I have a model InboxMessageHelper with relations like
'message', 'sender' and 'receiver' and I am using the following criteria to query data:
$model = new CActiveDataProvider('IndividualMessageHelper', array(
'criteria'=>array(
'condition'=>'receiver_id = '.Yii::app()->user->id,
'order'=>'message.created_at DESC',
'with'=>array('message', 'sender', 'receiver'),
'together'=>true,
),
));
I want to get all data (i.e. including the relations data) inside the controller and form a JSON, but the problem is that i cannot access related fields data. I can see that the data is available when I use
CVarDumper::dump()
when I try to encode $model->data then only data from current table gets encoded. How should I go about it?
I don't think CActiveDataProvider can be used in this way. You need to be working with the model. So you'll need something like this in your controller.
$models = IndividualMessageHelper::model()->findAll('receiver_id = '.Yii::app()->user->id);
foreach($models as $model){
$json[] = $model->getAttributes; //This won't get any model properties you've declared
yourself, only database columns
}
//Now get the related records and add them to the array.
array_push($json, $model->getRelated('message')->getAttributes());
array_push($json, $model->getRelated('sender')->getAttributes());
array_push($json, $model->getRelated('receiver')->getAttributes());
echo json_encode($json);
Related
I have a many-to-many relation setup using a junction table in MySQL. Table Article is related to Activity via table Article_Activity.
In model Article I have a relation setup like this
public function getActivities()
{
return $this->hasMany(Activity::className(), ['id' => 'activity_id'])
->viaTable('article_activity', ['article_id' => 'id']);
}
When rendering a view for one Article I would like to display a GridView of all Activities related to that Article.
The way most people seem to this is to create a ActiveDataProvider and insert a query into it that fetches related data but that feel a bit redundant since I have the relation setup in the model and there should be a way to get a dataprovider from that.
My question is: Is there a way to get a yii\data\ActiveDataProvider or yii\db\Query based on a instantiated models relation that can be used to display all related records in a GridView?
You can actually call it like this:
$dataProvider = new ActiveDataProvider([
'query' => $article->getActivities(),
]);
If you call the get method directly you get a yii\db\ActiveQueryInterface which is what you need to provide as query to the ActiveDataProvider.
When you call the activities attribute like $article->activities the ActiveQueryInterface is executed and you get the records from the query results.
Based on you Article model you have the activities relation that get multiple Activity related to you Article
a normal dataProvider like
$dataProvider = new ActiveDataProvider([
'query' => Article::find()->joinWith(['activities'])->
where(['your_column'=> $your_value]),
]);
return activeRecord whit also the related activities
you can refer to the related models inside the dataProvider
$models = $dataProvider->models; // this is a collection of model related to the dataProvider query
// so the firts model is $model=models[0]
then for each of this you can obtain acyivities
$activities = $model->activities;
or value
$my_activity_field = $model->activities->my_activity_field
How to pass $id to search scenario? Maybe in model look like this, so I can call like in controller like:
$model = new job('search',$id);
I think that you are trying to do a search. Search is one thing, a "scenario" is something else.
Scenarios are used in validation rules in order to be able to validate the same model in multiple ways depending from where you're inserting/adding OR searching data.
There's also a scenario called 'search' that is used by the model's search() method, but I tell you why:
There are a couple of ways to search for something in your database using Yii, I will mention two:
1) By using ClassName::model()->findCommandHere
And there are a couple of them:
ClassName::model()->findByPk($id);
ClassName::model()->findAll("id=$id");
ClassName::model()->findByAttributes(array('id'=>$id));
And so on, more here: http://www.yiiframework.com/doc/guide/1.1/en/database.ar#reading-record
2) By using the model's search() method
This way of finding data is mostly used for easily creating search pages and in combination with data grids.
If you generate CRUD code with the GII code generation tool it will generate all the parts for you, but I will explain each part how it works.
This code is from the blog demo found in Yii files:
In controller it defines a $model using Post class and 'search' as scenario.
$model=new Post('search');
if(isset($_GET['Post'])) // <- checks if there are search params in the URL
$model->attributes=$_GET['Post']; // <- assigns all search params masively to the model (later you'll see why)
$this->render('admin',array(
'model'=>$model,
));
The 'search' scenario here tells Yii what validation rules to use when assigning search parameters directly from $_GET (URL).
You can see that the params are assigned massively to reduce code written but $model->attributes=$_GET['Post'] it is the same as doing:
$model->title=$_GET['Post']['title'];
$model->status=$_GET['Post']['status'];
In the Post model you can find the validation rules for the search scenario. Tells Yii that it is safe to assign title and status fields in order to later use them in the search.
public function rules()
{
return array(
// ... //
array('title, status', 'safe', 'on'=>'search'),
);
}
Then also in the Post model you can see the search() method that will actually be used to get the data:
public function search()
{
$criteria=new CDbCriteria;
$criteria->compare('title',$this->title,true);
$criteria->compare('status',$this->status);
return new CActiveDataProvider('Post', array(
'criteria'=>$criteria,
'sort'=>array(
'defaultOrder'=>'status, update_time DESC',
),
));
}
The search method creates a "criteria" and applies the desired way of filtering using the values you have previously assigned to this model. See the $this->title it comes from the $model->attributes=$_GET['Post'] you used in the controller.
The criteria can be used directly on the model, such as Post::model()->findAll($criteria), but in this case the search() method uses something different, a "data provider".
The data provider is a good thing because it provides you a lot of tools in one place, it returns you the data, but also the pagination, and the sorting, so you don't have to manually define more code for that purposes (CPagination, CSort).
Finally, in the view admin.php in this case it will display the results using a grid view:
$this->widget('zii.widgets.grid.CGridView', array(
'dataProvider'=>$model->search(),
'filter'=>$model,
'columns'=>array(
array(
'name'=>'title',
'type'=>'raw',
'value'=>'CHtml::link(CHtml::encode($data->title), $data->url)'
),
array(
'name'=>'status',
'value'=>'Lookup::item("PostStatus",$data->status)',
'filter'=>Lookup::items('PostStatus'),
),
),
));
Now you can see that in the configuration of the grid it passes $model->search() method as the dataProvider that the grid should use.
Now the grid can access the rest of the dataProvider elements such as sort, pagination and display them on the page.
If you did not want to use the CGridView because it's a very basic table and you want to create your own html markup, you can also retrieve the dataProvider and its components one by one and place them in your HTML code and display data as you want:
$dataProvider=$model->search(); // get the dataprovider from search method
$models=$dataProvider->getData(); // actually get the data (rows) and assign them to a $models variable that you can put in a foreach loop
// show pagination somewhere
$this->widget('CLinkPager', array(
'pages' => $dataProvider->pagination,
));
// create sort links
echo $dataProvider->sort->link('title', 'Title');
So I hope it solves some of your doubts on how to use Yii for displaying/searching data.
I suggest you read the official manual: http://www.yiiframework.com/doc/guide/1.1/en/index
I also suggest to look at the API and so search there all the Yii components to see what methods and params they have: http://www.yiiframework.com/doc/api/
Also exploring the framework codebase manually is quite a good way to learn. If you don't know how CActiveDataProvider works, then find the CActiveDataProvider class file in the code and you'll see all the methods and properties that it uses, so do this for everything you don't understand how it works.
Also for beginners I recommend using a good IDE that auto-completes code and allows you to Ctrl+Click a class name and it will locate the original file where it was defined. I use NetBeans for PHP and after creating a project I add Yii framework files to the project's include paths that way NetBeans knows how to find the framework files for auto-complete and for ctrl+click.
Yes you can define scenario with parameter but that is included within the class constructor of the model
$model = new Job('search'); // creating a model with scenario search
If you wish to include more parameters, then you need to use createComponent -remember, all is a component
$model = Yii::createComponent(array('class'=>'Job','scenario'=>'search'));
I think this will simply do the job without needing any scenario
$model = new job;
$model->search($id);
But If I have failed to understand your problem then you can also try this
$model = new job('search');
$model->search($id);
Think of scenarios as a special variable that you can use in the model.
$userModel = new User("register");
$userModel->setId = 10;
which is the same
$userModel = new User();
$userModel->scenario = 10
$userModel->setId = 10;
And in your model
class Manufacturer extends CActiveRecord
{
// :
if ($this->scenario == 'register') ...
// :
}
I am not sure why I can't get the columns from my other tables via my relations. I was thinking is it because of my scope? After i had a default scope in my models, everything seems to be out of place, even if i use resetscope() at some places. Some sections I can't get to my relation columns; when that happens, I'd have to use Model::model->findbypk(n)->name.. that doesn't look pretty.
the id shows if i don't have the relations, but the name is blank when i put the relation name.
CHtml::listData(Model::model()->findAll(),'product_id','main.product_name'),
my model defaultscope is pretty basic:
return array(
'condition'=>'store_id1=:store_id OR store_id2=:store_id' ,
'params' => array(':store_id' => $store_id)
);
You can change the way you use your model like below:
Model::model()->with('main')->findAll();
I have two data sources
1) Database
2) Memcached or Totally different database
From 1st database I am getting the group members list IDs ( has more than 10 thousand rows) and after getting the list I query second database or hit Memcached to get the actual details
$connection = Yii::app()->db;
$sql = 'select user_id,joined_date from group_data where group_id=:group_id ';
$dataProvider = new CSqlDataProvider($sql, array(
'keyField' => 'user_id',
'sort' => array(
'attributes' => array(
'user_id',
'joined_date'
),
'defaultOrder'=>array(
'user_id ASC',
)
),
'pagination' => array(
'pageSize' => 30,
),
'totalItemCount' => $count,
));
$connection_master = Yii::app()->masterdb;
In the above data provider , I am not sure how to include my second database and get the actual user name since it is in other database .No direct connection from one database to other.
Here the problem is pagination is based on the first database table ( 30 records per page) and the actual data is from second table .
Any idea on how to implement this ?
Thank You
You could build your own custom data provider class. You can extend it from CDataProvider and implement the missing abstract methods:
abstract protected function fetchData();
abstract protected function fetchKeys();
abstract protected function calculateTotalItemCount();
Notice, that your storage model already comes very close to these methods: You store the IDs in one database and the actual data in another. So you can focus on one DB at a time in each of these methods.
You'd start with fetchKeys(). There you need to query the keys (=IDs) from your second database. Have a look at CSqlDataProvider::fetchData() to see how to use the CPagination object from $this->getPagination() to find out the current limit and offset. If you need sorting you'd also inspect the CSort object from $this->getSort() for current sort settings. With all that data available you should be able to build the query for the IDs on the currently requested result page.
Then in fetchData() you can obtain those keys through $this->getKeys(). You should not call fetchKeys() directly, because with getKeys() the result from fetchKeys() is "cached", so if it's called again later in the same request, there won't be another query. With that keys you now can easily query your main database. The result will represent the rows on the current page.
Finally you have to implement calculateTotalItemCount(). This is very easy again: Just query your second DB for the total number of items.
That's all you need to do - the code in the base CDataProvider class will take care of the rest and call your methods on demand.
i am using kohana ORM in order to get some results from the database. My problem is: even though i have consulted the documentation, i can't find a way to select only the column i am interested in. To be more explicit, i have:
$sale_stock = Model::factory('product_type')
->where('product_type_id','=', $id )
-> find_all();
var dumping it, it selects me all the "SELECT product_type.* from product_type where etc".
But i want to select only the 'stock' field from the salestock table. doing find('stock') instead find_all() returns a weired object... Where am i wrong, and how can i actually select only the column 'stock' using kohana orm?
thank you!
ORM methods find() and find_all() always select all table columns, so there is two ways to get specified fields:
Load full table rows and get columns
from it:
$sale_stock = Model::factory('product_type')
->where('product_type_id','=', $id )
-> find_all();
// get array of id=>stock values
$columns = $sale_stock->as_array('id', 'stock');
Create special method in model using
Query Builder:
// model Model_Product_Type
public function get_stocks($product_type_id)
{
return DB::select(array('stock'))
->from($this->_table_name)
->where('product_type_id', '=', $product_type_id)
->execute($this->_db);
}
I realise this isn't exactly what you're looking for, but I've pulled the following from the Kohana documentation ...
$articles = ORM::factory('article')->select_list('id', 'title');
foreach ($articles as $id => $title)
{
// Display a list of links
echo html::anchor('articles/'.$id, $title);
}
// Display a dropdown list
echo form::dropdown('articles', $articles);
You could think of it as a discount, two fields for the price of one.
It's common practice for ORMs to return a 'non-standard' object when partial model or merged model fields are requested. This prevents confusing operations using the original object (ie. how do you save an object when it contains only 2 of 8 fields, plus maybe some fields from another model?).
If you print_r the object, and give me an indication of how that looks ... it might be just what you want.
I know this is an old question, but i found maybe easier solution:
$sale_stock = ORM::factory('product_type')
->where( 'product_type_id','=', $id )
->find_all();
die($sale_stock->stock);