i have a behavior for my models, the behavior has beforeFind, beforeSave, in methods i override user_id, something like:
...
public functio beforeSave() {
$this->owner->user_id = Yii::app()->user->id
}
I have model User, how can i disable behavior for registration new user?
Saving code:
$user = new User();
$user->id = 1332;
$user->field1 = 'data';
$user->save();
but on save i have null in $user->id (because work behavior).
i tried
$user->disableBehaviors();
$user->detachBehavior();
Without result.
Maybe its not right way? I create behaviors for identify users in system (find only user something, save only with user id...), but that if i have new user with full previegies, i should again detach behaviors?
If condition can be changed in future I just pass it as callback parameter into behavior from model.
This give you a bit more control over the condition. Hence, behavior becomes more reusable - if it is used by several models this condition can be unique for each.
Example below is a bit simplified, but you should get the idea.
Behavior:
class SomeBehavior extends CActiveRecordBehavior
{
public $trigger;
public function beforeSave($event)
{
if(!call_user_func($this->trigger))
return;
// do what you need
$this->owner->user_id = Yii::app()->user->id;
}
}
Model:
class SomeModel extends CActiveRecord
{
public function behaviors()
{
$me=$this;
return array(
'some'=>array(
'class'=>'SomeBehavior',
'trigger'=>function() use($me){
return $me->scenario=='some-scenario';
}
)
);
}
}
Also I use PHP 5.3. So, I use closure for trigger callback.
If your PHP version is less than 5.3 - anything callable can be used instead. Check here http://www.php.net/manual/en/function.is-callable.php
Because of behavior is a method, you can declare your own logic inside.
The model knows about its scenario, so there is no problem to return different arrays for different conditions:)
Hope it be helpful for somebody.
You can check Yii::app()-user->isGuest to determine if the user is logged in or not. or you can just try looking for the null. Like this:
if (!Yii::app()->user->isGuest)
$this->owner->user_id = Yii::app()->user->id;
or
if (null !== Yii::app()->user->id)
$this->owner->user_id = Yii::app()->user->id;
Related
I m testing a class that saves a model. This model has a behavior where it saves the user of the record which I m going to insert, with this method
public function beforeSave($event) {
if (($this->getOwner()->getTableSchema()->getColumn($this->campoUsuarioCreacion)!==null))
$this->getOwner()->{$this->campoUsuarioCreacion} = Yii::app()->user->id;
if ($this->getOwner()->getTableSchema()->getColumn($this->campoUsuarioModificacion)!==null)
$this->getOwner()->{$this->campoUsuarioModificacion} = Yii::app()->user->id;
return parent::beforeSave();
}
But when I m testing, there is a problem with Yii::app()->user->id. I think that the problem is that no user is logged in. So, How can I solve the problem, without copying again the class with a harcoded user id? Is there a way to set the app user id?
Without broaching the issue of whether or not the user is logged in or Yii::app()->user->id is defined, you need to add this to the bottom of your function:
return parent::beforeSave();
eg:
public function beforeSave($event) {
if (($this->getOwner()->getTableSchema()->getColumn($this->campoUsuarioCreacion)!==null))
$this->getOwner()->{$this->campoUsuarioCreacion} = Yii::app()->user->id;
if ($this->getOwner()->getTableSchema()->getColumn($this->campoUsuarioModificacion)!==null)
$this->getOwner()->{$this->campoUsuarioModificacion} = Yii::app()->user->id;
return parent::beforeSave();
}
I have a resource in Laravel I have called artists with an ArtistsController. I would like to add filters to some of the pages, but not all. I know I can add a filter to all of the functions/views in the resource controller like so:
public function __construct()
{
$this->beforeFilter('auth', array('except' => array()));
}
How do I add the beforeAuth filter to only a certain view/function? I would like a user to be logged in in order to go the "index" view, but I would like a user to be able to go to the "show" pages without necessarily being logged in:
public function index()
{
$artists = Artist::all();
return View::make('artists.index', compact('artists'))
->with('artists', Artist::all())
->with('artists_new', Artist::artists_new());
}
public function show($id)
{
$artist = Artist::find($id);
return View::make('artists.show', compact('artist'))
->with('fans', Fan::all());
}
Is there a way to do this? Thank you.
Not sure if this helps but you could use the only key instead of the except (if I understand your question correctly).
$this->beforeFilter('auth', array('only' => array('login', 'foo', 'bar')));
Although that would still go in the constructor.
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 have a yii application. Data is validated properly. the $model->validate() returns true but data is not being saved. Is there any way that I know about the error. It does nothing. neither prints error nor any warning.
if (isset($_POST['Invoice'])) {
$model->validate();
$model->attributes = $_POST['Invoice'];
if (!$model->validate()) {
die(CVarDumper::dump($model->errors,10,true));
}
if ($model->save()) {
die("Data saved");
$this->redirect(array('view', 'id' => $model->id));
} else {
CVarDumper::dump($model->attributes,10,true);
CVarDumper::dump($model->errors,10,true);
}
}
if you override beforeSave or afterFind method in your model,
public function beforeSave() {
return true; //don't forget this
}
public function afterFind() {
return true; //don't forget this
}
make sure you return true for those function
If save() is returning true and there are no errors as such in your database and queries. Only thing, thats possible is you haven't marked some of the column safe for mass assignment via "$model->attributes".
Make sure the column you are trying to save are marked safe in the "rules" function in your model. You can mark columns safe via adding the following rule in "rules" function in the model.
array ( "column_name1, column_name2 ....." , "safe" )
I've just ran into something similar to this. Everything was validating correctly, and $model->save() was returning true, but no data was saved in the database.
The problem and solution was that I was creating the $model object like so:
$model = ClassName::model();
but you need to create the object like so:
$model = new ClassName;
If you have this problem, you replace this:
$model->save(false)
This solves your problem.
If you use $model->save(); the filters is running that is not good for you.
Fire up some logging and see what going on...
I got the same error when I was using reCaptcha. I just did this and it worked:
$model->scenario = NULL;
Make sure you do this AFTER validation.
I had the same issue, my mistake was with the post name in the controller, where I used $model->save. I had given wrong - if(isset($_POST['postname']))
If I am not wrong, you are doing an AR save() in the $model->save() method. You do not get any error, but the data is not saved as well.
If this is the case you would like to do a:
die(CVarDumper::dump($arObj->errors,10,true));
after the $arObj->save(); call. Most of the time this happens because of the Database rejecting the values provided for insert or update.
Also do not override your model constructor:
function __construct() { } // don't do this
The issue for me was that I had a property for the column name in the ActiveRecord class, so it wasn't saving.
You should not declare properties for column names as I guess the magic methods __get() and __set() are used to save data, I guess by checking if there are column changes when you click the save() method to avoid useless SQL queries. In my case, because the column was a user-declared property, it wasn't in the columns list and therefore changes to it were not detected.
Hope this helps other people
I'm writing a small webapp in Grails, and to make sure all users are authenticated I'm using the following filter:
class LoginFilters {
static filters = {
loginCheck(controller:'*', action:'*') {
before = {
if (session.user_id) {
request.user = User.get(session.user_id)
} else if (!actionName.equals("login")) {
redirect(controller: "login", action: "login")
return false
}
}
}
}
}
And all controller methods start with reading the user property of the request object:
def actionName = {
def user = request.user
...
}
The code above works, but I'd rather avoid the duplicate code in the all controller methods. Would it be possible for the filter to bind the user object to a variable named "user" instead of "request.user", that will be accessible from all controllers?
I understand that there might be scoping issues that makes this impossible, but the Grails framework seems to be able to create quite some magic under the hood, so I figured it might be worth asking.
Using the beforeInterceptor in a controller may help:
class LoginController {
def user
def beforeInterceptor = {
user = request.user
}
def index = {
render text:"index: ${user}"
}
def test = {
render text:"test: ${user}"
}
}
I think it generally not a good idea insert the user object into the request object every time:
The request lifetime is very short, so you might end up making round trips to caches or even worse to the database on each http-request to retrieve an object, that you might not even need and that get's deleted immideately afterwards. So if you must, better store the whole object in the session instead of just the id.
Generally, I'd suggest you write a AuthenticationService with a method isLoggedIn() that returns true when the user is authenticated and a method getLoggedInUser() that returns this object.
class AuthenticationService {
def transactional = false
boolean isLoggedIn() { return session.user_id }
def getLoggedInUser() { return User.get(session.user_id) }
}
Then you use the Filter for redirection if not authenticated, and maybe the Interceptor for storing the local reference user = authenticationService.loggedInUser. But also I don't think this the best way to go. I suggest you'd create an abstract AuthenticationAwareController as base class for all your controllers in src/groovy and there have the convenience method like user
class AuthenticationAwareController {
def authenticationService
def getUser() { return authenticationService.loggedInUser() }
}
This way, you can later change you mind about storing the user however you like and don't have to change your code. Also you benefit from Caches in Hibernate, that share already retrieved user object instances between different sessions, so db roundtrips are avoided.
You still should check the retrieved user object for validity or throw an AuthenticationException in case the retrieval does not succeed. (Maybe something like AuthenticationService.getLoggedInUser(failOnError = false).)
You can even make this Service/ControllerBase a small plugin an reuse that on every application or go directly with the spring security plugin... ;-)
I think you can do this but is it really worth the trouble? It seems to me your only advantage is typing "user" instead of "request.user". Not a big gain. Anyway, I think you could follow the instructions in "12.7 Adding Dynamic Methods at Runtime" of the User Guide. I think that if you created a dynamic method "getUser() {return request.user}" that the Groovy JavaBeans getter/setter access would allow you to simply reference "user" the way you want.
If you do add a dynamic method you might want to skip the filter and do it all in the dynamic method.