tags:

views:

72

answers:

2
+1  Q: 

Xslt for renaming a list

I have this XML:

<Root>
    <Tag>
        <Childlist>
            <I>1</I>
            <Dlist>2</Dlist>
            <Dlist>3</Dlist>
            <Dlist>4</Dlist>
        </Childlist>
        <Childlist>
            <I>11</I>
            <Dlist>2</Dlist>
            <Dlist>3</Dlist>
            <Dlist>4</Dlist>
        </Childlist>
    </Tag>
</Root>

and I want to convert this to:

<Root>
    <Tag>
        <Childlist>
            <item>
                <I>1</I>
                <Dlist>
                    <item>2</item>
                    <item>3</item>
                    <item>4</item>
                </Dlist>
            </item>
            <item>
                <I>11</I>
                <Dlist>
                    <item>2</item>
                    <item>3</item>
                    <item>4</item>
                </Dlist>
            </item>
        </Childlist>
    </Tag>
</Root>

using XSLT. As you can see the repeating tags in first XML gets replaced with tag and list tag is made as parent tag.

Can anybody please advise an xslt for this conversion?

Thanks in advance.

Edit: From comments.

I have a problem with this xls. If i add any tag below Dlist it gets appended to Dlist

There are going to be other preceding and following elements [from Dlist]

+1  A: 

This transformation:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:key name="kDlist" match="Dlist"
 use="generate-id((.|preceding-sibling::Dlist)[1])"/>

 <xsl:template match="node()|@*">
  <xsl:copy>
   <xsl:apply-templates select="node()|@*"/>
  </xsl:copy>
 </xsl:template>

 <xsl:template match="*[Childlist]">
  <xsl:copy>
   <xsl:copy-of select="@*"/>
   <Childlist>
    <xsl:apply-templates/>
   </Childlist>
  </xsl:copy>
 </xsl:template>

 <xsl:template match="Childlist|Dlist" name="makeItem">
  <Item>
   <xsl:apply-templates/>
  </Item>
 </xsl:template>

 <xsl:template match="Dlist[not(preceding-sibling::Dlist)]">
  <Dlist>
   <xsl:apply-templates select="key('kDlist', generate-id())" mode="copy"/>
  </Dlist>
 </xsl:template>

 <xsl:template match="Dlist" mode="copy">
  <xsl:call-template name="makeItem"/>
 </xsl:template>

 <xsl:template match="Dlist"/>
</xsl:stylesheet>

when applied on this XML document (the original text, corrected to be well-formed):

<Root>
    <Tag>
        <Childlist>
            <I>1</I>
            <Dlist>2</Dlist>
            <Dlist>3</Dlist>
            <Dlist>4</Dlist>
        </Childlist>
        <Childlist>
            <I>11</I>
            <Dlist>2</Dlist>
            <Dlist>3</Dlist>
            <Dlist>4</Dlist>
        </Childlist>
    </Tag>
</Root>

produces the wanted, correct result:

<Root>
    <Tag>
        <Childlist>
            <Item>
                <I>1</I>
                <Dlist>
                    <Item>2</Item>
                    <Item>3</Item>
                    <Item>4</Item>
                </Dlist>
            </Item>
            <Item>
                <I>11</I>
                <Dlist>
                    <Item>2</Item>
                    <Item>3</Item>
                    <Item>4</Item>
                </Dlist>
            </Item>
        </Childlist>
    </Tag>
</Root>
Dimitre Novatchev  2010-09-07 18:04:44
@Dimitre: +1 Good answer! This remains me http://stackoverflow.com/questions/3652194/wrap-group-of-xml-nodes/3653382 . So I'm adding one, too.
Alejandro  2010-09-07 20:04:03
@Dimitre: A minor: I think you should strip `Dlist` from pattern `Childlist|Dlist` just for clarity, anyway error recovery mechanism will select last empty template matching `Dlist`
Alejandro  2010-09-07 20:13:48
@Dimitre: Thanks for your reply. :)
Syam  2010-09-08 14:04:30
+1  A: 

Besides Dimitre's excellent answer, this stylesheet:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
    <xsl:template match="node()|@*">
        <xsl:copy>
            <xsl:apply-templates select="node()[1]|@*"/>
        </xsl:copy>
        <xsl:apply-templates select="following-sibling::node()[1]"/>
    </xsl:template>
    <xsl:template match="Tag">
        <Tag>
            <Childlist>
                <xsl:apply-templates select="node()[1]"/>
            </Childlist>
        </Tag>
    </xsl:template>
    <xsl:template match="*/Dlist[1]">
        <xsl:copy>
            <xsl:call-template name="makeItem"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="Childlist|Dlist" name="makeItem">
        <Item>
            <xsl:apply-templates select="node()[1]"/>
        </Item>
        <xsl:apply-templates select="following-sibling::node()[1]"/>
    </xsl:template>
</xsl:stylesheet>

Output:

<Root>
    <Tag>
        <Childlist>
            <Item>
                <I>1</I>
                <Dlist>
                    <Item>2</Item>
                    <Item>3</Item>
                    <Item>4</Item>
                </Dlist>
            </Item>
            <Item>
                <I>11</I>
                <Dlist>
                    <Item>2</Item>
                    <Item>3</Item>
                    <Item>4</Item>
                </Dlist>
            </Item>
        </Childlist>
    </Tag>
</Root>

Edit: With this input:

<Root>
    <Tag>
        <Childlist>
            <I>1</I>
            <Dlist>2</Dlist>
            <Dlist>3</Dlist>
            <Dlist>4</Dlist>
            <F>1</F>
        </Childlist>
        <Childlist>
            <I>11</I>
            <Dlist>2</Dlist>
            <Dlist>3</Dlist>
            <Dlist>4</Dlist>
            <F>11</F>
        </Childlist>
    </Tag>
</Root>

This stylesheet:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
    <xsl:template match="node()|@*">
        <xsl:copy>
            <xsl:apply-templates select="node()[1]|@*"/>
        </xsl:copy>
        <xsl:apply-templates select="following-sibling::node()[1]"/>
    </xsl:template>
    <xsl:template match="Tag">
        <Tag>
            <Childlist>
                <xsl:apply-templates select="node()[1]"/>
            </Childlist>
        </Tag>
    </xsl:template>
    <xsl:template match="*/Dlist[last()]" name="makeItem">
        <Item>
            <xsl:apply-templates select="node()[1]"/>
        </Item>
    </xsl:template>
    <xsl:template match="*/Dlist[1]">
        <xsl:copy>
            <xsl:call-template name="wrap"/>
        </xsl:copy>
        <xsl:apply-templates select="following-sibling::node()
                                         [not(self::Dlist)][1]"/>
    </xsl:template>
    <xsl:template match="Childlist|Dlist" name="wrap">
        <xsl:call-template name="makeItem"/>
        <xsl:apply-templates select="following-sibling::node()[1]"/>
    </xsl:template>
</xsl:stylesheet>

Output:

<Root>
    <Tag>
        <Childlist>
            <Item>
                <I>1</I>
                <Dlist>
                    <Item>2</Item>
                    <Item>3</Item>
                    <Item>4</Item>
                </Dlist>
                <F>1</F>
            </Item>
            <Item>
                <I>11</I>
                <Dlist>
                    <Item>2</Item>
                    <Item>3</Item>
                    <Item>4</Item>
                </Dlist>
                <F>11</F>
            </Item>
        </Childlist>
    </Tag>
</Root>

Note: This assumes Dlist elements are all next siblings. So Dlist[1] opens the new level and after that apply templates to next no Dlist node, and Dlist[last()] close the level not applying templates to next sibling.

Alejandro  2010-09-07 20:05:32
@Alejandro: You know I have a crush on fine-grained identity? :) +1.
Dimitre Novatchev  2010-09-07 20:38:06
@Dimitre: I also like it very much, ja! I should check if this pattern follows the restrictions for streaming in XSLT 2.1... I think it doesn't.
Alejandro  2010-09-07 20:45:57
@Alejandro: Could you please give and xsl to do the reverse conversion? Also could you suggest some good tutorials for xslt? :)
Syam  2010-09-08 13:45:21
@Syam: Reverse transformation is very straightforward. Use identity transform, strip `Childlist|Dlist` with empty template, transform `Childlist/Item` to `Childlist` and `Dlist/Item` to `Dlist`. First thing for learning XSLT is to grasp declarative paradigm (maybe Haskell, Miranda or some general paper) then to study XSLT and XPath specs, and last check FAQs http://www.dpawson.co.uk/xsl/ XSLT List http://www.mulberrytech.com/xsl/xsl-list/ SO Top Answers blogs, etc.
Alejandro  2010-09-08 14:26:24
@Alejandro: Thanks a Million.. :)
Syam  2010-09-08 17:01:18
@Syam: You're wellcome! Ask any time.
Alejandro  2010-09-08 17:38:40
@Alejandro: I have a problem with this xls. If i add any tag below Dlist it gets appended to Dlist.. Any idea!! Sorry to disturn you again.. :(
Syam  2010-09-09 14:12:48
@Syam: No problem, but good practice is to ask another question. You need to clarify if `Dlist` elements are going to be all siblings and there are going to be other preceding and following elements or there are goint to be other elements between `Dlist` elements. For both schemas, modifications are going to be minor but diferent.
Alejandro  2010-09-09 15:01:12
@Alejandro: There are going to be other preceding and following elements
Syam  2010-09-09 16:41:14
@Syam: Check my update.
Alejandro  2010-09-09 20:43:45

related questions

xslt and xpath: match conditionally upon current node value
How can I make an exact copy of a xml node's children with XSLT?
How to escape XML content with XSL to safely output it as JSON?
How do I convert a .docx to html using asp.net?
Using XSLT, How Do You Insert XML Into an existing XML node
How can I lay images out in a grid using XSL-FO?
How to validate an XML file against a schema using Visual Studio 2005
What is a good resource for learning XSL?
xslt: apply-templates in reverse order
How do you add an image in XSLT?
If you were programming a calendar in HTML would you use Table tags or Div tags?
How to Convert a StreamReader into an XMLReader object in .Net 2.0/C#
Can I create a value for a missing tag in XPath?
ASP.NET MVC vs. XSL
How to apply an XSLT Stylesheet in C#
XSLT dividing a list of nodes in half
XSLT nodesets Length
using a html entity in xslt (e.g. &nbsp;)
XSLT Find and Replace with Unique
XSL Dynamic Element Names
XSLT - Reverse Find in a string
VS2008 SP1 crashes when debugging an XSLT file
Is it "bad practice" to be sensitive to linebreaks in XML documents?
How do you use a variable in xsl when trying to select a node?
.Net XML comment into API Documentation