Paginate search results in Cakephp - php

Question

I made a report of the databases created in a server, showing some row counters results from similar tables on those databases, using a table named Host to change the database configuration..Now I'm trying to add a simple search form in the same view to show only the macthed results. But I having problems with pagination.
In the HostController:
<?php
class HostsController extends AppController {
public $paginate = array('limit' => 1); //just as a test
function index() {
$conditions = array();
if(!empty($_GET['host_name'])){
$asd=($_GET['host_name']);
$asd=strtoupper($asd);
$conditions = array(
'OR' => array(
'upper(Host.client_name) LIKE' => "%$asd%",
'upper(Host.contact_name) LIKE' => "%$asd%",
'upper(Host.contac_email) LIKE' => "%$asd%"));
}
$hosts = $this->Host->find('all', array('conditions'=>$conditions));
foreach ($hosts as $key => $host) {
App::import('model','Product');
$this->Product = new Product;
$this->Product->changeDataSource($host['Host']['db_name']);
if($this->changeDbSource($host['Host']['db_name'])) {
$count = $this->Product->query('SELECT count(id) as "Product__count" from products');
$hosts[$key]['Host']['count_products']= $count[0]['Product']['count'];
$options = array(
'conditions' => array(),
'order' => array(
'Product.created DESC'));
$product = $this->Product->find('first', $options);
$hosts[$key]['Host']['last_product'] = $product['Product'];
}
App::import('model','Sale');
$this->Sale = new Sale;
$this->Sale->changeDataSource($host['Host']['db_name']);
if($this->changeDbSource($host['Host']['db_name'])) {
$count = $this->Sale->query('SELECT count(id) as "Sale__count" from sales');
$hosts[$key]['Host']['count_sales']= $count[0]['Sale']['count'];
$options = array(
'conditions' => array(),
'order' => array(
'Sale.created DESC'));
$sale = $this->Sale->find('first', $options);
$hosts[$key]['Host']['last_sale'] = $sale['Sale'];
} }
$this->set(compact('hosts'));
$this->paginate ();}///<<<< Here is the problem?>
In the view:
<form method="GET" >
<input id="clientNameSearch" type="text" name="host_name"
value="<?php echo !empty($_GET['host_name']) ?
$_GET['host_name']:"";?>" placeholder="Client, Contact or email of the contact" >
<input type="submit" value="Search">
</form>
In the app controller(with a database config per row in hosts table):
function changeDbSource($database = 'default') {
$db = ConnectionManager::getInstance();
$connected = $db->getDataSource($database);
if($connected->isConnected()) {return true;} else {return false;}}
So, using just $this->paginate (); give me the WHOLE LIST of host table (or the search result) and doesnt paginate, only the leyend of the button result of
echo $this->Paginator->counter(array('format' => __());
in the view seems to work
I've been trying some variations on paginate() arguments, using conditions, definig a custom paginate() and others options suggested here with no positive result, probably because I'm new on this.
According to the documentation, I shoud be able to use:
<?php
if (!empty($_GET['host_name'])){
$asd=($_GET['host_name']);
$asd=strtoupper($asd);
$this->paginate = array(
'conditions' => array(
'OR' => array(
'upper(Host.client_name) LIKE' => "%$asd%",
'upper(Host.contact_name) LIKE' => "%$asd%",
'upper(Host.contac_email) LIKE' => "%$asd%")
),'limit' => 2);}//just a test
if(empty($_GET['host_name'])) {$this->paginate();}?>
Still, no results.
So, if anyone know how to make it work or any suggestion, I'm all ears.
Thanks in advance(sorry for the poor english and coding).
Solution:?
Since there was no anwsers..here is what I got:
I just put the next code before the foreach loops
if (!empty($_GET['host_name'])){
$getvar=($_GET['host_name']);
$getvar=strtoupper($getvar);
$this->paginate = array(
'conditions' => array(
'OR' => array(
'upper(Host.client_name) LIKE' => "%$getvar%",
'upper(Host.contact_name) LIKE' => "%$getvar%",
'upper(Host.contac_email) LIKE' => "%$getvar%",
'upper(Host.company_ruc) LIKE' => "%$getvar%")
),
'limit' =>2);}
$hosts=$this->paginate ();
And in the Index.php:
<?php
$args = $_GET;
unset($args['url']);
$paginator->options(
array(
'url' => $this->passedArgs + array('?'=>$args)
)
);?>
It shows what I need, in the way I need, but the sort option apparenly doesnt work..I guess I wont use it

Answer 1

Add the search form in your view:
echo $this->Form->create('Host');
echo $this->Form->text('keywords');
echo $this->Form->submit('Search');
echo $this->Form->end();
Add the search form method in your controller:
$settings = array();
// Simple search method
if(! empty($this->request->data['Host']['keywords'])){
$keywords = explode(' ',$this->request->data['Host']['keywords']);
foreach($keywords as $key=>$keyword){
$settings['conditions']['OR'][] = array('Host.foo Like' => '%'.$keyword.'%');
$settings['conditions']['OR'][] = array('Host.bar Like' => '%'.$keyword.'%');
}
}
$this->paginate = $settings;
$this->set('hosts', $this->Paginator->paginate('Host'));
(Replace Host.foo and Host.bar with the columns you want to search in your hosts table - you can add more if you need to).
The standard Cake paginator will work for you in your index view:
<table>
<thead>
<tr>
<th><?php echo $this->Paginator->sort('name'); ?></th>
</tr>
</thead>
<tbody>
<?php foreach ($hosts as $host): ?>
<tr>
<td><?php echo h($host['Host']['name']); ?> </td>
</tr>
<?php endforeach; ?>
</tbody>
</table>
Remember the Paginator->sort('heading') must match the name of the column in your table, if you want a different title to show on your page you can use the second - title - argument i.e. Paginator->sort('column_name', 'Title to Display');
Should get you rolling.