Loading exactly two modules in a particular position joomla [closed] - php

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I want to be able to assign multiple modules in a particular position but display only two of them at any given time in joomla. How do I be do that ?

After some research I found out that one can use the below code to do it.
$arrayofmoduleObj = JModuleHelper::getModules('position') // to get an array of module Object.
$options = array('style' => 'chrome');
foreach($arrayofmoduleObj as $moduleObject){
/* code to decide whether to print this module or not if yes */
echo JModuleHelper::renderModule($moduleObject,$options) //to render a module Object..
else continue;
}

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cake php multiple pagination on same view [closed]

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I want to show two listings from two differant models on the same ctp file.
How can i set two pagination without using ajax?
Its hardly requirement for my project.
Thanks in Advance.
It is simply not possible to use two paginators, with different models or with the same model. You need to use ajax pagination. The CakePHP standard does not allow two paginators on the same view.

Replace PHP with text [closed]

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I have some PHP that displays the users session name in the HTML and if the user is not logged in, then I would like it to display "User" instead.
Can I have an if statement in this case? I tried it myself but I got header errors.
This is what I've tried so far but each practical attempt just spits out more errors at me.
This is a different approach I have tried.
<?=$_SESSION['sess_user'] or ("User");?>!
<?php echo (isset($_SESSION['sess_user']) ? $_SESSION['sess_user'] : "User"); ?>
This will check if the session is set, if it is then echo the session, if it isn't then it will echo "User".

Link to external site in php [closed]

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I am trying to link to an external site.
where as $location outputs the URL (e.g. www.siteurl.com)
How do I make the outputted url a link and not just text
Thanks
Just echo out the anchor tags
echo 'Click';
Just try below code for other external site in new tab
echo 'Link name';

How can i store and fetch.php page in mysql [closed]

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I have a project in which i have to make a code in which when we click on menu bar and the option that we select in menu bar the page related to that option will display all this is to be done through my sql.Is there any way to do this task.
I don't understand exactly what you want and if there is a better way to do this, but don't put in your sql fields PHP code.
You can create a xml document instead, and use it better.PHP.net - XMLdocument

How to implement an event driven model in PHP? [closed]

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How could one build a system using PHP that tracks every event that occurs? Such as when someone comments on something, likes something or changes something.
The only thing that comes to mind right now is by the use of query strings when ever one invokes a command.
Try implementing the Observer pattern.

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