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.
Related
I'm using Cakephp 2.4.3 . I've read that "There are CakePHP plugins that are able to generate sitemaps for you. This way your sitemap.xml file will be created dynamically on demand and will always be up to date." . I've searched but all I find are from old cakephp version which is not useful as they only cause errors .
Is there still a good plugin for this?
Some plugins definitely exist:
https://github.com/sdevore/cakephp-sitemap-plugin
https://github.com/smarek/Croogo-Sitemap-2.0
https://github.com/webtechnick/CakePHP-Seo-Plugin
Are these the old, error-causing ones? As each CakePHP site can be radically different to the next, I'm not sure a one-size-fits-all solution will exist.
If you end up writing your own sitemap implementation, it'll depend mainly on whether your site has:
Lots of database-driven content with few controllers/actions (like a typical WordPress-style site)
Lots of controller/action driven content (more of a web application)
In the first case, you'd want to perform finds on your content, and inject the results into an xml template like this: http://bakery.cakephp.org/articles/masterkeedu/2008/08/26/automatically-generate-dynamic-sitemaps
For the second case, the following may help: a component I've used for development/testing, which lists all controllers and their methods:
<?php //File: app/Controller/Component/CtrlComponent.php
// Component rewritten for Cake2.x, original from : http://cakebaker.42dh.com/2006/07/21/how-to-list-all-controllers/
class CtrlComponent extends Component {
/**
* Return an array of Controllers and their methods.
* The function will exclude ApplicationController methods
* #return array
*/
public function get() {
$aCtrlClasses = App::objects('controller');
foreach ($aCtrlClasses as $controller) {
if ($controller != 'AppController') {
// Load the controller
App::import('Controller', str_replace('Controller', '', $controller));
// Load its methods / actions
$aMethods = get_class_methods($controller);
foreach ($aMethods as $idx => $method) {
if ($method{0} == '_') {
unset($aMethods[$idx]);
}
}
// Load the ApplicationController (if there is one)
App::import('Controller', 'AppController');
$parentActions = get_class_methods('AppController');
$controllers[$controller] = array_diff($aMethods, $parentActions);
}
}
return $controllers;
}
}
In reality, a full sitemap probably uses both methods, and you'll need to consider the difference between public and "private" areas of your site (excluding admin prefixes, for example)..
in my grails app I need to get some data from database and show it in a gsp page.
I know that I need to get data from controller, for example
List<Event> todayEvents = Event.findAllByStartTime(today)
gets all Event with date today
Now, how can I render it in a gsp page?How can I pass that list of Event objects to gsp?
Thanks a lot
You can learn many of the basic concepts using Grails scaffolding. Create a new project with a domain and issue command generate-all com.sample.MyDomain it will generate you a controller and a view.
To answer your question create a action in a controller like this:
class EventController {
//Helpful when controller actions are exposed as REST service.
static allowedMethods = [save: "POST", update: "POST", delete: "POST"]
def showEvents() {
List<Event> todayEvents = Event.findAllByStartTime(today)
[eventsList:todayEvents]
}
}
On your GSP you can loop through the list and print them as you wish
<g:each in="${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.
My app has sales listing functionality that will allow the user to add 1 or more photos for the product that they want to sell.
I'm attempting to use the upload/filestore_image of ATK with a Join table to create the relationship - my models:
class Model_Listing extends Model_Table {
public $entity_code='listing';
function init(){
parent::init();
$this->addField('name');
$this->addField('body')->type('text');
$this->addField('status');
$this->addField('showStatus')->calculated(true);
}
function calculate_showStatus(){
return ($this->status == 1) ? "Sold" : "For Sale" ;
}
}
class Model_listingimages extends Model_Table {
public $entity_code='listing_images';
function init(){
parent::init();
$this->addField('listing_id')->refModel('Model_Listing');
$this->addField('filestore_image_id')->refModel('Model_Filestore_Image');
}
}
In my page manager class I have added the file upload to the crud:
class page_manager extends Page {
function init(){
parent::init();
$tabs=$this->add('Tabs');
$s = $tabs->addTab('Sales')->add('CRUD');
$s->setModel('Listing',array('name','body','status'),array('name','status'));
if ($s->form) {
$f = $s->form;
$f->addField('upload','Add Photos')->setModel('Filestore_Image');
$f->add('FileGrid')->setModel('Filestore_Image');
}
}
}
My questions:
I am getting a "Unable to include FileGrid.php" error - I want the user to be able to see the images that they have uploaded and hoped that this would be the best way to do so - by adding the file grid to bottom of the form. - EDIT - ignore this question, I created a FileGrid class based on the code in the example link below - that fixed the issue.
How do I make the association between the CRUD form so that a submit will save the uploaded files and create entries in the join table?
I have installed the latest release of ATK4, added the 4 filestore tables to the db and referenced the following page in the documentation http://codepad.agiletoolkit.org/image
TIA
PG
By creating model based on Filestore_File
You need to specify a proper model. By proper I mean:
It must be extending Model_Filestore_File
It must have MasterField set to link it with your entry
In this case, however you must know the referenced ID when the images are being uploaded, so it won't work if you upload image before creating record. Just to give you idea the code would look
$mymodel=$this->add('Model_listingimages');
$mymodel->setMasterField('listing_id',$listing_id);
$upload_field->setModel($mymodel);
$upload_field->allowMultiple();
This way all the images uploaded through the field will automatically be associated with your listing. You will need to inherit model from Model_Filestore_File. The Model_Filestore_Image is a really great example which you can use. You should add related entity (join) and define fields in that table.
There is other way too:
By doing some extra work in linking images
When form is submitted, you can retrieve list of file IDs by simply getting them.
$form->get('add_photos')
Inside form submission handler you can perform some manual insertion into listingimages.
$form->onSubmit(function($form) uses($listing_id){
$photos = explode(',',$form->get('add_photos'));
$m=$form->add('Model_listingimages');
foreach($photos as $photo_id){
$m->unloadDdata()->set('listing_id',$listing_id)
->set('filestore_image_id',$photo_id)->update();
}
}); // I'm not sure if this will be called by CRUD, which has
// it's own form submit handler, but give it a try.
You must be careful, through, if you use global model inside the upload field without restrictions, then user can access or delete images uploaded by other users. If you use file model with MVCGrid you should see what files they can theoretically get access to. That's normal and that's why I recommend using the first method described above.
NOTE: you should not use spaces in file name, 2nd argument to addField, it breaks javascript.
I'm working on a small CakePHP application that is subject to the following constraint (awkward but out of my control): I need it to work on either of two identical databases, with the choice being based on URL. For example:
http://www.example.com/myapp/foo/action/param
http://www.example.com/myapp/bar/action/param
The first obvious solution is to have two identical CakePHP applications at myapp/foo and myapp/bar with different database configurations. This has a kludgy feel to it, though, so I'm trying to find an elegant way of creating a single application.
The approach I'm considering is this: Define routes such that myapp/foo and myapp/bar will be associated with the same controller. Then give my DATABASE_CONFIG class a constructor:
function __construct() {
$pathParts = explode('/', $_SERVER['REQUEST_URI']);
if (array_search('foo', $pathParts)) {
$this->default = $this->fooConfig;
} else if (array_search('bar', $pathParts)) {
$this->default = $this->barConfig;
}
}
(Where of course I've defined fooConfig and barConfig for the two databases.) I do have control over the URL, so I can be confident that there won't be extraneous occurrences of foo or bar in the URL.
My question is this: Is there a simpler, more elegant way of handling this odd situation? Maybe something in AppModel and/or AppController? Although I'm getting rid of duplicated code, I can't shake the feeling that I'm replacing one code smell with another.
There are a few ways to do this, here is one.
Write a sweet custom route in which you always match:
Router::connect('/:ds/*', array(), array('routeClass' => 'SweetDbRoute'));
Then have SweetDbRoutes set a class variable you can use everywhere, including in your model constructors. Then it should fail so you don't actually adjust the request.
App::import('SweetDbClass', array('file' => '/path/to/your/sweet_db_class.php'));
class SweetDbRoute extends CakeRoute {
// put your failing route code here, but use your SweetDbClass to track datasource ...
// see http://book.cakephp.org/view/1634/Custom-Route-classes
}
Then in your AppModel:
App::import('SweetDbClass', array('file' => '/path/to/your/sweet_db_class.php'));
class AppModel extends Model {
public function __construct($id = false, $table = null, $ds = null) {
$ds = SweetDbClass::$ds;
parent::__construct($id, $table, $ds);
}
}
So for example, after you perform an insert in one database, the two won't be "identical", right? Are these 2 DB somehow synced with each other? I don't know what do you need to do on those DB, but it's probably easier just to do 2 separate apps.
Yes, you can specify the DB configuration in the model: http://book.cakephp.org/view/922/Database-Configuration but you can't change it on-the-fly though (the models are not expected to change association to another table, I suppose). What you do is probably the only way.
I do have control over the URL, so I can be confident that there won't be extraneous occurrences of foo or bar in the URL
Yes, there can be "extraneous occurrences of foo or bar in the URL" :)) But it won't break your app.
In my CakePHP application, I have a model like this:
class Duck extends AppModel {
var $name = 'Duck';
function get_table_name() {
$tbl_name = //compute default table name for this model
}
}
I would like to write the function get_table_name() that outputs the default table name for the model. For the example above, it should output ducks.
EDIT:
Several people have pointed out the use of $this->table.
I did small testing and found out the following:
In the question as I have put above, $this->table indeed contains the table name.
However, actually, my code looked more like this:
class Duck extends Bird {
var $name = 'Duck';
function get_table_name(){
$tbl_name = //comput default table name for this model
}
}
class Bird extends AppModel {
}
In this case $this->table is empty string.
I went with this approach because I wanted to share some code between two of my models. Looks like this is not a good way to share code between models which need some common functionality.
You're looking for the Inflector class.
Inflector::tableize($this->name)
(tableize calls two Inflector methods to generate the table name: underscore() and pluralize())
Edit:
According to the source code, $this->table should contain the name of the table that CakePHP will use for the model, but in my experience this isn't always set. I'm not sure why.
To get the name of the table that the model is currently using, you can use: $this->table. If you don't manually change the model's table conventions, this may be the most useful in the case of CakePHP ever changing its conventions to use table names using something other than Inflector.
CakePHP's Inflector
function get_table_name() {
$tbl_name = Inflector::pluralize($this->name);
}
OR the tableize method
function get_table_name() {
$tbl_name = Inflector::tableize($this->name);
}
Edit
This also addresses the apparent "ghost" issue with $this->table in the Model.
Digging around in the __construct for Model I discovered two things:
Cake uses Inflector::tableize() to get the table name. This alone is enough to warrant using tableize over pluralize. You'll get consistent results.
$this->table is not set by the Model::__construct() unless $this->useTable === false AND $this->table === false.
It appears that if you know you haven't set $this->useTable to false you should be able to use this over $this->table. Admittedly though I only briefly scanned the source and I haven't really dug deep enough to say why $this->table isn't working sometimes.
To get the full table name for a model you have to take the table prefix into account.
$table = empty($this->table) ? Inflector::tableize($this->name) : $this->table;
$fullTableName = $this->tablePrefix . $table;
I used to use inflector to get the table name from model's name
$tableName = Inflector::pluralize(Inflector::underscore($model));
but this is not really universal, using useTable looks better, by default it will contain table's name by convention, and if you have a table that does not match the conventions, then you should manually specify it by useTable. So, in both cases the result will be correct
$this->User->useTable