My page tree looks like this:
Index:
- List
- SingleElement
- Folder
"Folder" contains content elements that i would like to display in a form of a list on page "List". Each element of this list should link to (different) single content element. How can I create such links, and display single content element in page "SingleElement".
2 different content elements are required to solve your questions.
Menu of content elements
To create a menu of content elements, go to your page List and add a content element "Special Menu" (inside the tab "Special elements"). As "Menu Type" please select Section Index and below select the page "List" in the "Selected Pages" field.
Reference of a content element
To reference a content element from one page at another page, use the content element "Insert record" which can be found in the tab "Special Element". At the field "Records" select the content elements you want to reuse.
Related
I've a page called Page where I'll add some blocks called a PageBlock. This two are both document types. Under the Page you could create Pageblocks and Pages. Inside my listview on the backoffice, I'll only show the Pages. In the content three, I'll only show the Pages.
How could I do that?
I've tried to find a filter but nothing found and I wouldn't write a plugin. I'm using Umbraco 7.6
Update: I know how to create an listview but how could I make a filter to show only the childeren of one document type?
I don't know if I understand the question correctly, but I think what you want to do is set the permission of Pages to allow Pageblocks as its children (On the permissions tab of Pages)
Now if you want to display the children of Pages as a listview go to List view (top right) of Pages and enable Yes - Enable list view
If you want to both be able to view children in the tree and in a listview, you can leave the above option unchecked and add a List View property to the Pages documenttype (as displayed in your screenshot)
I would like to share content (essentially blocks of html) between templates.
For example, suppose I have a common footer section with a graphic or two, text and links to 'about us', 'contact us' and so on.
I have different templates for different page layouts, but they all require this footer.
So far I can think of three ways :
Nesting templates : ie have a master one which has the footer content, then a child one for each layout, then the actual page template, but this could get tricky if some pages need a different footer
Using a Partial View to hold the footer content and using #Html.Partial() to pull in the partial view on the various templates
Create a settings document with the footer content and use Umbraco.Content() to fetch the footer property
Is any of these the recommended process (and are there any pitfalls?) or is there a better way?
I would normally do one of the following:
Have properties on the homepage for the footer links etc (in a separate tab) and pull in the values into the footer partial, this way you only have to set it once, rather than having it on every page
Have a Site Settings node at the same level as the home page and pull the values from there into the footer partial
That seems to be fairly standard from most of the Umbraco sites that I've worked on. I wouldn't have all of the properties on each page, unless you need a unique footer each page for some reason.
For example, lets say you add a tab called "Footer Settings" to the Home Page DocType with a single string property with the alias "copyRightNotice" and then you want to display that in a partial, your partial might look something like:
#inherits Umbraco.Web.Mvc.UmbracoTemplatePage
#{
var rootPage = Model.Content.AncestorOrSelf(1);
<h3>#rootPage.GetPropertyValue("copyRightNotice")</h3>
}
I've seen in Joomla documentation the way to detect if you are in Frontpage while creating a template in Joomla 2.5. This is the code:
<?php
$app = JFactory::getApplication();
$menu = $app->getMenu();
if ($menu->getActive() == $menu->getDefault()) {
echo 'This is the front page';
}
?>
This works when you are at home page (or clicking on Home menu), but I have an slider in home page, and I link in the slider to an article which is not in any menu item in the application. When I load this article the code above returns as I were at frontpage. I guess that if I doesn't click on any menu item, $menu->getActive() doesn't change.
Any suggestion?
Thanks in advance.
You can do one thing to solve this problem. Create a hidden menu of all the article links which are linked in the sliders.By creating hidden menu the link will be initialized and $menu->getActive() will work for all the links..Hopefully it will works for you .
I wish to add to the present answer and provide some clarifications.
In order for the menu selection to be detected the page requires to be assigned to a menu item. If this is a hidden menu than the link to such a page called My Page would be:
/index.php/my-page - “my-page” is the menu title alias for this item
However, if one wants to show the correct hierarchy in the breadcrumbs for the same page, then the menu hierarchy would have to be replicated in the hidden menu.
For example if My Page is under My Articles main menu item, then in the hidden menu you should add “My Articles” item, of the type: Menu Item Alias, which is in the Systems submenu of the menu type field. The My Page item should be a sub-menu item of My Articles.
The “My Articles” menu item in the hidden menu must have a different menu alias than the same one in the main menu hence the new link to My Page would look like:
/index.php/my-articles2/my-page
To create a hidden menu, one simply creates another menu. It does not matter whether one creates a module for it or not, but if one does then one just should not assign any position to that module.
I have non-used elements on my page. In list view I can see them. But now I want to create a page translation for this element. But in the list view there is no "Create new page translation" selection box. How can I create a localized element?
This is what I see? Where is the 2nd option?
In List module click the edit icon (pen symbol) of the content element. At the very top you will see a Language dropdown menu. Any language overlays that do not yet exist are marked with "[NEW]" (see Portuguese in the screenshot below). If you select one of those, the corresponding overlay record is immediately created and a you are presented with a form to fill in its content.
This is my scenario
I have an view(page) with list of items, a user could select single or multiple items from this page and click on a "Add to Group" button. Then a modal dialog(JQuery dialog) will be shown, from that he could select group, then press the add button. Which causes the items selected in the parent page is added to that particular group.
So, which is the best way to pass the selected items to the modal pop-up?
Though the query string ? - what happens if the no:of items selected is large, will the url support that much characters
Keep the list in the parent page in a javascript variable and return the selected group from the modal pop-up?
Or is there is any other better option?
Thanks,
Rajeesh
2a. There is no "parent" page; modal dialog is part of the same page as selected items. Therefore on "OK" function from the dialog you can go through the checked items and do whatever you want, including POSTing to the server. It's not clear from your post whether "adding items to the group" happens on the server or client