Hi Im new to Umbraco, but i have been trying to make a tree menu using default XSLT without success.
Im having two different top menus and i want my left tree submeny starting from level 3 frum the tree structure in Umbraco.
I have tested XSLT "list subfolder from current position" but the parent of the tree strucure will be lost. I have also tested to list the whole site using the default site script. When im using this I can´´t figure out how i can use it to list the tree structure starting from level 3 on different trees. (depending in wich item from menu 2 that is selected) I have tested to list subfolder from current position starting from level3, but then the parents are also lost.
Menu1 Menu 1 Menu1 Menu 1
Menu2 Menu2 Menu 2 Menu 2
Menu3
Submenu3
Menu3
Submenu3
Menu3
Submenu3
is there any one that has the solution to my proble?
I have you tried the Navigation prototype and set it to start at level 3. An see if this can solve your problem.
<xsl:variable name="level" select="3"/>
<xsl:template match="/">
<!-- The fun starts here --> <ul> <xsl:for-each select="$currentPage/ancestor-or-self::* [@isDoc and @level=$level]/* [@isDoc and string(umbracoNaviHide) != '1']"> <li> <a href="{umbraco.library:NiceUrl(@id)}"> <xsl:if test="$currentPage/ancestor-or-self::*/@id = current()/@id"> <!-- we're under the item - you can do your own styling here --> <xsl:attribute name="class">selected</xsl:attribute> </xsl:if> <xsl:value-of select="@nodeName"/> </a> </li> </xsl:for-each> </ul>
I can´t get this navigation to work. No one that has managed to get this type of navigation to workand that has an exmaple? Menus below leevel 3 will not be displayed.
When i open a menu, i want the parent of menu3 to be displayed same time as its children are.
Anyone out there who has the solution to this problem?
<!-- Input the documenttype you want here --> <!-- Typically '1' for topnavigtaion and '2' for 2nd level --> <!-- Use div elements around this macro combined with css --> <!-- for styling the navigation --> <xsl:variable name="level" select="3"/>
<xsl:template match="/">
<!-- The fun starts here --> <ul> <xsl:for-each select="$currentPage/ancestor-or-self::* [@isDoc and @level=$level]/* [@isDoc and string(umbracoNaviHide) != '1']"> <li> <a href="{umbraco.library:NiceUrl(@id)}"> <xsl:if test="$currentPage/ancestor-or-self::*/@id = current()/@id"> <!-- we're under the item - you can do your own styling here --> <xsl:attribute name="class">selected</xsl:attribute> </xsl:if> <xsl:value-of select="@nodeName"/> </a> </li> </xsl:for-each> </ul>
create tree menu starting from level 3
Hi Im new to Umbraco, but i have been trying to make a tree menu using default XSLT without success.
Im having two different top menus and i want my left tree submeny starting from level 3 frum the tree structure in Umbraco.
I have tested XSLT "list subfolder from current position" but the parent of the tree strucure will be lost. I have also tested to list the whole site using the default site script. When im using this I can´´t figure out how i can use it to list the tree structure starting from level 3 on different trees. (depending in wich item from menu 2 that is selected) I have tested to list subfolder from current position starting from level3, but then the parents are also lost.
Menu1 Menu 1 Menu1 Menu 1
Menu2 Menu2 Menu 2 Menu 2
Menu3
Submenu3
Menu3
Submenu3
Menu3
Submenu3
is there any one that has the solution to my proble?
BR Gustav
Hi Gustav,
I have you tried the Navigation prototype and set it to start at level 3. An see if this can solve your problem.
Hope this helps,
/Dennis
HI Dennis and thanks for your help!
The navigation prototype will only list items from level 3 and not it´s children. Any suggestion what im doing wrong?
/gustav
Hi Gustav.
Try to see if this post can pull you in the right direction.
http://our.umbraco.org/forum/developers/xslt/27134-display-the-parent-and-child-nodes-in-a-navigation
Hope this helps you,
/Dennis
I can´t get this navigation to work. No one that has managed to get this type of navigation to workand that has an exmaple? Menus below leevel 3 will not be displayed.
When i open a menu, i want the parent of menu3 to be displayed same time as its children are.
Anyone out there who has the solution to this problem?
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xsl:stylesheet [ <!ENTITY nbsp " "> ]>
<xsl:stylesheet
version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:msxml="urn:schemas-microsoft-com:xslt"
xmlns:umbraco.library="urn:umbraco.library" xmlns:Exslt.ExsltCommon="urn:Exslt.ExsltCommon" xmlns:Exslt.ExsltDatesAndTimes="urn:Exslt.ExsltDatesAndTimes" xmlns:Exslt.ExsltMath="urn:Exslt.ExsltMath" xmlns:Exslt.ExsltRegularExpressions="urn:Exslt.ExsltRegularExpressions" xmlns:Exslt.ExsltStrings="urn:Exslt.ExsltStrings" xmlns:Exslt.ExsltSets="urn:Exslt.ExsltSets" xmlns:Examine="urn:Examine"
exclude-result-prefixes="msxml umbraco.library Exslt.ExsltCommon Exslt.ExsltDatesAndTimes Exslt.ExsltMath Exslt.ExsltRegularExpressions Exslt.ExsltStrings Exslt.ExsltSets Examine ">
<xsl:output method="xml" omit-xml-declaration="yes" />
<xsl:param name="currentPage"/>
<!-- Input the documenttype you want here -->
<!-- Typically '1' for topnavigtaion and '2' for 2nd level -->
<!-- Use div elements around this macro combined with css -->
<!-- for styling the navigation -->
<xsl:variable name="level" select="3"/>
<xsl:template match="/">
<!-- The fun starts here -->
<ul>
<xsl:for-each select="$currentPage/ancestor-or-self::* [@isDoc and @level=$level]/* [@isDoc and string(umbracoNaviHide) != '1']">
<li>
<a href="{umbraco.library:NiceUrl(@id)}">
<xsl:if test="$currentPage/ancestor-or-self::*/@id = current()/@id">
<!-- we're under the item - you can do your own styling here -->
<xsl:attribute name="class">selected</xsl:attribute>
</xsl:if>
<xsl:value-of select="@nodeName"/>
</a>
</li>
</xsl:for-each>
</ul>
</xsl:template>
</xsl:stylesheet>
BR
Gustav
is working on a reply...