

Hi Rimvis,

Created my xslt but am facing some trouble to get date and time inside my xslt

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
 <xsl:template match="/">
 <xsl:apply-templates select="Stockpile" />
 </xsl:template>
 <xsl:template match="Stockpile">
<Message>
<MessageHeader>
<Reference> 
<Identifier>
"MessageResponse_[b]2008-05-19T09:18:01.807+10:00[/b]_"<xsl:value-of select="Header/Reference/Identifier"/>
</Identifier>
[b]<ReferenceDate>"2008-05-19T09:18:01.807+10:00"</ReferenceDate>[/b]
</Reference> 
  <Sender>
 <Identifier><xsl:value-of select="Header/Receiver/Identifier"/></Identifier>
</Sender>
<Receiver>
<Identifier>"BusService"</Identifier>
</Receiver>
</MessageHeader>
<MessageBody>      
 <MessageResponse> 
 <OriginalReference>                                                
  <Identifier>
<xsl:value-of select="Header/Reference/Identifier"/>
</Identifier>        
  <ReferenceDate>
</ReferenceDate>      
 </OriginalReference>                                               
<ResponseCode>"accepted"</ResponseCode>                               
<Comments>                                                          
   <Comment type="informational">"automatically approved"</Comment> 
</Comments>                                                         
 <xsl:text>
</xsl:text>
 </MessageResponse>      
</MessageBody>           
</Message>
  </xsl:template>
</xsl:stylesheet>

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some one help me how can i get the system date and time inside my xslt.

Thanks in Advance

ASKER CERTIFIED SOLUTION

Rimvis

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