Usually we check the find operation by checking the sql-dump. But it is not possible if query operation done through ajax. How can we check any find operation whick cakePhp done in backgoround and Any way to print it in pr()?
If you give $this->layout='ajax' in the controller code, then it would not show the query.So if you want to view the query , just set the layout to your normal layout (only for testing) and check the query and then revert the code.
So for finding queries you have to follow these steps
1- Now go to cake folder and find this path cake\libs\model\datasources\dbo\dbo_mysql.php
2- find the function _execute. This function is responsible for all the queries what cakephp execute from its functions.
3- debug the $sql variable like
function _execute($sql) {
echo $sql;
return mysql_query($sql, $this->connection);
}
4- die the code after the cakephp background retrieval functions. like
$referral = $this->find('first', array(
'conditions' => array('Referral.id' => $id)
));
die();
5- If this is ajax call for that you can view the queries in firebug.
Note: The last displayed query will be your desire query before die
Related
I am trying to find records from CallForwardingCondition model using following line of code:
$this->loadModel('CallForwardingCondition');
$this->set('callForwardingCondition', $this->CallForwardingCondition->findByType('list'));
In SQL Dump following query is done when page is reloaded:
SELECT `CallForwardingCondition`.`type`, `CallForwardingCondition`.`description` FROM `vpbx`.`call_forwarding_condition` AS `CallForwardingCondition` WHERE `CallForwardingCondition`.`type` = 'list' LIMIT 1
How can I direct Cakephp to findByType which will result in following query?
SELECT `CallForwardingCondition`.`type`, `CallForwardingCondition`.`description` FROM `vpbx`.`call_forwarding_condition` AS `CallForwardingCondition` WHERE `CallForwardingCondition`.`type` LIKE '%' LIMIT 10
For CakePHP 2.x you need to use find('all') and pass it the required conditions and limit:-
$result = $this->CallForwardingCondition->find('all',[
'conditions' => ['CallForwardingCondition.type Like' => '%'],
'limit' => 10
);
findByType is a special find method that will only return the first record matching the type passed as the find method's parameter which is why it isn't returning what you want. You can read more about the findBy magic functions in the official docs.
try this:
$result = $this-> CallForwardingCondition ->find('all',['limit'=>10,'conditions'=>['CallForwardingCondition.type Like'=>'%'])->toArray();
I don't think you can't use findBy with limit. If you want to use limit, you must use findAllBy<fieldName>. Even findAllBy<fieldName>, you can't use LIKE.
This is findAllBy from cakephp
findAllBy<fieldName>(string $value, array $fields, array $order, int $limit, int $page, int $recursive)
So if you want to covert it, you must do the following way;
$this->CallForwardingCondition->findAllByType('something',['CallForwardingCondition.*'],['CallForwardingCondition.id'=>'desc'],'10');
This is findBy from cakephp doc.
findBy<fieldName>(string $value[, mixed $fields[, mixed $order]]);
Hope this help for you.
Hello everyone I want to show some values from the database in a joomla article in the following manner how can I do it ???
I've created the database in the following manner
One way to include php code in your article is to use an extension like
http://extensions.joomla.org/extension/sourcerer
This will allow you to put some php code that will make a query. Doing this in a Joomla way it would look like:
$db = JFactory::getDbo();
$query = $db->getQuery(true)
$query
->select($db->quoteName(array('ques', 'options', 'ans')))
->from($db->quoteName('#__yourtablename'))
->where($db->quoteName('id') . ' = ' . $idOfQuestionYouWantToGrab);
$db->setQuery($query);
$result = $db->loadAssoc();
Now you can show your question:
<?php echo $result['ques']; ?>
then somehow you will need to visualize questions separating them them with some regular expression, since your fields aren't structured well :/
Same with the answers. Then you will need to make some validation with JavaScript to see if answer is the correct one.
I manually set the CakePHP Pagination values in my Usergals Controller like so, so as to Paginate a related Model (TreasuresUsergal) on the view of Usergal. Here is a simplified snippet from the UsergalController:
public function view($id = null) {
$this->loadModel('TreasuresUsergal');
$options['joins'] = array(
array('table' => 'treasures',
'alias' => 'Treasure2',
'type' => 'LEFT',
'conditions' => array(
'Treasure2.id = TreasuresUsergal.treasure_id',
)
)
);
$options['conditions']=array('TreasuresUsergal.usergal_id'=>$id);
$options['fields']=array('Treasure2.*','TreasuresUsergal.ord','TreasuresUsergal.comments');
$options['order']=array('TreasuresUsergal.ord'=>'asc');
$options['limit']=$limit;
$this->Paginator->settings = $options;
$treasures=$this->Paginator->paginate('TreasuresUsergal');
$this->set('treasures',$treasures);
}
So in the above example, $id is the value passed to the view function from the URL. There is a live example of this here:
http://collections.centerofthewest.org/usergals/view/20
As you can see, it works just fine for a single page. However, today I tested the Paginator in the view and discovered the "next" button does not work. The Counter, sorting, and Page numbers all load correctly - but anytime the actual named parameter "page:n" is passed (when n is greater than 1) I get a Not Found page with the following error:
Not Found
Error: The requested address '/usergals/view/20/page:2?url=%2Fusergals%2Fview%2F20' was not found on this server.
I must be missing something simple - I have experimented with the routes a little, but haven't been able to figure it out. Or perhaps I am missing some Paginator options? Or does it think its OutOfBounds when its not?
UPDATE / WORKAROUND
After some messing around, I have devised this workaround. Not as nice as I'd like, but here is the basic idea (error handling, etc can be added)
First, I added a check in beforeFilter to see if page paramter was set. If so, I change it to 'p' parameter and redirect.
I did this here because otherwise I had problems with the Not Found exception (see notes at bottom). So, in beforeFilter:
if (isset($this->params['named']['page'])){
$newurl=$this->params['named'];
$pg=$newurl['page'];
unset($newurl['page']);
$newurl['p']=$pg;
$this->redirect(array('action' => 'view/'.$this->params['pass'][0])+$newurl);
}
Then, in the 'view' function of the same controller, I added this along with the other Paginator options:
if (isset($this->params['named']['p'])) $options['page']=$this->params['named']['p'];
With this, the standard Paginator behavior seems to work fine in the view. Prev, next, etc.
If anyone has a better suggestion, I would love to hear it. I don't like the idea of having to redirect, but it works for now.
It's worth noting that adding this code (even just to experiment) - caused all of my pagination counts to stop working. The query in debug was correct, but the displayed counts were wrong:
try {
$this->Paginator->paginate();
} catch (NotFoundException $e) {
}
i am working on a cakephp 2.x .want to add a condition into my save query .. for example i want to implement this query
INSERT INTO 'this' where condition is 'this'
right now i am doing this
$count = $this->Message->find('count', array(
'conditions' => array('Message.mobileNo' => $mobileNo,
'Message.body'=>$body
)));
if($count>0){
echo "already have a message";
}else
{
$this->Message->create();
$this->Message->save($this->request->data);
}
at times now i am first checking through count and then saving into the database ... can i add condition into my save so i dont have to query two times into database just to accomplish one task
This is not really CakePHP question rather than MySQL. But you can't do this since the INSERT query doesn't have conditional query.
There are 2 ways of doing it:
As Mark in the comment said use validation. The validation although apply to single field, so it will be quite tricky to do it.
Use beforeValidate() or beforeSave() callbacks in your model to check this and if they return false the save operation wont be executed.
You can put UNIQUE index to your table so it won't allow insertion of phone+message together.
I would go with the method 2.
try the beforeSave methode in your Model
You can use the following code in a better way:
$conditions = array('Message.mobileNo' => $mobileNo,
'Message.body'=>$body);
if ($this->Message->hasAny($conditions)){
echo "already have a message";
}
else{
$this->Message->create();
$this->Message->save($this->request->data);
}
I searched a lot but I couldn't find on How to use the find('all') in Views as used in Rails, but here I'm getting the error "Undefined property: View::$Menu [APP\Lib\Cake\View\View.php, line 804]"
'Menu' is the model which I'm using to fetch data from the menus table.
I'm using the below code in views:
$this->set('test',$this->Menu->find('all'));
print_r($test);
Inside your Menu model create a method, something like getMenu(). In this method do your find() and get the results you want. Modify the results as you need and like to within the getMenu() method and return the data.
If you need that menu on every page in AppController::beforeFilter() or beforeRender() simply do
$this->set('menu', ClassRegistry::init('Menu')->getMenu());
If you do not need it everywhere you might go better with using requestAction getting the data using this method from the Menus controller that will call getMenu() from the model and return the data. Setting it where you need it would be still better, if you use requestAction you also want to cache it very likely.
TRY TO NOT RETRIEVE DATA WITHIN VIEW FILE. VIOLATION OF MVC RULE
try this in view file:
$menu = ClassRegistry::init('Menu');
pr($menu->find('all'));
In AppHelper ,
Make a below function
function getMenu()
{
App::import('Model', 'Menu');
$this->Menu= &new Menu();
$test = array();
$test = $this->Menu->find('all');
return $test;
}
Use above function in view like :
<?php
$menu = $html->getMenu();
print_r($menu);
?>
Cakephp not allow this .
First create the reference(object) of your model using ClassRegistry::init('Model');
And then call find function from using object
$obj = ClassRegistry::init('Menu');
$test = $obj->find('all');
echo ""; print_r($test); `
This will work.