I am using ORM in FuelPHP to fetch some data on a table view I created in my data base.
I followed the instruction that is given here but I'm getting an error as seen on the title above. Here is a screenshot of the error for reference (http://prntscr.com/72ssqc).
Here is the code:
http://pastebin.com/ips5VCzV
Here is a screenshot of the view Table:
http://prntscr.com/72st1e
Your view definition is wrong, it misses the 'columns' array, as the message explains.
protected static $_views = array(
'hugot_summary' => array(
'columns' => array(
'id',
'user_id',
'photo_id',
'hugot',
'url',
'comment_count',
'upvotes',
'created_at',
'updated_at'
),
),
);
Related
I'm having some hard time with CGridView, where one field is the foreign key to another table.
There is a table called Person which contains an id_scholarity
And a table Scholarity where id_scholarity is PK. I want to show the description of scholarity and not the id number.
Gii has created the relations:
In Scholarity model:
return array(
'person' => array(self::HAS_MANY, 'PERSON', 'ID_SCHOLARITY'),
);
In Person Model
return array(
'id_scholarity' => array(self::BELONGS_TO, 'SCHOLARITY', 'ID_SCHOLARITY'),
);
And finally my grid (in views/person/admin.php)
$this->widget('zii.widgets.grid.CGridView', array(
'id'=>'person-grid',
'dataProvider'=>$model->search(),
'filter'=>$model,
'columns'=>array(
'NAME',
array('name'=>'ID_SCHOLARITY', 'value'=>'$data->ID_SCHOLARITY->DESCRIPTION'),
array(
'class'=>'CButtonColumn',
),
),
));
The page just gets blank (by the way, how can I make yii show errors?).
What I'm doing wrong?
try
'columns'=>array(
'NAME',
array('value'=>'$data->id_scholarity->DESCRIPTION'),
array(
'class'=>'CButtonColumn',
),
),
when you access other table with arrow operator you have to access it using relation name not the attribute name. In your code relation name is id_scholarity but you are using ID_SCHOLARITY .
I have three tables
1. tbl_employee: id(PK), name, position_id(FK), type_id(FK)
2. tbl_position: id, position
3. tbl_type : id, type
I want to display records in field position like what sql does below.
SELECT tbl_employee.name, tbl_position.position
FROM tbl_employee, tbl_position
WHERE tbl_employee.position_id = tbl_position.id AND tbl_position.position LIKE '%Designer';
In my EmployeeController
public function actionIndex(){
$dataProvider=new CActiveDataProvider('Employee', array(
'criteria'=>array(
'select'=>'t.name, tbl_position',
'with'=>array(
'position'=>array('select'=>'position'),
'type'=>array('select'=>'type'),
),
'condition'=>"tbl_position.position LIKE '%Designer'",
),
));
$this->render('index',array('dataProvider'=>$dataProvider,));
}
and in my models
public function relations(){
// NOTE: you may need to adjust the relation name and the related
// class name for the relations automatically generated below.
return array(
'position' => array(self::BELONGS_TO, 'Position', 'position_id'),
'type' => array(self::BELONGS_TO, 'Type', 'type_id'),
);
}
and this is my view
<?php $this->widget('zii.widgets.CDetailView', array(
'data'=>$model,
'attributes'=>array(
'id',
'name',
array(
'name'=>'position_id',
'value'=>CHtml::encode($model->position->position),
),
array(
'name'=>'type_id',
'value'=>CHtml::encode($model->type->type),
)
),
)); ?>
The error I've got is:
Active record "Employee" is trying to select an invalid column "tbl_position".
Note, the column must exist in the table or be an expression with alias.
How can I access position field in tbl_position by using join?What is the correct syntax for achieving that purpose.
Many thanks
'select'=>'t.name, position.position',
'condition'=>"position.position LIKE '%Designer'",
or
$model = TblEmployee::model()->with('position','type')->findall(array("condition"=>"position.position LIKE '%Designer'"));
Here I find the solution, but actually your answer was very close
public function actionIndex(){
$dataProvider=new CActiveDataProvider('Employee', array(
'criteria'=>array(
'select'=>'t.name',
'with'=>array(
'position'=>array('select'=>'position'),
'type'=>array('select'=>'type'),
),
'condition'=>"position.position LIKE '%Designer'",
),
));
$this->render('index',array('dataProvider'=>$dataProvider,));
}
I know we can show a gridview with a model and it's search method and filter the results, but can we make a gridview with another dataprovider and another model like this and filter its results? Does filter needs to be a part of dataprovider?
$attr = Yii::app()->request->getParam($name);
$model = new User('search');
$model->unsetAttributes();
$model->setAttributes($attr);
$this->widget('zii.widgets.grid.CGridView', array(
'dataProvider' => $myDataProvider,
'filter' => $model,
'columns' => array(
array(
'name' => 'username',
'type' => 'raw',
'value' => 'CHtml::encode($data->username)'
),
array(
'name' => 'email',
'type' => 'raw',
),
),
));
The above code doesn't work and I need to add a filter on a previously made data provider.
Btw $attr has a valid data, but grid is not filtered.
$model doesn't affect $myDataProvider since the data provider is not obtained using this model.
$model->search() returns a CActiveDataProvider which contains a CDbCriteria instance. Different CDbCriteria can be combined using mergeWith(). So if you would like the data to be filtered using the values from the $model
...
$model->setAttributes($attr);
$newDataProvider=$model->search();
$myDataProvider->criteria->mergeWith($newDataProvider->criteria);
$this->widget('zii.widgets.grid.CGridView', array(
...
Filter does not need to be a part of dataprovider, but data provider needs to take the model into account, if you want to use it for filtering.
The way this is done by default is to create the data provider using search method on your model, which sets conditions of your data provider based on model values, like so:
'dataProvider' => $model->search()
There is nothing preventing you from creating different data provider, for example:
'dataProvider' => $model->createAnotherDataProvider()
And in your User model:
public function createAnotherDataProvider() {
{
// create your second data provider here
// with filtering based on model's attributes, e.g.:
$criteria = new CDbCriteria;
$criteria->compare('someAttribute', $this->someAttribute);
return new CActiveDataProvider('User', array(
'criteria' => $criteria,
));
}
I have two tables: user and asset and there is user_asset table that makes many_many relation
Now I want to show only the assets related with logged-in user in CGridView (user_id comes from Yii::app()->user->id)
How to write a criteria that makes it possible?
1) You need to define your relations in both of the model
user:
'assets' => array(self::MANY_MANY, 'Asset', 'user_asset(user_id, asset_id)'),
asset:
'users' => array(self::MANY_MANY, 'Asset', 'user_asset(asset_id, user_id)'),
2) You Create a dataProvider that will fetch only the recquired datas (define it in a model (best) or a controller bu not in a view):
$dataProvider=new CActiveDataProvider('Asset', array(
'criteria'=>array(
'with'=>array(
'users'=>array(
'condition' => 'id = ' . Yii::app()->user->id
),
),
),
...
));
3) You give to your CGridView the good provider:
$this->widget('zii.widgets.grid.CGridView', array(
'dataProvider'=>$dataProvider,
...
));
The code is given as example, you should adapt it to your need!
I got it work, adding this inside the Model class's search method:
$criteria->with=array(
'users'
);
$criteria->together = true;
$criteria->condition = "users_users.user_id='".Yii::app()->user->id."'";
... and inside CActiveDataProvider declaration:
'criteria'=>array(
'with' => 'users',
'together'=>true,
'condition'=>'users_users.user_id = '.Yii::app()->user->id,
),
Maybe not the best way but does what needed. users_users is sql table alias by Yii, I got a hint from error message :)
I am new with Yii, Sorry if my question might be stupid, I am using CGridView to show some fields of my database in a table:
<?php echo CHtml::link('Advanced Search','#',array('class'=>'search-button')); ?>
<div class="search-form" style="display:none">
<?php $this->renderPartial('_search',array(
'model'=>$model,
)); ?>
</div><!-- search-form -->
<?php
$this->widget('zii.widgets.grid.CGridView', array(
'id'=>'show-grid',
'dataProvider'=>$model->search(),
'filter'=>$model,
'columns'=>array(
'id',
'title',
'brief',
'tbl_season_id',
'on_season',
array(
'name'=>'status',
'value'=>'Lookup::item("NewsStatus",$data->status)',
'filter'=> Lookup::items('NewsStatus'),
),
array(
'class'=>'CButtonColumn',
),
),
)); ?>
</div>
I want to replace some of the values that are shown, for example, the on_season field is binary and in the table the values are 0 or 1, I want to change this values to Yes and NO,
And tbl_season_id is a foreign key form another table, I would like to get the name of the season and put it instead of the id which is not understandable by the users.
You can refer this wiki article for customizing your column values to your heart's content
Yii Documentataion: cgridview-render-customized-complex-datacolumns
Just remember that the value property can be an expression string, which is later evaluated for each data of rows. So you have a method call there which can dynamically calculate any value for you depending on the current values of that row.
Let you have two table
1): clients 2): projects
Relation is a client has many projects.
in model
and clients model the relation is (the Client table model name is Client)
class Client extends CActiveRecord{
}
and relations() method ;
return array(
'projects' => array(self::HAS_MANY, 'Projects', 'clients_id'),
);
project model the relation is (the project table model name is Projects)
class Projects extends CActiveRecord{
}
and relations() method ;
return array(
'clients' => array(self::BELONGS_TO, 'Client', 'clients_id')
);
Now
now you can use the following to get the client_name replace the client_id in proejct table
in CGridView
'dataProvider' => $model->search(),
'columns' => array(
'id',
'project_name',
array(
'name' => 'client Name',
'value' => '$data->clients->name', //where name is Client model attribute
),
)
and
Project view page in CDetailView you can use the following
'data' => $model,
'attributes' => array(
'id',
'project_name',
array(
'name'=>'Client Name',
'value'=>$model->clients->name ,
),
)
if you have client relation with company table (a company has many clients )
client model
'company' => array(self::BELONGS_TO, 'Company', 'company_id'),
you can also get the company name by following method in
index (CGridview)
array(
'name' => 'client Name',
'value' => '$data->clients->company->name', //where name is company model attribute
),
and in view CDetailView
array(
'name'=>'Client Name',
'value'=>$model->clients->company->name ,
),