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.
Related
I'm trying to set up easy test data in my Acceptance tests:
public function shouldUseAFakeAccountHolder(AcceptanceTester $I) {
$I->have(AccountHolder::class);
// ...
}
I've copied the example code from the Codeception documentation and modified it with my entity names (as well as fixing the bugs).
<?php
public function _beforeSuite()
{
$factory = $this->getModule('DataFactory');
// let us get EntityManager from Doctrine
$em = $this->getModule('Doctrine2')->_getEntityManager();
$factory->_define(AccountHolder::class, [
'firstName' => Faker::firstName(),
// Comment out one of the below 'accountRole' lines before running:
// get existing data from the database
'accountRole' => $em->getRepository(AccountRole::class)->find(1),
// create a new row in the database
'accountRole' => 'entity|' . AccountRole::class,
]);
}
The relationship using existing data 'accountRole' => $em->getRepository(AccountRole::class)->find(1) always fails:
[Doctrine\ORM\ORMInvalidArgumentException] A new entity was found through the relationship 'HMRX\CoreBundle\Entity\AccountHolder#accountRole' that was not configured to cascade persist operations for entity: HMRX\CoreBundle\Entity\AccountRole#0000000062481e3f000000009cd58cbd. To solve this issue: Either explicitly call EntityManager#persist() on this unknown entity or configure cascade persist this association in the mapping for example #ManyToOne(..,cascade={"persist"}). If you cannot find out which entity causes the problem implement 'HMRX\CoreBundle\Entity\AccountRole#__toString()' to get a clue.
If I tell it to create a new entry in the related table 'accountRole' => 'entity|' . AccountRole::class, it works, but then it adds rows to the table when it should be using an existing row. All the role types are known beforehand, and a new random role type makes no sense because there's nothing in the code it could match to. Creating a duplicate role works, but again it makes so sense to have a separate role type for each user since roles should be shared by users.
I've had this error before in Unit tests, not Acceptance tests, when not using Faker / FactoryMuffin, and it's been to do with accessing each entity of the relationship with a different instance of EntityManager. As soon as I got both parts using the same instance, it worked. I don't see how to override the native behaviour here though.
It works (at least in Codeception 4.x) by using a callback for the existing relation:
<?php
public function _beforeSuite()
{
$factory = $this->getModule('DataFactory');
$em = $this->getModule('Doctrine2')->_getEntityManager();
$factory->_define(AccountHolder::class, [
'firstName' => Faker::firstName(),
'accountRole' => function($entity) use ($em) {
$em->getReference(AccountRole::class)->find(1);
},
]);
}
I've found it here: https://github.com/Codeception/Codeception/issues/5134#issuecomment-417453633
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 $result now will return a json array of person, and each person will have attribute fkCountry:{...}
I am using Phalcon and have a model Order that has a one-to-many relationship with model OrderAddress. I access those addresses through the following function:
public function getAddresses($params = null) {
return $this->getRelated("addresses", array(
"conditions" => "[OrderAddress].active = 'Y'"
));
}
The OrderAddress model has a public property errors that I do not want persisted to the database. The problem I am having is that everytime I access the getAddresses function, it reloads the object from MySQL which completely wipes the values that I set against that property.
I really only want the OrderAddress models to be loaded once, so that each call to getAddresses doesn't make another trip to the DB- it just iterates over the collection that was already loaded.
Is this possible?
I suppose there's no such option in phalcon, so it has to be implemented in your code.
You could create an additional object property for cached addresses, and return it if it's already been initialized:
protected $cachedAddresses = null;
public function getAddresses($params = null) {
if ($this->cachedAddresses === null) {
$this->cachedAddresses = $this->getRelated("addresses", array(
"conditions" => "[OrderAddress].active = 'Y'"
));
}
return $this->cachedAddresses;
}
This could be a quick solution, but it will be painful to repeat it if you have other relations in your code. So to keep it DRY, you could redefine a 'getRelated' method in base model so it would try to return cached relations, if they already were initialized.
It may look like this:
protected $cachedRelations = [];
public function getRelated($name, $params = [], $useCache = true) {
//generate unique cache object id for current arguments,
//so different 'getRelated' calls will return different results, as expected
$cacheId = md5(serialize([$name, $params]));
if (isset($this->cachedRelations[$cacheId]) && $useCache)
return $this->cachedRelations[$cacheId];
else {
$this->cachedRelations[$cacheId] = parent::getRelated($name, $params);
return $this->cachedRelations[$cacheId];
}
}
Then, you can leave 'getAddresses' method as is, and it will perform only one database query. In case you need to update cached value, pass false as a third parameter.
And, this is completely untested, but even if there're any minor errors, the general logic should be clear.
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
I'm trying to load, edit and save a record with CakePHP 2.0 but I get a generic error during the save method that don't help me to understand where is the problem.
if I try with debug($this->User->invalidFields()); I get an empty array, but I get false from $this->User->save() condition.
Here is the controller action where I get the error:
public function activate ($code = false) {
if (!empty ($code)) {
// if I printr $user I get the right user
$user = $this->User->find('first', array('activation_key' => $code));
if (!empty($user)) {
$this->User->set(array (
'activation_key' => null,
'active' => 1
));
if ($this->User->save()) {
$this->render('activation_successful');
} else {
// I get this error
$this->set('status', 'Save error message');
$this->set('user_data', $user);
$this->render('activation_fail');
}
debug($this->User->invalidFields());
} else {
$this->set('status', 'Account not found for this key');
$this->render('activation_fail');
}
} else {
$this->set('status', 'Empty key');
$this->render('activation_fail');
}
}
When I try the action test.com/users/activate/hashedkey I get the activation_fail template page with Save error message message.
If I printr the $user var I get the right user from cake's find method.
Where I'm wrong?
I think the problem may be in the way you're querying for the User record. When you do this:
$user = $this->User->find('first', array('activation_key' => $code));
The variable $user is populated with the User record as an array. You check to ensure it's not empty, then proceed; but the problem is that $this->User hasn't been populated. I think if you tried debug($this->User->id) it would be empty. The read() method works the way you're thinking.
You could try using the ID from that $user array to set the Model ID first, like so:
if (!empty($user)) {
$this->User->id = $user['User']['id']; // ensure the Model has the ID to use
$this->User->set(array (
'activation_key' => null,
'active' => 1
));
if ($this->User->save()) {
...
Edit: Well another possible approach is to use the $user array instead of modifying the current model. You said that you get back a valid user if you debug($user), so if that's true you can do something like this:
if (!empty($user)) {
$user['User']['activation_key'] = null;
$user['User']['active'] = 1;
if ($this->User->save($user)) {
...
This method works in the same way as receiving form data from $this->request->data, and is described on the Saving Your Data part of the book.
I'm curious though if there's another part of your setup that's getting in the way. Can other parts of your app write to the database properly? You should also check to make sure you aren't having validation errors, like their example:
<?php
if ($this->Recipe->save($this->request->data)) {
// handle the success.
}
debug($this->Recipe->validationErrors);