XML to XSL

hello,
can anyone please help me? I have a dtd and a xml file, how can I get a xsl or xslt using xml spy. I would like to be able to make an html file at the end. Thank you.

---------------------dtd --------------

<?xml version="1.0" encoding="UTF-8"?>
<!ELEMENT THINGS (STUFF, MORESTUFF+, LOTSOFSTUFF)>
<!ELEMENT LOTSOFSTUFF (stuffa, stuffb, stuffc)>
<!ELEMENT MORESTUFF (stuffa, stuffb, stuffc, stuffd?)>
<!ELEMENT STUFF (stuffa, stuffb, stuffc)>
<!ELEMENT stuffa (#PCDATA)>
<!ELEMENT stuffb (#PCDATA)>
<!ELEMENT stuffc (#PCDATA)>
<!ELEMENT stuffd EMPTY>

----------------------------------------------

----------------------xml--------------------

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE THINGS SYSTEM "C:\STUFF.dtd">
<THINGS>
<STUFF>
<stuffa>STUFF1</stuffa>
<stuffb>STUFF2</stuffb>
<stuffc>STUFF3</stuffc>
</STUFF>
<MORESTUFF>
<stuffa>MORE STUFF A</stuffa>
<stuffb>MORE STUFF B</stuffb>
<stuffc>MORE STUFF C</stuffc>
<stuffd/>
</MORESTUFF>
<LOTSOFSTUFF>
<stuffa>LOTS OF STUFF A </stuffa>
<stuffb>LOTS OF STUFF B</stuffb>
<stuffc>LOTS OF STUFF C</stuffc>
</LOTSOFSTUFF>
</THINGS>

--------------------------------------------------

Oct 21 '07 #1

Subscribe Post Reply

1369

JosAH

11,448

Expert 8TB

Your question doesn't belong in the XML Articles section, it belongs in the XML
Forums section instead. I'll move your question for you.

kind regards,

Jos

Oct 21 '07 #2

Dököll

2,364

Expert 2GB

hello,
can anyone please help me? I have a dtd and a xml file, how can I get a xsl or xslt using xml spy. I would like to be able to make an html file at the end. Thank you.

---------------------dtd --------------

<?xml version="1.0" encoding="UTF-8"?>
<!ELEMENT THINGS (STUFF, MORESTUFF+, LOTSOFSTUFF)>
<!ELEMENT LOTSOFSTUFF (stuffa, stuffb, stuffc)>
<!ELEMENT MORESTUFF (stuffa, stuffb, stuffc, stuffd?)>
<!ELEMENT STUFF (stuffa, stuffb, stuffc)>
<!ELEMENT stuffa (#PCDATA)>
<!ELEMENT stuffb (#PCDATA)>
<!ELEMENT stuffc (#PCDATA)>
<!ELEMENT stuffd EMPTY>

----------------------------------------------

----------------------xml--------------------

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE THINGS SYSTEM "C:\STUFF.dtd">
<THINGS>
<STUFF>
<stuffa>STUFF1</stuffa>
<stuffb>STUFF2</stuffb>
<stuffc>STUFF3</stuffc>
</STUFF>
<MORESTUFF>
<stuffa>MORE STUFF A</stuffa>
<stuffb>MORE STUFF B</stuffb>
<stuffc>MORE STUFF C</stuffc>
<stuffd/>
</MORESTUFF>
<LOTSOFSTUFF>
<stuffa>LOTS OF STUFF A </stuffa>
<stuffb>LOTS OF STUFF B</stuffb>
<stuffc>LOTS OF STUFF C</stuffc>
</LOTSOFSTUFF>
</THINGS>

--------------------------------------------------

What are you trying to do?

Oct 21 '07 #3

seanh

What are you trying to do?

I am trying to create an xml transformation to a stylesheet using xml spy

the ouput that I would like to get is something like:

STUFF
stuff1
stuff2
stuff3

MORESTUFF
morestuffa
morestuffb
morestuffc

so I can view it in html.

thanks

Oct 22 '07 #4

seanh

Your question doesn't belong in the XML Articles section, it belongs in the XML
Forums section instead. I'll move your question for you.

kind regards,

Jos

Sorry...and thank you for moving my message to the right spot.

Oct 22 '07 #5

jkmyoung

2,057

Expert 2GB

Expand|Select|Wrap|Line Numbers

 
<?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" version="1.0" encoding="UTF-8" indent="yes"/>

    <xsl:template match="THINGS">

        <html>

            <head>

                <title/>

            </head>

            <body>

            <xsl:for-each select="*"><!-- header element -->

                <b><xsl:value-of select="name()"/></b><br/>

                <xsl:for-each select="*"> <!-- data element -->

                    <xsl:value-of select="."/><br/>

                </xsl:for-each>

            </xsl:for-each>

            </body>

        </html>

    </xsl:template>

</xsl:stylesheet>

Oct 22 '07 #6

seanh

Expand|Select|Wrap|Line Numbers

<?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" version="1.0" encoding="UTF-8" indent="yes"/>

 <xsl:template match="THINGS">

 <html>

 <head>

 <title/>

 </head>

 <body>

 <xsl:for-each select="*">

 <xsl:value-of select="name()"/> 

 <xsl:for-each select="*"> 

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

 </xsl:for-each>

 </xsl:for-each>

 </body>

 </html>

 </xsl:template>

</xsl:stylesheet>

woooow!!! THANK YOU, THANK YOU, THANK YOU MR.GURU jkmyoung.

I have one last question, if I add in the DTD
<!Element Image (#PCDATA)>

and in the xml file
<Image>picture.jpg</Image>

only the character data comes out "picture.jpg". but I would like the
picture to display, I tried putting in the dtd:

<!ELEMENT STUFF (stuffa, stuffb, stuffc, graphic)>

<!ELEMENT graphic EMPTY>

<!ATTLIST graphic image ENTITY #REQUIRED >
<!NOTATION JPG SYSTEM "Graphic">
<!ENTITY mypicture SYSTEM "picture1.jpg" NDATA JPG>

and in the xml file:

<graphic image="mypicture" />

but the image doesn't display.

my question is how do I display different images so I can view it in html

thank you

Oct 22 '07 #7

jkmyoung

2,057

Expert 2GB

instead of

Expand|Select|Wrap|Line Numbers

 <xsl:for-each select="*"> <!-- data element -->

    <xsl:value-of select="."/><br/>

</xsl:for-each>

use:

Expand|Select|Wrap|Line Numbers

 <xsl:for-each select="*"> <!-- data element -->

    <xsl:apply-templates/><br/>

</xsl:for-each>

and have

Expand|Select|Wrap|Line Numbers

 
<xsl:template match="Image">

  <img src="{.}"/>

</xsl:template>

OR

<xsl:template match="Graphic">

  <img src="{@image}"/>

</xsl:template

Oct 26 '07 #8

seanh

instead of

Expand|Select|Wrap|Line Numbers

<xsl:for-each select="*"> 

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

</xsl:for-each>

use:

Expand|Select|Wrap|Line Numbers

<xsl:for-each select="*"> 

 <xsl:apply-templates/> 

</xsl:for-each>

and have

Expand|Select|Wrap|Line Numbers

<xsl:template match="Image">

 <img src="{.}"/>

</xsl:template>

OR

<xsl:template match="Graphic">

 <img src="{@image}"/>

</xsl:template

WOW, Thank you for understanding my questions guru JKMYOUNG. Thank you so much for your time and response and thank you for sharing your gift of knowledge with others.

-Sean

Oct 26 '07 #9

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