XML / XSL

Re: XSL to get only elements which have no childs

19-Sep-17 2:15

Something like this should work:

XML

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:template match="Root_Element">
        <xsl:element name="{name()}">
            <xsl:apply-templates select="*|@*"/>
        </xsl:element>
    </xsl:template>
    
    <xsl:template match="*">
        <xsl:choose>
            <xsl:when test="count(*) = 0">
                <xsl:element name="{name()}">
                    <xsl:apply-templates select="@*"/>
                </xsl:element>
            </xsl:when>
            <xsl:otherwise>
                <xsl:apply-templates select="*" />
            </xsl:otherwise>
        </xsl:choose>
    </xsl:template>
    
    <xsl:template match="@*">
        <xsl:attribute name="{name()}">
            <xsl:value-of select="."/>
        </xsl:attribute>
    </xsl:template>
</xsl:stylesheet>

"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer

MrKBA19-Sep-17 3:02

MrKBA

19-Sep-17 3:02

Yes great, thank you.

share me best practices to enhance xslt related functionality in production

Member 1178590912-Jul-17 1:50

Member 11785909

12-Jul-17 1:50

I have already gone through MSDN and multiple good known resources.
but I don't have time to elaborate and go deep in it.
I just want to give me best practices library for deeply working on xslt, XML resolver, and class library where I can implement my actual business logics using c#.Net and TPL.

Re: share me best practices to enhance xslt related functionality in production

Richard MacCutchan12-Jul-17 3:58

Re: share me best practices to enhance xslt related functionality in production

12-Jul-17 3:58

Member 11785909 wrote:
I don't have time

You may find that no one here does either.

jschell1-Sep-17 12:24

jschell

1-Sep-17 12:24

Apparently you were correct.

Re: share me best practices to enhance xslt related functionality in production

Richard MacCutchan1-Sep-17 21:42

1-Sep-17 21:42

No surprise there. Sigh | :sigh:

Transform XML

TML6-Jul-17 9:15

TML

6-Jul-17 9:15

I am trying to transform an XML file through code using XSL, but for some reason, the file output is displaying the entire XML as a string at the top of the document and then, just under the XML string, the document is parsed correctly by the XSL stylesheet. I would like to learn why this is happening. Here is the VB transform code:

        Dim ds As New DataSet
        ds.ReadXml(Server.MapPath("FormResults/" & FormID.ToString() & ".xml"))


        Dim DocXSL As New XslCompiledTransform
        Dim sw As New StringWriter()
        ds.WriteXml(sw)
        ds.Dispose()

        Dim DocXML As New XmlDocument
        DocXML.LoadXml(sw.ToString())


        'The GetXSLContent function returns the XLS template to be used
        DocXSL.Load(GetXSLContent(DFLID))
        DocXSL.Transform(DocXML, Nothing, sw)
        Dim result As String = sw.ToString()
        sw.Close()
        sw = Nothing

Me.My_Literal.Text = result.ToString()


    Private Function GetXSLContent(ByVal DFLID As Integer) As String
        Dim str As String = String.Empty
        str = Server.MapPath("FormLetters/" & DFLID.ToString() & "_DataTemplate.xslt")
        Return str
    End Function

Here is the XML

<?xml version="1.0" encoding="utf-8" standalone="yes"?>
<Results>
  <Fields ID="48">
    <Department>I.T.</Department>
    <Extension>2267</Extension>
    <Location>Primary Office Bldg</Location>
    <Name_of_User>David</Name_of_User>
    <Date>9/12/2014</Date>
    <Time>04:49</Time>
    <Details1>This is a test. asdf sasdf sa</Details1>
    <Details2>This is only a test.</Details2>
    <Details3>This is a test.</Details3>
    <Details4>This is only a test.</Details4>
    <Details5>This is a test.</Details5>
    <Details6>This is only a test.</Details6>
    <Details7>This is a test.</Details7>
    <Details8>This is only a test.</Details8>
  </Fields>
</Results>

Here is the XSLT

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="html" indent="yes" encoding="utf-8" omit-xml-declaration="yes"/>
    <xsl:template match="/">
        <div class="rndc" style="left:50%;right:50%;margin:auto;width:90%;border: 1px solid #e0e0e0;padding: 0px 5px 0px 5px">
            <xsl:for-each select="Results/Fields">
                <div style="text-align:left;vertical-align:top;padding: 0px 5px 0px 5px;width:96%;white-space:pre-wrap;display:inline-block">
<table style='border-collapse: collapse; left: 50%; right: 50%; margin: auto; width: 80%'>
    <tr>
        <td class="txtlrg" style="vertical-align:top;text-align:left" colspan="2"> <xsl:value-of select="Department"/> </td>
    </tr>
    <tr>
        <td style="vertical-align:top;text-align:right">User Name:</td>
        <td><xsl:value-of select="Name_of_User"/></td>
    </tr>
    <tr>
        <td style="vertical-align:top;text-align:right">Extension:</td>
        <td><xsl:value-of select="Extension"/></td>
    </tr>
    <tr>
        <td style="vertical-align:top;text-align:right">Location:</td>
        <td><xsl:value-of select="Location"/></td>
    </tr>
    <tr>
        <td style="vertical-align:top;text-align:right">Date & Time:</td>
        <td style="white-space:pre"><xsl:value-of select="Date"/> <xsl:value-of select="Time"/></td>
    </tr>
    <tr>
        <td style="vertical-align:top;text-align:right">Details:</td>
        <td style="white-space:pre"><xsl:value-of select="Details1"/><br />
        <xsl:value-of select="Details2"/><br />
        <xsl:value-of select="Details3"/><br />
        <xsl:value-of select="Details4"/><br />
        <xsl:value-of select="Details5"/></td>
    </tr>
</table>                </div>
                <div style="page-break-before: always;"></div>
            </xsl:for-each>
        </div>
    </xsl:template>
</xsl:stylesheet>

Here's the output:
I.T. 2267 Gov. Center Vickie 9/12/2014 04:49 This is a test. asdf sasdf sa This is only a test. This is a test. This is only a test. This is a test. This is only a test. This is a test. This is only a test.

I.T.
User Name: David
Extension: 2267
Location: Primary Office Bldg
Date & Time: 9/12/2014 04:49
Details: This is a test. asdf sasdf sa
This is only a test.
This is a test.
This is only a test.
This is a test.

Thanks in advance!

Re: Transform XML

Richard Deeming6-Jul-17 10:40

saving a file to the internet.

6-Jul-17 10:40

TML wrote:

Dim sw As New StringWriter()
ds.WriteXml(sw)
...
DocXSL.Transform(DocXML, Nothing, sw)
Dim result As String = sw.ToString()

You've declared a single StringWriter instance. You've written the XML to it. Then you've written the results of the transformation to it.

It's hardly surprising that it then contains both the XML and the results of the transformation. Smile | :)

After loading the XML, you either need to create a new StringWriter instance, or you need to clear the existing one.

VB.NET

...

DocXML.LoadXml(sw.ToString())

' Create a new instance:
' sw = New StringWriter()

' OR: Clear the existing one:
sw.GetStringBuilder().Clear()

...

"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer

Re: Transform XML

TML6-Jul-17 10:50

TML

6-Jul-17 10:50

If it was a snake, it would have bit me. Thanks Richard!

felixjrz198531-May-17 21:54

felixjrz1985

31-May-17 21:54

i want to save an xml file that will look like:
"http://www.codeproject.com/script/Membership/Uploads/1234567/update.xml"

i really need help how to do this.
When i do this in google drive the link to the file do not contain .xml extension.

Richard MacCutchan31-May-17 22:32

31-May-17 22:32

Link does not go anywhere. You need to explain your problem in proper detail.

felixjrz19851-Jun-17 0:43

felixjrz1985

1-Jun-17 0:43

iam sorry i did not explain very well.
it is not the actual link...
i mean i want to store my update.xml file somewhere on the remote location and access its link as something link www.website.com/update.xml and not as www.googledrive.com/id?=xskdhfIsksdhf

it is because i want to compare the two files update.xml in the local drives, and the www.website.com/update.xml from from remote location.

Thank you for your response.

Richard MacCutchan1-Jun-17 1:03

1-Jun-17 1:03

Sorry I still do not understand what your problem is, or what this has to do with XML/XSL.

Gerry Schmitz1-Jun-17 3:20

Gerry Schmitz

1-Jun-17 3:20

Do you have access to a domain / web site called "www.website.com"?

If you did, you would typically put "update.xml" in you website's "root directory"; or you would use some form of "redirection" to resolve the supplied address (e.g. MVC routing).

Is that clear?

"(I) am amazed to see myself here rather than there ... now rather than then".
― Blaise Pascal

XSLT passing a variable to starts-with

Terry Perez11-Sep-17 3:41

Terry Perez

11-Sep-17 3:41

Just make a google spreadsheet, that's all.

bjmallon13-Feb-17 14:03

bjmallon

13-Feb-17 14:03

I've got an XML file full of user details including 'FirstName', 'Surname' and 'StreetAddress'.

Currently my XSL file has a template (SurnameAlphabetGrouping) that groups users by the first character of the Surname element's value. I've only provided the process for 'A' to save space. All 26 letters are processed.

XML

<xsl:template name="SurnameAlphabetGrouping">
    <xsl:variable name="AMatch" select="/Users/User[starts-with(Surname, 'A')]"/>
    <xsl:variable name="ACount" select="count($AMatch)"/>
    <xsl:if test="$ACount > 0">
      <h2>
        <a href="javascript:void(0)" class="dsphead" onclick="dsp(this)">
          + A (<xsl:value-of select="$ACount"/>)
        </a>
      </h2>
      <div class="dspcont">
        <table border="1" cellpadding="1" cellspacing="1">
          <tr>
            <xsl:call-template name="HeadingRow"/>
          </tr>

          <xsl:for-each select="/Users/User[starts-with(Surname, 'A')]">
            <xsl:sort select="Surname"/>
            <xsl:call-template name="UserRow"/>
          </xsl:for-each>
        </table>
      </div>
    </xsl:if>
  </xsl:template>

That's all well and good. But now that I've got it working that way, I really want to make it smarter (i.e. cut down the code), so I'm trying to build a template that can take the required character as a parameter.

I've got a variable (Letters) that carries tokens for the 26 letters of the alphabet:

XML

<xsl:variable name="Letters">
    <xsl:call-template name="tokenise">
      <xsl:with-param name="string" select ="'A B C D E F G H I J K L M N O P Q R S T U V W X Y Z'"/>
      <xsl:with-param name="delimeters" select="' '"/>
    </xsl:call-template>
  </xsl:variable>

I'm then attempting to process each token, passing it to the template as a parameter:

XML

<xsl:for-each select="msxsl:node-set($Letters)/token">
  <xsl:variable name="currToken" select ="./node()[1]"/>
  <xsl:call-template name="ListSurnameStartsWith">
    <xsl:with-param name="character" select="$currToken"/>
  </xsl:call-template>
</xsl:for-each>

This is the template as it stands at the moment:

XML

<xsl:template name="ListSurnameStartsWith">
    <xsl:param name="character" select="''"/>
    <xsl:variable name="charMatch" select="/Users/User[starts-with(Surname, $character)]"/>    
    <xsl:variable name="charCount" select="count($charMatch)"/>
    
      <xsl:if test="$charCount > 0">
      <h2>
        <a href="javascript:void(0)" class="dsphead" onclick="dsp(this)">
          + <xsl:value-of select="$character"/> (<xsl:value-of select="$charCount"/>)
        </a>
      </h2>
      <div class="dspcont">
        <table border="1" cellpadding="1" cellspacing="1">
          <tr>
            <xsl:call-template name="HeadingRow"/>
          </tr>

          <xsl:for-each select="/Users/User[starts-with(Surname, $character)]">
            <xsl:sort select="Surname"/>
            <xsl:call-template name="UserRow"/>
          </xsl:for-each>
        </table>
      </div>
    </xsl:if>
  </xsl:template>

'currToken' is getting the value I want. It's landing in the 'ListSurnameStartsWith' template just fine, but 'charMatch' just comes back blank all the time, meaning I can't get a meaningful count to test whether the output needs to be generated. The later '

Richard Deeming14-Feb-17 2:32

14-Feb-17 2:32

Are you absolutely certain that the character parameter is getting the correct string value?

I've just tried the following simplified example, and it works as expected:

XML

<?xml version="1.0"?>
<bookstore>
    <book>
         <title>The Weather Pattern</title>
         <author>Weather Man</author>
         <price>100.00</price>
    </book>
    <book>
         <title>Weaving Patterns</title>
         <author>Weaver</author>
         <price>150.00</price>
    </book>
    <book>
         <title>Speech Pattern</title>
         <author>Speaker</author>
         <price>15.00</price>
    </book>
    <book>
         <title>Writing Style</title>
         <author>Writer</author>
         <price>1500.00</price>
    </book>
</bookstore>

XML

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:template match="/">
        <root>
            <xsl:apply-templates select="/bookstore">
                <xsl:with-param name="character" select="'W'" />
            </xsl:apply-templates>
        </root>
    </xsl:template>
    
    <xsl:template match="bookstore">
        <xsl:param name="character" select="''" />
        <xsl:variable name="charMatch" select="book[starts-with(title, $character)]"/>    
        <xsl:variable name="charCount" select="count($charMatch)"/>
        
        <result>
            <char><xsl:value-of select="$character" /></char>
            <count><xsl:value-of select="$charCount" /></count>
            <books>
                <xsl:for-each select="$charMatch">
                    <xsl:copy-of select="." />
                </xsl:for-each>
            </books>
        </result>
    </xsl:template>
</xsl:stylesheet>

"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer

bjmallon16-Feb-17 18:45

bjmallon

16-Feb-17 18:45

Richard,

Thanks for your reply.

Yes, if I pass 'A' to the template, I get the correct answer. So that's a good thing.

Now, given a tokenised string stored in a variable (Letters):

XML

<xsl:variable name="Letters">
    <xsl:call-template name="tokenise">
        <xsl:with-param name="string" select ="'A B C D E F G H I J K L M N O P Q R S T U V W X Y Z'"/>
        <xsl:with-param name="delimeters" select="' '"/>
    </xsl:call-template>
</xsl:variable>

And trying to process each token through the template:

XML

<xsl:for-each select="msxsl:node-set($Letters)/token">
  <xsl:variable name="currToken">
    <xsl:value-of select="./node()[1]"/>
  </xsl:variable>
  <xsl:call-template name="ListSurnameStartsWith">
    <xsl:with-param name="character" select="$currToken"/>
  </xsl:call-template>
</xsl:for-each>

What should I be using inside the 'select' attribute to ensure that i get 'A' and not some other value.

Some options I've tried, debugging in Visual Studio 2013, are:

XML

<xsl:variable name="currToken">
  <xsl:value-of select="."/>
</xsl:variable>

This gives me:

$currToken	{Dimension:[1]}
    [1]	/
        node()[1]	A

'select="./node()"' gives me the same answer.
'select="./node()[1]"' gives me the same answer.

I haven't had any help from Google, as I can't seem to ask the right question, or nobody's doing this.

What do I need to put into the select to actually get the character A.

Richard Deeming17-Feb-17 1:48

17-Feb-17 1:48

Have you tried:

./text()[1]

If that doesn't work, can you post your "tokenise" template?

"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer