fluid - Expanding the fluid_styled_content menu in TYPO3 7 -

-

i looking way expand fluid_styled_content menu in typo3 7, , not sure how proceed. example type1 menu has following code:

<ce:menu.directory pageuids="{pageuids}" as="pages"> <f:if condition="{pages}">     <ul class="ce-menu ce-menu-1">         <f:for each="{pages}" as="page">             <li>                 <f:link.page pageuid="{page.uid}">                     {page.title}                 </f:link.page>             </li>         </f:for>     </ul> </f:if> </ce:menu.directory>

now i'm looking way expand menu shows subpages when available.

if page has subpages, show list subpages (as nested ul tree).

how can done fluid?

i not know view helper using (<ce:menu.directory />). guess takes list of page uid´s , gets there children (menu.directory) , put them array/objectstorage ({pages}) - create array of selected page uid´s , answer not correct.

you can use view helper <f:debug>{variablename}</f:debug> debug have.

try this:

<ce:menu.directory pageuids="{pageuids}" as="pages">     <f:if condition="{pages}">         <ul class="ce-menu ce-menu-1">             <f:for each="{pages}" as="page">                 <li>                     <f:link.page pageuid="{page.uid}">                         {page.title}                     </f:link.page>                     <ce:menu.directory pageuids="{page.uid}" as="subpages">                         <f:if condition="{subpages}">                             <ul class="level-2">                                 <f:for each="{subpages}" as="subpage">                                     <li>                                         <f:link.page pageuid="{subpage.uid}">                                             {subpage.title}                                         </f:link.page>                                     </li>                                 </f:for>                             </ul>                         </f:if>                     </ce:menu.directory>                 </li>             </f:for>         </ul>     </f:if> </ce:menu.directory>

if above dosen´t work try dump {page} in debug tag , update question output.


Comments

Post a Comment

Popular posts from this blog

Hatching array of circles in AutoCAD using c# -

-
i beginner in autocad plugins .i have created array of circle using following code. now, have hatch circles in array .how that??how hatch circles @ once? moreover, how can return hatch method , use hatch objects.sorry language.thanks in advance. [assembly: commandclass(typeof(beamsection.class1))] namespace beamsection { public class class1 { //class1 obj = new class1(); [commandmethod("beamatsupport")] public void addlightweightpolyline() { // current document , database document acdoc = application.documentmanager.mdiactivedocument; database accurdb = acdoc.database; promptpointresult ppr = acdoc.editor.getpoint("\nselect starting point "); var ucs = acdoc.editor.currentusercoordinatesystem; point3d startingpt = ppr.value.transformby(ucs); list<double> radius = new list<double>(); radius.add(0.5);
Read more

ios - UITEXTFIELD InputView Uipicker not working in swift -

-
hello trying pop uipickerview view programmatically when user clicks on textfield. have tried doesn't doing anything. isn't working. nothing happening when click textfield class userprofiletableviewcontroller: uitableviewcontroller,uitextfielddelegate,uipickerviewdelegate,uipickerviewdatasource { var itempicker: uipickerview! = uipickerview() @iboutlet weak var gendertxtfield: uitextfield! var gender = ["male","female"] override func viewdidload() { super.viewdidload() gendertxtfield.delegate = self itempicker!.delegate = self itempicker!.datasource = self itempicker!.backgroundcolor = uicolor.blackcolor() self.gendertxtfield.inputview = itempicker } func numberofcomponentsinpickerview(pickerview: uipickerview) -> int{ return 1 } // returns # of rows in each component.. func pickerview(pickerview: uipickerview, numberofrowsincomponent component:
Read more

Python Pig Latin Translator -

-
so have been working on @ codecademy.com while trying learn python. don't know if blind , can't see why won't work or what. here's code won't let me use. function print variable if had typed in , print "empty" if didn't type anything. print "welcome english pig latin translator!" original = raw_input("what's name?") if len(original) > 0 print original else print "empty" if len(original) > 0 else the above lines contains syntax errors. try: if len(original) > 0: else: note added colons, ( : ).
Read more