I am working in a project in which i have to create MenuBar and pages from database. Which should be appear to front user. I have created MenuItem Model,Controller as well as View. My MenuItems also appearing in front end user. But now i want when i click on any menu item it's corresponding page's contents should appear. That thing making me very confuse how to link page with menu items. I am new to cakephp so please somebody explain me how to do this. Thanks in advance for any suggestion.
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I am trying to develop one application template from Google search in my local system-using or git, here we are displaying the Users list normally, but how can I get/list the same users list on button click instead of getting directly displaying on UI. Is there any configuration is required to integrate with service/controller with html form button ? Please help me and thanks in advance.
I have a Drupal website with 3 menus (a top menu, a main menu and a footer menu), and any content I publish belongs to one of those 3. My clients wants me to show the correct menu as an accordion on the sidebar, as a navigational support. I've looked all over drupal modules and can't find anything similar. Any thoughts?
I found it. it's called Menu Block
I am facing a problem in my EXTJS application.
Basically there is a grid which displays all the projects. In my main menu which is a different file, I have "Add Project" link. On clicking this, Add Project modal window opens irrespective of any main window which may be project list or dashboard currently displaying in browser.
Now I want if the main page is the Project List in the browser and I open the Add Project modal, When I will click Add button, I want the grid / store in the project list page should reload or refresh.
I am not getting any way to get a reference to the project store in the project list page from the Add Project modal window.
Can any one please help me on this.
Thanks in Advance... :)
You can use Ext.getStore('your_store_name'); to get a reference to your store.
The sencha grid tutorial covers this: http://docs-origin.sencha.com/extjs/4.2.2/#!/guide/grid
I am building a WPF Standalone application. It consists of a menu on the left of the window and when a menu option is selected I want a page showing content related to the option selected shown on the right hand side of the window.
If a user selects a different menu option then the new view will be loaded.
If the user selects a previously selected menu option then that view will be shown but it will have held it's state. So if has a result set for customers with "bob" as a first name then when I view it again the result set will still be showing.
My questions is what is the best way to implement this.
Pages are not suitable as the app will not be viewed in a browser.
I could not see a Master Page -> content layout approach.
So is it a case of loading custom controls?
I am using the MVVM pattern.
Any help would be appreciated.
You could use TabControl with TabStripPlacement value set to "Left" - is that what you are looking for?
I managed to implement something similar detailed here using "workspaces". http://msdn.microsoft.com/en-us/magazine/dd419663.aspx
I have a normal article page with a huge image shown as a small one. What i want to do is, when the user click on the image, a new page is rendered and the image is shown at full size of the page (no menus, header, footer, etc.. just page). If the user click again, he will be redirected to the previous page.
How can i do this with joomla? I mean, i could create an html page, but i don't know how can i access it.
Any help would be appreciated,
RR
All you need to do is add this to any link you want to display without all of the module positions -
?tmpl=component
This tells Joomla to load just the component part of the page without anything else around it.
Not sure that anything specifically in Joomla is going to allow you to do that. The only way you can achieve that directly in Joomla would be using a light box type of plugin. When the user clicks the image, it will open up a full size window of the image. When the user click on the image again, it will exit the light box.
Otherwise, you can just edit the source code of the page, and change it to <img src="http://www.pathroyourimage.com/image.png"> . That way when a user clicks the image, they go directly to the picture.