yii button to create db entry - php

Question

I would like to create a button that can be used to populate a table in my db with a single click.
I am just not sure what I need to do here to make this happen. Can I assign a method to be executed by a button? Or just have values picked up in my controller? Below is something like what I want to execute but through a button.
public function addInterest($interest)
{
$interest->UserId=Yii::app()->user->id;
$interest->ItemId=$this->ItemId;
return $interest->save();
}
**Additional details in response to Jaison Justus
With this implementation I am using controller and view from Model A (ItemId) where the button is to be displayed. Then there is Model B (UserId). Taking the info from Model A (ItemId) and Model B (UserId) I want to populate Model C ($interest) with that ItemId and UserId upon clicking a button. Looks like CJuiButton might provide a means to build it from being as then I can disable/hide the button after selected once. I am just not familiar with using buttons other than on a form where user input in collected, as links, or to provide pop up messages.
The code above currently sits in Model A model. With the code below in Model A controller everything works to populate Model C if I use a form and collect input. Since I do not require any input other then selecting the button from the user the form has nothing to put into it and therefore I know I can not use if(isset($_POST['Interest'])) as I have below.
public function actionView($id) {
$items=$this->loadModel($id);
$interest=$this->newInterest($items);
$this->render('view', array(
'model' => $items,
'interest' => $interest,
));
}
protected function newInterest($items)
{
$interest=new Interest;
if(isset($_POST['Interest']))
{
$interest->attributes=$_POST['interest'];
if($items->addInterest($interest))
$this->refresh();
}
return $interest;
}
In response to VarioN
Here is an attempt at using ajax. However this does not work and gives an Error 500 when ran. Is my controller action appropriate for what I am trying to do here?
Controller
public function actionAddInterest() {
$connection = yii::app()->db;
$sql1 = "INSERT INTO interest (UserId, ItemId)
VALUES(".Yii::app()->user->id.",".$this->ItemId.")";
$connection->createCommand($sql1)->execute();
}
View
<?php
echo CHtml::ajaxLink(
'Add Interest',
array('/item/addInterest'),
array('update'=>'#req_res')
);
?>

Answer 1

Looking at your question I see that you don't understand how MVC in Yii works.
Look at this 15 minutes screencast (Yii Tour - 3rd Stop - CRUD County) and after you will be able to create such button in any way you need (try use Gii and than customize it in your way - it's the easiest way).
Updated:
Seems that you need AJAX request. You can add CHtml::ajaxButton() in your view.
It will work this way:
User push the button, button do request (with JavaScript) to your
site without reloading the page and invisible for user.
Your controller action will serve this request: it can make some things (for ex., save data to db) and output data that your JavaScript possibly will display to user.
Than your JavaScript get answer and can make some changes on the page
(for example, hide button or show text got from request).
You can look at simple example with ajax here
If you needn't to submit form info with your button you can user ajaxLink. Example for it is here
There are a lot of examples with ajax and yii in the internet and at yii forum. Try to find them it may be very helpful.
Ask questions if you would have any.
Second update:
First, try to do your sql query simplier:
"INSERT INTO interest (UserId, ItemId) VALUES (1, 2)"
Than enable logging of mysql queries to log: at config/main.php add "trace" to "levels"
'components'=>array(
'log'=>array(
'class'=>'CLogRouter',
'routes'=>array(
array(
'class'=>'CFileLogRoute',
'levels'=>'error, warning, trace',
),
Now you can try to press an AJAX link and look at the protected/runtime/log.txt and determine the problem.
Additional info to AJAX requests
All that outputs your ajax scripts can be viewed by browser's features:
At Chrome: press F12, go to Network, press an ajax-link and look at request response.
At Firefox with addon "Firebug".
With this you can determine whether a request is done or not.