in my grails app I need to get some data from database and show it in a gsp page.
I know that I need to get data from controller, for example
List<Event> todayEvents = Event.findAllByStartTime(today)
gets all Event with date today
Now, how can I render it in a gsp page?How can I pass that list of Event objects to gsp?
Thanks a lot
You can learn many of the basic concepts using Grails scaffolding. Create a new project with a domain and issue command generate-all com.sample.MyDomain it will generate you a controller and a view.
To answer your question create a action in a controller like this:
class EventController {
//Helpful when controller actions are exposed as REST service.
static allowedMethods = [save: "POST", update: "POST", delete: "POST"]
def showEvents() {
List<Event> todayEvents = Event.findAllByStartTime(today)
[eventsList:todayEvents]
}
}
On your GSP you can loop through the list and print them as you wish
<g:each in="${eventsList}" var="p">
<li>${p}</li>
</g:each>
Good luck
I am not sure if this is really what you meant, because in that case I suggest you to read some more on the grails :), but anyway, for your case you can use render, redirect as well but here I am taking simplest way:
In your controller you have:
def getAllElements(){
List<Event> todayEvents = Event.findAllByStartTime(today)
[todayEvents :todayEvents ]
}
and then in the GSP(I assume you know about grails conventions, as if you don't specify view name, it will by default render gsp page with the same name as the function in the controller, inside views/):
<g:each in="${todayEvents}" var="eventInstance">
${eventInstance.<propertyName>}
</g:each>
something like this.
Related
I have an API like example
I have used cakephp HTTP client to get data, below my attempted code
public index()
{
$http = new Client();
$response = $http->get('https://restcountries.eu/rest/v2/all');
// $json = $response->getJson(); //also tried usgin json
$countries = $this->paginate($response);
$this->set(compact('countries '));
}
I am trying to apply pagination with this country data then fetch it in view with pagination.
After tried above code , I have gotten below error
Argument 1 passed to Cake\Datasource\Paginator::extractData() must be an instance of Cake\Datasource\RepositoryInterface, instance of Cake\Http\Client\Response given, called in \myapp\vendor\cakephp\cakephp\src\Datasource\Paginator.php on line 176
How can I get my desire result ?
You have probably need to implement a class who extend RepositoryInterface.
class JsonSource implements Cake\Datasource\RepositoryInterface
{ ... }
public index() {
$http = new Client();
$response = $http->get('https://restcountries.eu/rest/v2/all');
$src = new JsonSource();
$src->fromResponse($response);
$countries = $this->paginate($src);
$this->set(compact('countries ')); }
Is a bit tedious, because you need to define Json like a datasource.
The default pagination only supports querying tables (repositories), or operating on pre-built query instances.
To extend on #Zeppi's answer. You basically have three somewhat straightforward options here:
Create custom query/repository implementations as hinted by #Zeppi.
This can indeed be quite a lot of work though, so you might want to look into alternatively implementing it with the help of plugins, for example muffin/webservice, which does most of the hard work of implementing the required interfaces.
Or create a custom paginator that actually accepts and works on array data.
Or use what is widely know as a "datatable", that is a JavaScript based table in the frontend that paginates the data, for example jQuery DataTables.
we are using subdirectories in our projects no separete views and controllers but in models we didn’t learn yet. Recently I’ve found this https://github.com/cakephp/cakephp/issues/60451 and actually routes and plugins we are already using, we just want to separete our models like this:
Model
-Entity
–Financial
—Money.php
-Table
–Financial
—MoneyTable.php
I’ve tryed put like this then controller is not able to find his model. How can I do to organize it, and make it work?
Things that we've tried:
Use $this->setAlias('TableModel');
Call in controller:
$this->TableModel = $this->loadModel('Subfolder/TableModel');
didn't work for SQL build, and other classes.
CakePHP uses the TableRegister to load models. That class can be configured to use a class that implements the LocatorInterface, and CakePHP uses the TableLocator as the default.
The only thing you can do is configure your own LocatorInterface instance in your bootstrap.php. You would have to create your MyTableLocator and have it change the className for tables to point to subdirectories. What rules for this class name rewritting are used is purely up to you.
bootstrap.php:
TableRegister::setTableLocator(new MyTableLocator());
MyTableLocator.php:
class MyTableLocator extends TableLocator {
protected function _getClassName($alias, array $options = [])
{
if($alias === 'Subfolder/TableModel') {
return TableModel::class;
}
return parent::_getClassName($alias, $options);
}
}
The above isn't working code.
I'm just demonstrating what the function is you need to override, and that you need logic in place to return a different class name.
You can check if the $alias contains the / character, and if so. Return a class name by extracting the subfolder name from the $alias. Take a look at the TableLocator to see how it's using the App::className function.
I can't find any info about using $this->Cookie in view cells.
When i wrote code like this, error will arise:
namespace App\View\Cell;
use Cake\View\Cell;
class CityCell extends Cell {
public function display() {
$this->Cookie->config('encryption', false);
$cookie = $this->Cookie->read('city');
}
}
and the error is:
Error: Call to a member function read() on null
So can we use cookie in view cells ?
Thank You.
That of course won't work, view cells do not support the use of components, altough they can be thought of like "mini-controllers", they are not actually controllers in the terms of CakePHPs MVC system.
Depending on whether the cookies are encrypted, you can either use the request object to fetch them in your cell
$this->request->cookie('cookieName')
or
$this->request->cookies
see also API > \Cake\Network\Request::cookie()
or, in case they are encrypted, you have grab them via the Cookie component, and then for example pass them down from your controller to the view, and finally into the cell like
controller
public function controllerAction() {
// ...
$this->set('cookie', $this->Cookie->read('cookieName'));
}
cell
public function display($cookie) {
// ...
}
view
$this->cell('CellName', ['cookie' => $cookie]);
see also Cookbook > Views > View Cells > Passing Arguments to a Cell
I have followed the basics from w3schools: http://www.w3schools.com/angular/angular_sql.asp
I now have controller with this line:
$http.get("getjson.php?q=2")
(let's say it's making a json based on select * from someplace where country=2)
How can I get this dynamic so that I can use the same angular module for all the x countries I want to list? I guess copying the module x times and hard code the country number isn't the best option...
You can crate service that implements your get request
angular.service('someService', function() {
this.getJson = function(qParam){
$http.getjson('getjson.php',{q:qParam});
}
})
you can use it in controller like this
angular.controller('mainController',function(someService){
someService.getjson(2);
})
I need to share data between a component and helper. I'm converting my self-made payment service formdata generator to a CakePHP plugin and I'd like to be able to fill in the payment data from the controller(using a component) and use a helper to print out the data.
Everything I've tried so far have felt a little too hacky, so let me ask you: Is there any elegant way to pass data from a component to a helper?
edit:
I solved this particular situation by adding the original formadata class instance to ClassRegistry during the component initialization. This way the helper too can access the instance using ClassRegistry.
However, this only works for objects, so the question remains open.
Having a similar problem, I found this solution to work best for me.
You could use the helper's __construct method in pair with $controller->helpers array.
Since the Helper::_construct() is called after the Component::beforeRender, you can modify the $controller->helpers['YourHelperName'] array to pass the data to your helper.
Component code:
<?php
public function beforeRender($controller){
$controller->helpers['YourHelperName']['data'] = array('A'=>1, 'B'=>2);
}
?>
Helper code:
<?php
function __construct($View, $settings){
debug($settings);
/* outputs:
array(
'data' => array(
'A' => (int) 1,
'B' => (int) 2
)
)
*/
}
?>
I am using CakePHP 2.0, so this solution should be tested for earlier versions.
Is there any elegant way to pass data from a component to a helper?
Yes, the same way you pass any data to the helper. In your view.
Inside your component I would do something like the following. The beforeRender() action is a CakePHP component callback.
public function beforeRender(Controller $controller) {
$yourVars = 'some data';
$goHere = 'other stuff';
$controller->set(compact('yourVars', 'goHere'));
}
Then in your view you can pass the data off to your helpers just like normal.
// view or layout *.ctp file
$this->YourHelper->yourMethod($yourVars);
$this->YourHelper->otherMethod($goHere);
In addition to what #Vanja, you can also do this just prior to instantiating a new view in your controller:
// In your controller method
// must be set prior to instantiating view
$this->helpers['YourHelperName']['paramsOrAnyName'] = ['var' => $passed_var];
$_newView = new View($this);
$return_result = $_newView->render($element_to_view, $layout);