I'm developing structure/template in 5.2 SP4, was trying to get page name where the Web Content is added.

Search around and found out one of the way is digging in $request. After examine the values store in $request, $request.attributes.LAYOUT.name might be a way to pull it off. But I was unable to make it work, $request.attributes.LAYOUT returns the layout object though.

Tried getName(), name(), get("name") too, but no prevail, am I doing anything wrong? Or perhaps there's a different way to get the page name?

Thanks.

Hi Chee,

I'm developing structure/template in 5.2 SP4, was trying to get page name where the Web Content is added.

Search around and found out one of the way is digging in $request. After examine the values store in $request, $request.attributes.LAYOUT.name might be a way to pull it off. But I was unable to make it work, $request.attributes.LAYOUT returns the layout object though.

Tried getName(), name(), get("name") too, but no prevail, am I doing anything wrong? Or perhaps there's a different way to get the page name?



Page name nothing but Layout name.you can achieve by this way. Try this in Template

#set ($theme_display = $themeDisplay)
#set( $pageName = $themeDisplay.getLayout().getName())

$pageName ..........by printing this you can get current pageName.

Thanks
Anil

Hello,

I am having the same problem (5.2.3), and the above solution isn't working for me. Adding $request.attributes.LAYOUT to my template does display the layout values; however, I'm trying to get at the value for the layout field named 'column-2'. Here's the display of my LAYOUT data:

LAYOUT: {plid=99999, groupId=99999, companyId=99999, privateLayout=true, layoutId=999, parentLayoutId=1, name=Assets, title=, description=, type=portlet, typeSettings=layout-template-id=my template 118_INSTANCE_8Cs6_column-1=118_INSTANCE_8Cs6, 118_INSTANCE_Csu6_column-1=118_INSTANCE_Csu6, column-2=56_INSTANCE_p7Tc,56_INSTANCE_7uwL,56_INSTANCE_Pdv5 , hidden=false, friendlyURL=/assets, iconImage=false, iconImageId=0, themeId=, colorSchemeId=, wapThemeId=, wapColorSchemeId=, css=, priority=6, dlFolderId=0}

I have tried so MANY combinations of getting just the column-2 field value that I'm now feeling pretty dense... I really need to have this done like 'yesterday', so I'd very much appreciate any assistance.

Thanks in advance.

Same issue with Liferay 6.05.

Can't get the friendly url...

Hi,

To access the layout object, you can take the layout id from the following code, and then call the layout service and access the full object.

##take layout id
#set ($layoutId = $request.get("theme-display").get("plid"))

##get the service for layout
#set($layoutService = $serviceLocator.findService("com.liferay.portal.service.LayoutLocalService"))

##convert the layout id into long
#set ($layoutLong = $getterUtil.getLong($layoutId))

##take a layout object
#set($layout = $layoutService.getLayout($layoutLong))


To use service locator in vm, you need to override the following property in portal-ext.properties.

journal.template.velocity.restricted.variables=

Thanks,
Felix

I was needing to print the Current Layout ID in my portal_normal.vm

After asking around was able to display with

$layout.getLayoutType().getLayoutTemplateId()

hopefully this helps.

Hi All,

#set ($layoutId = $request.attributes.LAYOUT)) --> This line working fine with template.

But

#set ($layoutId = $request.attributes.LAYOUT.anyproperties)) --- Does not working

Can anyone justify ? is this a bug in LIferay 6 sp1.

Thanks,
Sagar Vyas

Looks like $request.attributes.LAYOUT is just a string. So you cannot call any properties on it.