I have two models that I would like to present in a single Yii Controller/View now I know how to make it so that it can Create from both however I resorted to a "hack" for the update side and I would appreciate some help making it right.
I have two models Recipes and RecipeSteps that in defined as a relation in the Recipe controller. RecipeSteps has a column "recipe_id" which is a foreign key and the Primary key of Recipes. Both tables are innoDB and have relations setup. RecipeSteps has ON DELETE CASCADE and ON UPDATE RETAIN set.
This is my actionUpdate from RecipesController.php
public function actionUpdate($id)
{
$model=$this->loadModel($id);
$recipeSteps=$this->loadModelSteps($id);
// Uncomment the following line if AJAX validation is needed
// $this->performAjaxValidation($model);
if(isset($_POST['Recipes']) && isset($_POST['RecipeSteps']))
{
$model->attributes=$_POST['Recipes'];
$recipeSteps->attributes=$_POST['RecipeSteps'];
if($model->save())
$recipeSteps->recipe_id = $model->recipe_id;
$recipeSteps->save();
$this->redirect(array('view','id'=>$model->recipe_id));
}
$this->render('update',array(
'model'=>$model,
'recipeSteps'=>$recipeSteps,
));
}
And this is my loadModel.
public function loadModel($id)
{
$model=Recipes::model()->findByPk($id);
if($model===null)
throw new CHttpException(404,'The requested page does not exist.');
return $model;
}
And here is the second one that I made for the other Model that I would like to condense into a single model.
public function loadModelSteps($id)
{
$recipeSteps=RecipeSteps::model()->findByPk($id);
if($recipeSteps===null)
throw new CHttpException(404,'The requested page does not exist.');
return $recipeSteps;
}
So what I would like to do is figure out how to only have one "LoadModel" that works for both models in actionUpdate.
try this on php 5.3.0 (Class Constants)
public function loadModel($id,$modelName){
$o=call_user_func(array($modelName,'model'));
$model = $o->findByPk($id);
if($modelByPk===null)
throw new CHttpException(404,'The requested page does not exist.');
return $modelByPk;
}
Related
I'm struggling with something for several days. I'm using API which stores data about food products.
What I would like to do is :
A user is looking for a product (using the bar code )
We search the product in the database
If nothing is found, then we reach the API to retrieve the information and we save them in the database
And then we process as usual with the loaded entity
Is there a way (event or other) to add a behavior in case of the entity does not exists in the DB ?
Thanks in advance
untested code to illustrated the flow.
in your repository:
public function findByEAN(string $ean): Product
{
$product = $this->findOnyBy(['ean' => $ean]);
if ($product === null) {
throw new EntityNotFoundException(sprintf('No Product with EAN %s', $ean);
}
return $product;
}
in your controller action:
try {
$product = $this->getEntityManager()->getRepository(Product::class)->findByEan($request->get('ean'));
} catch (EntityNotFoundException $e) {
$product = '' ; // fetch product via api and store in DB
}
return $product;
I am having an issue inserting a record into the database. I am a beginner with the Yii framework, so I may have made some stupid mistakes.
This is from the SiteController
public function actionCreatePost(){
$model = new PostForm();
$post = new Post();
if ($model->load(Yii::$app->request->post()) && $model->validate()) {
$post->body = $model->body;
$post->title = $model->title;
$post->save();
return $this->redirect('index');
}else {
return $this->render('createPost', ['model' => $model]);
}
}
This is from the Post class
public function behaviors()
{
return [
[
'class' => TimestampBehavior::className(),
'createdAtAttribute' => 'created_at',
'updatedAtAttribute' => 'updated_at',
'value' => new Expression('NOW()'),
],
[
'class' => BlameableBehavior::className(),
'createdByAttribute' => 'id_author',
]
];
}
The issue is that you have created a PostForm class for the form (which is correct) but you are then trying to load the response into the Post class - a completely different class. This won’t work without modification.
If you have a look at the response:
var_dump(Yii:$app->request->post());
You will see the form data is located within the PostForm key. Yii will therefore only load the data into the PostForm class.
The correct solution is therefore to create a savePost() function within the PostForm eg:
public function savePost(){
$model = new Post();
$model->propertyABC = $this->propertyABC
...etc...
$model->save();
So the action would appear as follows:
$model = new PostForm();
If($model->load(Yii::$app->request->post()) && $model->validate()){
$model->savePost();
The other option is to rename the key from PostForm to Post. Yii will then load the data but this is not the best approach as it is a bit obscure.
Hope that helps
I would guess the issue is with the validation.
I can see several issues I will point out. First, I cannot figure out why are you creating a new PostForm, loading the data in it and verifying it, just to dump some values in a new Post and save it. Are there some functions, you are running in the PostForm model, that are triggered by load or verify? If that is not the case, I would suggest dropping one of the models, and using only the other. Usually, that is the Form model. It serves as a link between the ActiveForm and the model handling everything. You can do everything in the createPost() function in the Form model, and then in the controller it will look like
if ($model->load(Yii::$app->request->post())) {
$model->save();
return $this->redirect('index');
}
Second of all, you can dump post->getErrors() before the save to see if there are any errors with the validation. What you can also do, is call $post->save(false) instead. If you pass false to it, it will not trigger $post->validate(), and some errors can be neglected. Please, let me know if there is anything unclear.
My question is related to controller in Voyager admin panel. For example I created a table with migration . It's name was "groups" and then I created BREAD and added it to menu in Voyager.
I created a folder that it's name is "groups" in \resources\views\vendor\voyager andthen I created two file to override the view.
But I do not know where the controller is . I created controller with php artisan make:controller GroupsController. I guess this controller is not related to voyager controllers.
I want to change the index or create method and pass some data to views in controller but I do not know where it is.
I created a controller in \vendor\tcg\voyager\src\Http\Controllers that it's name is VoyagerGroupsController.php but when I create class and index method in it , it does not work.
How can I create controller for "groups" and pass the data to the view?
Whenever we create a table in voyager, Voyager calls it datatype. And for all tables / datatypes created by us, Voyager users only one controller VoyagerBreadController.php located at **vendor\tcg\voyager\src\Http\Controllers**.
For example, if I create a table named brands. Laravel will use controller VoyagerBreadController.
But where are the routes which use or point to this controller. Routes are located in file vendor\tcg\voyager\routes\voyager.php. In this file, find the following lines:
try {
foreach (\TCG\Voyager\Models\DataType::all() as $dataTypes) {
Route::resource($dataTypes->slug, $namespacePrefix.'VoyagerBreadController');
}
} catch (\InvalidArgumentException $e) {
throw new \InvalidArgumentException("Custom routes hasn't been configured because: ".$e->getMessage(), 1);
} catch (\Exception $e) {
// do nothing, might just be because table not yet migrated.
}
In my version, these lines are between line No. 29 to 37.
As you can see, above code is fetching all our datatypes and creating a resouce route for our tables / datatypes.
Now, if I want to override this route and create a route to use my own controller for a particular action. For example, if I want to create a route for brands/create url. I can do this by simply adding following line (my route) below above code (i.e. after line 37):
Route::get('brands/create', function(){return 'abc';})->name('brands.create');
or you can do the same by adding following line in routes\web.php after Voyager::routes();
Route::get('brands/create', function(){return 'abc';})->name(**'voyager.brands.create'**);
Because it's now it's using your App controller not a Voyager controller so you have to override your full controller
like
In config/voyager.php add
'controllers' => [
'namespace' => 'App\\Http\\Controllers',
],
Create new controller like MyBreadController.php into App/controller
<?php
namespace App\Http\Controllers;
class MyBreadController extends \TCG\Voyager\Http\Controllers\Controller
{
//code here
}
app/Providers/AppServiceProvider.php
use TCG\Voyager\Http\Controllers\VoyagerBreadController;
use App\Http\Controllers\MyBreadController;
public function register()
{
$this->app->bind(VoyagerBreadController::class, MyBreadController::class);
//
}
I added Route::get('groups', 'GroupsController#index') as you said in routes/web.php
like this
Route::group(['prefix' => 'admin'], function () {
Voyager::routes();
Route::get('groups', 'GroupsController#index');
});
and added these lines in index method
public function index(Request $request){
// GET THE SLUG, ex. 'posts', 'pages', etc.
$slug = $this->getSlug($request);
// GET THE DataType based on the slug
$dataType = DataType::where('slug', '=', $slug)->first();
// Check permission
Voyager::can('browse_'.$dataType->name);
// Next Get the actual content from the MODEL that corresponds to the slug DataType
$dataTypeContent = (strlen($dataType->model_name) != 0)
? app($dataType->model_name)->latest()->get()
: DB::table($dataType->name)->get(); // If Model doest exist, get data from table name
$view = 'voyager::bread.browse';
if (view()->exists("voyager::$slug.browse")) {
$view = "voyager::$slug.browse";
}
return view($view, compact('dataType', 'dataTypeContent'));
}
But getSlug method does not work. This error will be shown
ErrorException in GroupsController.php line 23:
Trying to get property of non-object
I guess after overriding Controlles getSlug() does not work and I have to set the slug manually
$slug = 'groups';
I've set up Yii2 REST API with custom actions and everything is working just fine. However, what I'm trying to do is return some data from the API which would include database relations set by foreign keys. The relations are there and they are actually working correctly. Here's an example query in one of the controllers:
$result = \app\models\Person::find()->joinWith('fKCountry', true)
->where(..some condition..)->one();
Still in the controller, I can, for example, call something like this:
$result->fKCountry->name
and it would display the appropriate name as the relation is working. So far so good, but as soon as I return the result return $result; which is received from the API clients, the fkCountry is gone and I have no way to access the name mentioned above. The only thing that remains is the value of the foreign key that points to the country table.
I can provide more code and information but I think that's enough to describe the issue. How can I encode the information from the joined data in the return so that the API clients have access to it as well?
Set it up like this
public function actionYourAction() {
return new ActiveDataProvider([
'query' => Person::find()->with('fKCountry'), // and the where() part, etc.
]);
}
Make sure that in your Person model the extraFields function includes fKCountry. If you haven't implemented the extraFields function yet, add it:
public function extraFields() {
return ['fKCountry'];
}
and then when you call the url make sure you add the expand param to tell the action you want to include the fkCountry data. So something like:
/yourcontroller/your-action?expand=fKCountry
I managed to solve the above problem.
Using ActiveDataProvider, I have 3 changes in my code to make it work.
This goes to the controller:
Model::find()
->leftJoin('table_to_join', 'table1.id = table_to_join.table1_id')
->select('table1.*, table_to_join.field_name as field_alias');
In the model, I introduced a new property with the same name as the above alias:
public $field_alias;
Still in the model class, I modified the fields() method:
public function fields()
{
$fields = array_merge(parent::fields(), ['field_alias']);
return $fields;
}
This way my API provides me the data from the joined field.
use with for Eager loading
$result = \app\models\Person::find()->with('fKCountry')
->where(..some condition..)->all();
and then add the attribute 'fkCountry' to fields array
public function fields()
{
$fields= parent::fields();
$fields[]='fkCountry';
return $fields;
}
So $result now will return a json array of person, and each person will have attribute fkCountry:{...}
I have a table which has only two column key-value. I want to create a form which allow user insert 3 pair of key-value settings.
Do I need pass 3 different models to the view? Or is there any possible way to do this?
Check out this link:
http://www.yiiframework.com/doc/guide/1.1/en/form.table
This is considered best form in Yii for updating for creating multiple models.
In essence, for creation you can create a for loop generate as many inputs a you wish to have visible, and in your controller loop over the inputs to create new models.
View File:
for ( $settings as $i=>$setting ) //Settings would be an array of Models (new or otherwise)
{
echo CHtml::activeLabelEx($setting, "[$i]key");
echo CHtml::activeLabelEx($setting, "[$i]key");
echo CHtml::error($setting, "[$i]key");
echo CHtml::activeTextField($setting, "[$i]value");
echo CHtml::activeTextField($setting, "[$i]value");
echo CHtml::error($setting, "[$i]value");
}
Controller actionCreate:
$settings = array(new Setting, new Setting, new Setting);
if ( isset( $_POST['Settings'] ) )
foreach ( $settings as $i=>$setting )
if ( isset( $_POST['Setttings'][$i] ) )
{
$setting->attributes = $_POST['Settings'][$i];
$setting->save();
}
//Render View
To update existing models you can use the same method but instead of creating new models you can load models based on the keys in the $_POST['Settings'] array.
To answer your question about passing 3 models to the view, it can be done without passing them, but to validate data and have the correct error messages sent to the view you should pass the three models placed in the array to the view in the array.
Note: The example above should work as is, but does not provide any verification that the models are valid or that they saved correctly
I'm going to give you a heads up and let you know you could potentially make your life very complicated with this.
I'm currently using an EAV patterned table similar to this key-value and here's a list of things you may find difficult or impossible:
use CDbCriteria mergeWith() to filter related elements on "value"s in the event of a search() (or other)
Filtering CGridView or CListView
If this is just very straight forward key-value with no related entity aspect ( which I'm guessing it is since it looks like settings) then one way of doing it would be:
create a normal "Setting" CActiveRecord for your settings table (you will use this to save entries to your settings table)
create a Form model by extending CFormModel and use this as the $model in your form.
Add a save() method to your Form model that would individually insert key-value pairs using the "Setting" model. Preferably using a transaction incase a key-value pair doesn't pass Settings->validate() (if applicable)
optionally you may want to override the Form model's getAttributes() to return db data in the event of a user wanting to edit an entry.
I hope that was clear enough.
Let me give you some basic code setup. Please note that I have not tested this. It should give you a rough idea though.:
Setting Model:
class Setting extends CActiveRecord
{
public function tableName()
{
return 'settings';
}
}
SettingsForm Model:
class SettingsForm extends CFormModel
{
/**
* Load attributes from DB
*/
public function loadAttributes()
{
$settings = Setting::model()->findAll();
$this->setAttributes(CHtml::listData($settings,'key','value'));
}
/*
* Save to database
*/
public function save()
{
foreach($this->attributes as $key => $value)
{
$setting = Setting::model()->find(array('condition'=>'key = :key',
'params'=>array(':key'=>$key)));
if($setting==null)
{
$setting = new Setting;
$setting->key = $key;
}
$setting->value = $value;
if(!$setting->save(false))
return false;
}
return true;
}
}
Controller:
public function actionSettingsForm()
{
$model = new Setting;
$model->loadAttributes();
if(isset($_POST['SettingsForm']))
{
$model->attributes = $_POST['SettingsForm'];
if($model->validate() && $model->save())
{
//success code here, with redirect etc..
}
}
$this->render('form',array('model'=>$model));
}
form view :
$form=$this->beginWidget('CActiveForm', array(
'id'=>'SettingsForm'));
//all your form element here + submit
//(you could loop on model attributes but lets set it up static for now)
//ex:
echo $form->textField($model,'fieldName'); //fieldName = db key
$this->endWidget($form);
If you want further clarification on a point (code etc.) let me know.
PS: for posterity, if other people are wondering about this and EAV they can check the EAV behavior extention or choose a more appropriate DB system such as MongoDb (there are a few extentions out there) or HyperDex