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Using the first transformation from my answer to your "Looping templates"
question, and editing it a little we get the following: <xsl:stylesheet version="1.0" xmlns  sl="http://www.w3.org/1999/XSL/Transform"><xsl:template match="/"> <html> <body> <xsl:apply-templates/> </body> </html> </xsl:template> <xsl:template match="*[not(self::node) and node]"> <table border="1"> <tr> <xsl:apply-templates select="node"/> </tr> </table> </xsl:template> <xsl:template match="node"> <tr> <td ><xsl:value-of select="name()"/></td> <td><xsl:value-of select="@no"/></td> <xsl:if test="node"> <tr> <td>   </td> <td> <table border="1"> <xsl:apply-templates select="node"/> </table> </td> </tr> </xsl:if> </tr> </xsl:template> </xsl:stylesheet> When applied on this source.xml: <struct> <node level="1" no="1"> <node level="2" no="2" /> <node level="2" no="3"> <node level="3" no="4"/> </node> <node level="2" no="5" /> </node> <node level="1" no="6"> <node level="2" no="7"> <node level="3" no="8"/> <node level="3" no="9"/> </node> <node level="2" no="10" /> <node level="2" no="11" /> </node> </struct> the wanted result is produced (as displayed by a browser): node 1 node 2 node 3 node 4 node 5 node 6 node 7 node 8 node 9 node 10 node 11 Hope this helped. Dimitre Novatchev. FXSL developer, XML Insider, http://fxsl.sourceforge.net/ -- the home of FXSL Resume: http://fxsl.sf.net/DNovatchev/Resume/Res.html "Ruthless" <ruthless@NO_SPAM.poczta.onet.pl> wrote in message news:bsre5q$9ms$... > hello. > > All XML and XSLT are processed by preprocessor as a trees. > > How can i simply display my XML as some kind of tree. > > given xml: > > <struct> > <node level="1" no="1"> > <node level="2" no="2" /> > </node> > <node level="1" no="5"> > <node level="2" no="8" /> > </node> > </struct> > > given xslt: > <xsl:template match="struct"> > <html> > <body> > <xsl:for-each select="node"> > <xsl:sort select="@level"/> > <blockquote><pre> > <xsl:value-of select="name()"/> <br/> > <xsl:value-of select="@no"/> <br/> > <xsl:value-of select="@level"/> > </pre></blockquote> > </xsl:for-each> > </body> > </html> > </xsl:template> > > i'd like with a little help of <blockquote> do indentation > and as a result sth like this: > > node 1 > node 2 > node 3 > node 4 > node 2 > node 1 > > but my xslt does it linear and the nodes of all levels are in the same > position: > [..] > node 4 > node 2 > node 1 > > thanx in advance > greetings R > > > --- > Outgoing mail is certified Virus Free. > Checked by AVG anti-virus system (http://www.grisoft.com). > Version: 6.0.554 / Virus Database: 346 - Release Date: 03-12-20 > > |
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thanks again ;D
greetings R Użytkownik "Dimitre Novatchev" <> napisał w wiadomości news:bsrl4i$l82u$... > Using the first transformation from my answer to your "Looping templates" > question, and editing it a little we get the following: > > > > <xsl:stylesheet version="1.0" > xmlns sl="http://www.w3.org/1999/XSL/Transform">> > <xsl:template match="/"> > <html> > <body> > <xsl:apply-templates/> > </body> > </html> > </xsl:template> > > <xsl:template match="*[not(self::node) and node]"> > <table border="1"> > <tr> > <xsl:apply-templates select="node"/> > </tr> > </table> > </xsl:template> > > <xsl:template match="node"> > <tr> > <td ><xsl:value-of select="name()"/></td> > <td><xsl:value-of select="@no"/></td> > <xsl:if test="node"> > <tr> > <td>   </td> > <td> > <table border="1"> > <xsl:apply-templates select="node"/> > </table> > </td> > </tr> > </xsl:if> > </tr> > </xsl:template> > </xsl:stylesheet> > > > When applied on this source.xml: > > <struct> > <node level="1" no="1"> > <node level="2" no="2" /> > <node level="2" no="3"> > <node level="3" no="4"/> > </node> > <node level="2" no="5" /> > </node> > <node level="1" no="6"> > <node level="2" no="7"> > <node level="3" no="8"/> > <node level="3" no="9"/> > </node> > <node level="2" no="10" /> > <node level="2" no="11" /> > </node> > </struct> > > > the wanted result is produced (as displayed by a browser): > > > > node 1 > > > node 2 > > node 3 > > > node 4 > > > > node 5 > > > > node 6 > > > node 7 > > > node 8 > > node 9 > > > > node 10 > > node 11 > > > > Hope this helped. > > Dimitre Novatchev. > FXSL developer, XML Insider, > > http://fxsl.sourceforge.net/ -- the home of FXSL > Resume: http://fxsl.sf.net/DNovatchev/Resume/Res.html > > > > > > > "Ruthless" <ruthless@NO_SPAM.poczta.onet.pl> wrote in message > news:bsre5q$9ms$... > > hello. > > > > All XML and XSLT are processed by preprocessor as a trees. > > > > How can i simply display my XML as some kind of tree. > > > > given xml: > > > > <struct> > > <node level="1" no="1"> > > <node level="2" no="2" /> > > </node> > > <node level="1" no="5"> > > <node level="2" no="8" /> > > </node> > > </struct> > > > > given xslt: > > <xsl:template match="struct"> > > <html> > > <body> > > <xsl:for-each select="node"> > > <xsl:sort select="@level"/> > > <blockquote><pre> > > <xsl:value-of select="name()"/> <br/> > > <xsl:value-of select="@no"/> <br/> > > <xsl:value-of select="@level"/> > > </pre></blockquote> > > </xsl:for-each> > > </body> > > </html> > > </xsl:template> > > > > i'd like with a little help of <blockquote> do indentation > > and as a result sth like this: > > > > node 1 > > node 2 > > node 3 > > node 4 > > node 2 > > node 1 > > > > but my xslt does it linear and the nodes of all levels are in the same > > position: > > [..] > > node 4 > > node 2 > > node 1 > > > > thanx in advance > > greetings R > > > > > > --- > > Outgoing mail is certified Virus Free. > > Checked by AVG anti-virus system (http://www.grisoft.com). > > Version: 6.0.554 / Virus Database: 346 - Release Date: 03-12-20 > > > > > > > --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.554 / Virus Database: 346 - Release Date: 03-12-20 |
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I thought about my example and thought about what might happen if e.g.
element node will be inside some other hierarchy. For instance - i took my doc.xml and created family.xml I took <node> as <person>. I've created generations, mariages, singles, and children(mariages) for new recursive generations. Still does node has its own hierarchy inside generation hierarchy(i hope i made myself clear  )I tried to branch(display) only the <person> given xml: <?xml version="1.0" encoding="iso-8859-2"?> <?xml-stylesheet type="text/xsl" href="style.xsl"?> <tree> <generation level="1"> <mariage> <person> <first_name>Aaa</first_name> <last_name>Qwq</last_name> </person> <person> <first_name>Bbb</first_name> <last_name>aaa</last_name> </person> <children> <generation level="2"> <mariage> <person> <first_name>MMM</first_name> <last_name>qqq</last_name> </person> <person> <first_name>P</first_name> <last_name>K</last_name> </person> <children/> </mariage> <single> <person> <first_name>P</first_name> <last_name>ww</last_name> </person> </single> </generation> </children> </mariage> <single> <person> <first_name>P</first_name> <last_name>ww</last_name> </person> </single> </generation> </tree> and yours the stylesheet(a bit modified): <xsl:stylesheet version="1.0" xmlns sl="http://www.w3.org/1999/XSL/Transform"><xsl:template match="/"> <html> <body> <xsl:apply-templates/> </body> </html> </xsl:template> <xsl:template match="*[not(self:  erson) and person]"><table border="1"> <tr> <xsl:apply-templates select="person"/> </tr> </table> </xsl:template> <xsl:template match="person"> <tr> <td ><xsl:value-of select="name()"/></td> <td><xsl:value-of select="last_name"/></td> <td><xsl:value-of select="first_name"/></td> <xsl:if test="person"> <tr> <td>   </td> <td> <table border="1"> <xsl:apply-templates select="person"/> </table> </td> </tr> </xsl:if> </tr> </xsl:template> </xsl:stylesheet> I tried to display all the persons - but only top generation(level='1') was displayed I don't know how to ommit all those unnecessary(from my point of view, as far as i'm only interested in persons) elements and simply display only persons as a branching table. thanks in advance you already helped me a lot with the understanding the XSLT greetings R --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.554 / Virus Database: 346 - Release Date: 03-12-20 |
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"Ruthless" <ruthless@NO_SPAM.poczta.onet.pl> wrote in message news:bssk8l$9a1$... > I thought about my example and thought about what might happen if e.g. > element node will be inside some other hierarchy. > > For instance - i took my doc.xml and created family.xml > > I took <node> as <person>. I've created generations, mariages, singles, and > children(mariages) for new recursive generations. > > Still does node has its own hierarchy inside generation hierarchy(i hope i > made myself clear )> > I tried to branch(display) only the <person> > > given xml: > > <?xml version="1.0" encoding="iso-8859-2"?> > <?xml-stylesheet type="text/xsl" href="style.xsl"?> > <tree> > <generation level="1"> > <mariage> > <person> > <first_name>Aaa</first_name> > <last_name>Qwq</last_name> > </person> > <person> > <first_name>Bbb</first_name> > <last_name>aaa</last_name> > </person> > <children> > <generation level="2"> > <mariage> > <person> > <first_name>MMM</first_name> > <last_name>qqq</last_name> > </person> > <person> > <first_name>P</first_name> > <last_name>K</last_name> > </person> > <children/> > </mariage> > <single> > <person> > <first_name>P</first_name> > <last_name>ww</last_name> > </person> > </single> > </generation> > </children> > </mariage> > <single> > <person> > <first_name>P</first_name> > <last_name>ww</last_name> > </person> > </single> > > </generation> > </tree> > > and yours the stylesheet(a bit modified): > > <xsl:stylesheet version="1.0" > xmlns sl="http://www.w3.org/1999/XSL/Transform">> > <xsl:template match="/"> > <html> > <body> > <xsl:apply-templates/> > </body> > </html> > </xsl:template> > > <xsl:template match="*[not(self: erson) and person]">> <table border="1"> > <tr> > <xsl:apply-templates select="person"/> > </tr> > </table> > </xsl:template> > > <xsl:template match="person"> > <tr> > <td ><xsl:value-of select="name()"/></td> > <td><xsl:value-of select="last_name"/></td> > <td><xsl:value-of select="first_name"/></td> > <xsl:if test="person"> > <tr> > <td>   </td> > <td> > <table border="1"> > <xsl:apply-templates select="person"/> > </table> > </td> > </tr> > </xsl:if> > </tr> > </xsl:template> > </xsl:stylesheet> > > I tried to display all the persons - but only top generation(level='1') was > displayed > > I don't know how to ommit all those unnecessary(from my point of view, as > far as i'm only interested in persons) elements and simply display only > persons as a branching table. > > thanks in advance > you already helped me a lot with the understanding the XSLT Dear R, Yes, this is possible. I started with a transformation, which produces a nice hierarchical html display of all elements in the xml tree. Then I masked all elements, whose name is not the same as the value of a special xsl aram.Here's the result: <xsl:stylesheet version="1.0" xmlns sl="http://www.w3.org/1999/XSL/Transform"><xsl aram name="pNodeName" select="'person'"/><xsl:template match="/"> <html> <body> <xsl:apply-templates/> </body> </html> </xsl:template> <xsl:template match="*"> <tr> <td > <xsl:if test="name() = $pNodeName"> <xsl:value-of select="name()"/> </xsl:if> <xsl:text> </xsl:text> </td> <td> <xsl:if test="name() = $pNodeName"> <xsl:variable name="vNodeNum"> <xsl:number count="*" level="multiple"/> </xsl:variable> <xsl:value-of select="$vNodeNum"/> </xsl:if> <xsl:text> </xsl:text> </td> <xsl:if test="*"> <tr> <td>   </td> <td> <table border="1"> <xsl:apply-templates select="*"/> </table> </td> </tr> </xsl:if> </tr> </xsl:template> </xsl:stylesheet> Hope that this helped. Dimitre Novatchev. FXSL developer, XML Insider, http://fxsl.sourceforge.net/ -- the home of FXSL Resume: http://fxsl.sf.net/DNovatchev/Resume/Res.html |
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once again thanks
greetings R Użytkownik "Dimitre Novatchev" <> napisał w wiadomości news:bt4k5c$3bp61$... > > "Ruthless" <ruthless@NO_SPAM.poczta.onet.pl> wrote in message > news:bssk8l$9a1$... > > I thought about my example and thought about what might happen if e.g. > > element node will be inside some other hierarchy. > > > > For instance - i took my doc.xml and created family.xml > > > > I took <node> as <person>. I've created generations, mariages, singles, > and > > children(mariages) for new recursive generations. > > > > Still does node has its own hierarchy inside generation hierarchy(i hope i > > made myself clear )> > > > I tried to branch(display) only the <person> > > > > given xml: > > > > <?xml version="1.0" encoding="iso-8859-2"?> > > <?xml-stylesheet type="text/xsl" href="style.xsl"?> > > <tree> > > <generation level="1"> > > <mariage> > > <person> > > <first_name>Aaa</first_name> > > <last_name>Qwq</last_name> > > </person> > > <person> > > <first_name>Bbb</first_name> > > <last_name>aaa</last_name> > > </person> > > <children> > > <generation level="2"> > > <mariage> > > <person> > > <first_name>MMM</first_name> > > <last_name>qqq</last_name> > > </person> > > <person> > > <first_name>P</first_name> > > <last_name>K</last_name> > > </person> > > <children/> > > </mariage> > > <single> > > <person> > > <first_name>P</first_name> > > <last_name>ww</last_name> > > </person> > > </single> > > </generation> > > </children> > > </mariage> > > <single> > > <person> > > <first_name>P</first_name> > > <last_name>ww</last_name> > > </person> > > </single> > > > > </generation> > > </tree> > > > > and yours the stylesheet(a bit modified): > > > > <xsl:stylesheet version="1.0" > > xmlns sl="http://www.w3.org/1999/XSL/Transform">> > > > <xsl:template match="/"> > > <html> > > <body> > > <xsl:apply-templates/> > > </body> > > </html> > > </xsl:template> > > > > <xsl:template match="*[not(self: erson) and person]">> > <table border="1"> > > <tr> > > <xsl:apply-templates select="person"/> > > </tr> > > </table> > > </xsl:template> > > > > <xsl:template match="person"> > > <tr> > > <td ><xsl:value-of select="name()"/></td> > > <td><xsl:value-of select="last_name"/></td> > > <td><xsl:value-of select="first_name"/></td> > > <xsl:if test="person"> > > <tr> > > <td>   </td> > > <td> > > <table border="1"> > > <xsl:apply-templates select="person"/> > > </table> > > </td> > > </tr> > > </xsl:if> > > </tr> > > </xsl:template> > > </xsl:stylesheet> > > > > I tried to display all the persons - but only top generation(level='1') > was > > displayed > > > > I don't know how to ommit all those unnecessary(from my point of view, as > > far as i'm only interested in persons) elements and simply display only > > persons as a branching table. > > > > thanks in advance > > you already helped me a lot with the understanding the XSLT > > Dear R, > > Yes, this is possible. > > I started with a transformation, which produces a nice hierarchical html > display of all elements in the xml tree. > > Then I masked all elements, whose name is not the same as the value of a > special xsl aram.> > Here's the result: > > <xsl:stylesheet version="1.0" > xmlns sl="http://www.w3.org/1999/XSL/Transform">> > <xsl aram name="pNodeName" select="'person'"/>> > <xsl:template match="/"> > <html> > <body> > <xsl:apply-templates/> > </body> > </html> > </xsl:template> > > <xsl:template match="*"> > <tr> > <td > > <xsl:if test="name() = $pNodeName"> > <xsl:value-of select="name()"/> > </xsl:if> > <xsl:text> </xsl:text> > </td> > <td> > <xsl:if test="name() = $pNodeName"> > <xsl:variable name="vNodeNum"> > <xsl:number count="*" level="multiple"/> > </xsl:variable> > <xsl:value-of select="$vNodeNum"/> > </xsl:if> > <xsl:text> </xsl:text> > </td> > <xsl:if test="*"> > <tr> > <td>   </td> > <td> > <table border="1"> > <xsl:apply-templates select="*"/> > </table> > </td> > </tr> > </xsl:if> > </tr> > </xsl:template> > </xsl:stylesheet> > > > Hope that this helped. > > > Dimitre Novatchev. > FXSL developer, XML Insider, > > http://fxsl.sourceforge.net/ -- the home of FXSL > Resume: http://fxsl.sf.net/DNovatchev/Resume/Res.html > > --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.554 / Virus Database: 346 - Release Date: 03-12-20 |
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