I have a sidebar defined in my main layout which most of the time will display the login form. After the user is logged in I need to remove that form and replace it with user data. I also need to change that sidebar when viewing the support section to show the sub-sections.
Do I need to move the element loading to each view or is there another way?
Thanks in advance,
Denis
Bottom line is you're gonna need an if($supportpage){} elseif($loggedin){} else{} block. If you don't want to put it in your layout file you could create an element for each option and then set() the correct one from the app_controller:
if ($supportpage) $sidebar = 'support';
elseif ($loggedin) $sidebar = 'loggedin';
else $sidebar = 'notloggedin';
$this->set(compact($sidebar));
And then put $this->element($sidebar) in your layout.
Related
Lets assume I have created my own custom view for a Link content type. When the user adds a 2sxc Content app to a Pane, then picks the Content Type (Link) then my custom View, when it first starts up, how can I detect that a) the View does not use a Demo item vs. b) the View uses a demo item and is showing the Demo item vs. c) its not the first time and there is a real user added Content (Entity) in place?
I have done stuff like this for the a) case:
var link = AsDynamic(Data["Default"]).First();
then checked if it was null, but it looks like my View code never executes and instead I just see, "No demo item exists for the selected template."
If I do assign a demo, is there a more elegant way to know that the Entity I am handed as Content.First() or Data["Default"]).First() is a Demo item and now a user created Entity? Currently I am hard-coding the EntityId in the template and testing for that.
The template system does not render the template if there is no demo item (unless it's a template without a content-type).
When we need this, we have two ways
give the demo item a unique value in one of the fields and check for that in the template
check the demo-item ID on GUID and check for that (Content.EntityGuid == ...)
IsDemoItem property added in 2sxc 10.06
Dynamic Entity
If a Content Editor "Hides" the only Content Item, the anonymous user will then see a Demo Item where the item was. This is confusing and unexpected from the Content Editor's point of view (as well as the public/anonymous user). If anyone else runs in to it, here is the simple code snippet to add to the start of your view. Basically, if the current user is not logged in and the item to display is a demo item, exit the View w/o displaying anything.
if(!Request.IsAuthenticated) {
if(Content.IsDemoItem ?? false) {
return;
}
}
Best to put it near the start of your first #{} Razor block.
Note: this will not throw an error in 2sxc prior to 10.6.x (because of the "?? false"), but it will not work either.
I want to remove the sidebar from the specific page and all its subsequent pages in Drupal 7
My code is mention below.code is in mytheme_preprocess_node(&$variables) function
if ($variables['type'] === 'project'){
$node = $variables['node'];
if($node->type=='project'){
//print_r($node);
echo $node->type;
unset($page['sidebar_second']);
}
why don't you create a tpl file for that specific content type and remove the sidebar from there ? just an idea
Try restricting the block in Blocks UI or with the Context module.
You can restrict that sidebar content in admin panel itself. login as admin and configure that sidebar block to display only on perticular url.
Working on Drupal, I have a page with a form made with "Webform" module, containing several fields (text fields and sliders) and a "Submit" button.
When the user enters the information and presses the "Submit" button, another page is loaded with custom code into it.
The new page is devided into 2 parts - the first one contains new information(based on the user input from the previous page); the second one contains (block) the same form, used in the previous page.
Is there a way to load the values, filled in the form from the first page into the new page?
First of all you would need to make a custom module, with the help of
hook_form_alter
You would need to store the previous form's information in cookies with prefix
Drupal_visitor_
and then display it in the new page like:
$form['submitted']['FirstName']['#default_value'] =
$_COOKIE[$firstname];
Thanks
I have a problem here.
I bind a store on the pagingtoolbar which has more than one page. For example I change current page to the page 2, then I change the store content which has only one page by a search form . The grid loads the collect data, but the input item still shows that it's in page 2, where I want it to show 1 after I call the search event.
I don't want to use store.loadPage(1) because this will cause one more exchange between the server, can anyone helps me?
Try this:
grid.store.currentPage = 1;
grid.down('pagingtoolbar').onLoad();
I've been trying to get this to work for 4 days now, with no luck.
I have a very simple jquery mobile app.
The app has a header, content and footer.
The footer is being generated dynamically on the 'pagecreate' event because it is always the same and I don't want to have its HTML in every page.
So I do something like this:
$(document).delegate('[data-role="page"]', 'pagecreate', function (e) {
DrawHeader($(this));
DrawFooter($(this));
SetFooterEvents($(this));
SetActiveFooter($(this));
});
DrawHeader() and DrawFooter() simply prepent the header div to the page and append the footer div.
SetFooterEvents() sets the onclick events of the footer navbar buttons and SetActiveFooter() is SUPPOSED to set the ui-btn-active to the current active footer link.
To do this, I've added the data-active-footer attribute to the page div and the data-name attribute to the navbar elements. I'm searching for the current element according to the data-active-footer in the page and apply the ui-btn-active class.
function SetActiveFooter(page) {
page.children('div[data-role="footer"]')
.find('a[data-name="' + page
.attr("data-active-footer") + '"]').addClass("ui-btn-active");}
So far so good.
Now, say I've changed to a page and the navbar is lit (it has successfully recieved the ui-btn-active class), and I'm clicking on the previous page, the lit item in the navbar doesnt change back!
If i click on the the page again (ie: changed to second page [corrent lit], changed back to first page [second page is still lit], then clicked on first page again) it does light the navbar button.
What I found out was that jqm also changed the navbar of the previous page when I'm changing the navbar of the current page in the 'pagecreate' event.
I've tried to overwrite this behaviour using the 'pageshow' event, that is, trying to apply the ui-btn-active class to the current element in the navbar but the problem is that $(this) and e.currentTarget objects in the 'pageshow' event DO NOT CONTAIN THE FOOTER ELEMENT!!!
$(".ui-page").live('pageshow', function (e) {
alert($(this).children('div').length); // returns 2!
alert($(this).children('div[data-role="footer"]').length); //returns 0
alert($(e.currentTarget).children('div').length); // returns 2!
alert($(e.currentTarget).children('div[data-role="footer"]').length); //returns 0});
Any ideas?!
Thanks.
Before delving into more detail, please try adding .ui-state-persist together with .ui-btn-active
This makes sure active buttons stay active when you changePage and the footer is the same. Also make sure, all your footers have the same data-id attribute.
On a sidenote: check the latest blog post about upcoming features for jqm 1.1 - it will include a fetch link utility, which allows to ajax-update portions of a page. So you could use this functionality to grab and insert a footer on every page. I'm trying the same right now with a login form, which I need on every page.
Have you tried "ui-state-persist"?
<div data-role="navbar" data-iconpos="top">
<ul>
<li>Home</li>
<li>Favorite</li>
</ul>
</div>
I still dont know why but jqm moves the footer from page to page, eventhough I assign a new footer to each page.
Maybe because I set the same ID to all of them.
Anyhow, I used this workaround to solve the problem:
On the 'pagebeforeshow' event, I set the button I want active to all the footers in the documents. I've set a special data-name attribute to each navbar button, I give it the 'ui-btn-active' class after removing it from the rest of the items.
var $footers = $(document).find('div[data-role="footer"]');
$footers.find('a').removeClass("ui-btn-active");
$footers.find('a[data-name="' + page
.attr("data-active-footer") + '"]').addClass("ui-btn-active");