I'm back again for another question, i'm trying and trying. But i can't get it fixed.
This is my issue;
I have a database table, with a ProvinceID this can alter from 1 to 12. But it's an ID of the province. The provinces are stored with the value pName in the table Provinces
I have the following code to alter the table, and join it with the preferences table
$veiling = ORM::factory('veilingen')
->select('veilingvoorkeur.*')
->join('veilingvoorkeur', 'LEFT')
->on('veilingen.id', '=', 'veilingvoorkeur.vId')
->find_all();
$this->template->content = View::factory('veiling/veilingen')
->bind('veiling', $veiling);
It displays correctly, in the view i have;
echo '<div class="row">';
foreach($veiling as $ve)
{
echo $ve->provincie;
}
?>
</div>
it displays the provincie id; but i want to add a function to it; So it will be transformed to a province name. Normally i would create a functions.php file with a function getProvince($provinceId)
Do a mysql query to grab the pName value from Provinces and that is the job. But i'm new to kohana. Is there an option to turn the province id to province['pName'] during the ORM selection part, or do i have to search for another solution. Which i can't find :(
So. Please help me on the road again.
Thanx in advance.
Kind regards,
Kevin
Edit: 1:08
I've tried something, and it worked. I used ORM in the view file, for adding a function;
function provincie($id){
$provincie = ORM::factory('provincies', $id);
return $provincie->pName;
}
But i'm not glad with this way of solution, is there any other way? Or am i have to use it this way?
Check out Kohana's ORM relationships. If you use this, you could access the province name by simply calling:
$ve->provincie->name;
The relationship definition could look something like:
protected $_belongs_to = array(
'provincie' => array(
'model' => 'Provincie',
'foreign_key' => 'provincie_id'
),
);
Note that if you have a table column called provincie the relationship definition I give above will not work. You'd either have to change the table column to provincie_id or rename the relationship to e.g. provincie_model.
The output you describe when you do echo $ve->provincie; suggests that you store the ID in a column called provincie, thus the above applies to you.
I'd personally go for the first option as I prefer accessing IDs directly with an _id suffix and models without any suffix. But that's up to you.
Edit: If you use Kohanas ORM relationships you could even load the province with the initial query using with() e.g:
ORM::factory('veiling')->with('provincie')->find_all();
This could save you hundreds of extra queries.
Related
Cursos has a many-to-many relationship with Cursos and i want to get only the primaryKey from the related Cursos.
$curso = Curso::with(['thumb','interna','area','cursos_relacionados' => function($query){
$query->pluck('curso_id');
}])->find($curso_id);
The problem: Column 'curso_id' in field list is ambiguous.
I could do this:
DB::table('cursos_has_cursos_relacionados')->select('curso_relacionado_id')
->where('curso_id','=',$curso_id)->pluck('curso_relacionado_id')->toArray();
But i would like to know if its possible and how to add a alias to the with() method and use the first sample of code.
UPDATE
This is the relation
public function cursos_relacionados(){
return $this->belongsToMany('App\Model\Curso','cursos_has_cursos_relacionados','curso_id','curso_relacionado_id');
}
The problem: Column 'curso_id' in field list is ambiguous.
You need to specify table: 'thumb.curso_id' or the table you want.
Unfortunately, i couldn't find anything about setting a alias on the with() method. But, if anyone finds himself here, loking for the same anwser. Try the following code, it should work just fine
$curso = Curso::with(['thumb.media_root','interna.media_root','objetivos_media.media_root','area'])->find($curso_id);
$curso->cursos_relacionados = DB::table('cursos_has_cursos_relacionados')->select('curso_relacionado_id')->where('curso_id','=',$curso_id)->pluck('curso_relacionado_id')->toArray();
I am using PHP Yii framework's Active Records to model a relation between two tables. The join involves a column and a literal, and could match 2+ rows but must be limited to only ever return 1 row.
I'm using Yii version 1.1.13, and MySQL 5.1.something.
My problem isn't the SQL, but how to configure the Yii model classes to work in all cases. I can get the classes to work sometimes (simple eager loading) but not always (never for lazy loading).
First I will describe the database. Then the goal. Then I will include examples of code I've tried and why it failed.
Sorry for the length, this is complex and examples are necessary.
The database:
TABLE sites
columns:
id INT
name VARCHAR
type VARCHAR
rows:
id name type
-- ------- -----
1 Site A foo
2 Site B bar
3 Site C bar
TABLE field_options
columns:
id INT
field VARCHAR
option_value VARCHAR
option_label VARCHAR
rows:
id field option_value option_label
-- ----------- ------------- -------------
1 sites.type foo Foo Style Site
2 sites.type bar Bar-Like Site
3 sites.type bar Bar Site
So sites has an informal a reference to field_options where:
field_options.field = 'sites.type' and
field_options.option_value = sites.type
The goal:
The goal is for sites to look up the relevant field_options.option_label to go with its type value. If there happens to be more than one matching row, pick only one (any one, doesn't matter which).
Using SQL this is easy, I can do it 2 ways:
I can join using a subquery:
SELECT
sites.id,
f1.option_label AS type_label
FROM sites
LEFT JOIN field_options AS f1 ON f1.id = (
SELECT id FROM field_options
WHERE
field_options.field = 'sites.type'
AND field_options.option_value = sites.type
LIMIT 1
)
Or I can use a subquery as a column reference in the select clause:
SELECT
sites.id,
(
SELECT id FROM field_options
WHERE
field_options.field = 'sites.type'
AND field_options.option_value = sites.type
LIMIT 1
) AS type_label
FROM sites
Either way works great. So how do I model this in Yii??
What I've tried so far:
1. Use "on" array key in relation
I can get a simple eager lookup to work with this code:
class Sites extends CActiveRecord
{
...
public function relations()
{
return array(
'type_option' => array(
self::BELONGS_TO,
'FieldOptions', // that's the class for field_options
'', // no normal foreign key
'on' => "type_option.id = (SELECT id FROM field_options WHERE field = 'sites.type' AND option_value = t.type LIMIT 1)",
),
);
}
}
This works when I load a set of Sites objects and force it to eager load type_label, e.g. Sites::model()->with('type_label')->findByPk(1).
It does not work if type_label is lazy-loaded.
$site = Sites::model()->findByPk(1);
$label = $site->type_option->option_label; // ERROR: column t.type doesn't exist
2. Force eager loading always
Building on #1 above, I tried forcing Yii to always to eager loading, never lazy loading:
class Sites extends CActiveRecord
{
public function relations()
{
....
}
public function defaultScope()
{
return array(
'with' => array( 'type_option' ),
);
}
}
Now everything always works when I load Sites, but it's no good because there are other models (not pictured here) that have relations that point to Sites, and those result in errors:
$site = Sites::model()->findByPk(1);
$label = $site->type_option->option_label; // works now
$other = OtherModel::model()->with('site_relation')->findByPk(1); // ERROR: column t.type doesn't exist, because 't' refers to OtherModel now
3. Make the reference to the base table somehow relative
If there was a way that I could refer to the base table, other than "t", that was guaranteed to point to the correct alias, that would work, e.g.
'on' => "type_option.id = (SELECT id FROM field_options WHERE field = 'sites.type' AND option_value = %%BASE_TABLE%%.type LIMIT 1)",
where %%BASE_TABLE%% always refers to the correct alias for table sites. But I know of no such token.
4. Add a true virtual database column
This way would be the best, if I could convince Yii that the table has an extra column, which should be loaded just like every other column, except the SQL is a subquery -- that would be awesome. But again, I don't see any way to mess with the column list, it's all done automatically.
So, after all that... does anyone have any ideas?
EDIT Mar 21/15: I just spent a long time investigating the possibility of subclassing parts of Yii to get the job done. No luck.
I tried creating a new type of relation based on BELONGS_TO (class CBelongsToRelation), to see if I could somehow add in context sensitivity so it could react differently depending on whether it was being lazy-loaded or not. But Yii isn't built that way. There is no place where I can hook in code during query buiding from inside a relation object. And there is also no way I can tell even what the base class is, relation objects have no link back to the parent model.
All of the code that assembles these queries for active records and their relations is locked up in a separate set of classes (CActiveFinder, CJoinQuery, etc.) that cannot be extended or replaced without replacing the entire AR system pretty much. So that's out.
I then tried to see if I can create "fake" database column entries that would actually be a subquery. Answer: no. I figured out how I could add additional columns to Yii's automatically generated schema data. But,
a) there's no way to define a column in such a way that it can be a derived value, Yii assumes it's a column name in way too many places for that; and
b) there also doesn't appear to be any way to avoid having it try to insert/update to those columns on save.
So it really is looking like Yii (1.x) just does not have any way to make this happen.
Limited solution provided by #eggyal in comments: #eggyal has a suggestion that will meet my needs. He suggests creating a MySQL view table to add extra columns for each label, using a subquery to look up the value. To allow editing, the view would have to be tied to a separate Yii class, so the downside is everywhere in my code I need to be aware of whether I'm loading a record for reading only (must use the view's class) or read/write (must use the base table's class, does not have the extra columns). That said, it is a workable solution for my particular case, maybe even the only solution -- although not an answer to this question as written, so I'm not going to put it in as an answer.
OK, after a lot of attempts, I have found a solution. Thanks to #eggyal for making me think about database views.
As a quick recap, my goal was:
link one Yii model (CActiveRecord) to another using a relation()
the table join is complex and could match more than one row
the relation must never join more than one row (i.e. LIMIT 1)
I got it to work by:
creating a view from the field_options base table, using SQL GROUP BY to eliminate duplicate rows
creating a separate Yii model (CActiveRecord class) for the view
using the new model/view for the relation(), not the original table
Even then there were some wrinkles (maybe a Yii bug?) I had to work around.
Here are all the details:
The SQL view:
CREATE VIEW field_options_distinct AS
SELECT
field,
option_value,
option_label
FROM
field_options
GROUP BY
field,
option_value
;
This view contains only the columns I care about, and only ever one row per field/option_value pair.
The Yii model class:
class FieldOptionsDistinct extends CActiveRecord
{
public function tableName()
{
return 'field_options_distinct'; // the view
}
/*
I found I needed the following to override Yii's default table data.
The view doesn't have a primary key, and that confused Yii's AR finding system
and resulted in a PHP "invalid foreach()" error.
So the code below works around it by diving into the Yii table metadata object
and manually setting the primary key column list.
*/
private $bMetaDataSet = FALSE;
public function getMetaData()
{
$oMetaData = parent::getMetaData();
if (!$this->bMetaDataSet) {
$oMetaData->tableSchema->primaryKey = array( 'field', 'option_value' );
$this->bMetaDataSet = TRUE;
}
return $oMetaData;
}
}
The Yii relation():
class Sites extends CActiveRecord
{
// ...
public function relations()
{
return (
'type_option' => array(
self::BELONGS_TO,
'FieldOptionsDistinct',
array(
'type' => 'option_value',
),
'on' => "type_option.field = 'sites.type'",
),
);
}
}
And all that does the trick. Easy, right?!?
I'm building a application where admins can add multiple users to groups. In the Add action of the UserController i've:
$this->set('groups', $this->User->Group->find('list'));
The output should be a selectbox with the different group currently existing in the groups table.
Groups table:
'id' 'group' 'comments'
Model (user):
public $belongsTo = array('Group');
UserController:
$this->set('groups', $this->User->Group->find('list'));
View (Users/add)
echo $this->Form->input('Group_id');
What am I doing wrong?
In the Group class add this line
var $displayField = 'group';
This will tell the Cake that, for this class, it should treat group as if it was the name field in a standard model and find('list') will produce the results you want.
It looks correct, I assume you have group_id field in users table. You may want to try couple of things.
Change the case, e.g. Group_id to group_id
Change the display of select box as follow
echo $this->Form->input('group_id', $groups);
the case of the chars is pretty important in programming.
not only for files but also for class names and in your case the field names.
otherwise your code will not work correctly:
echo $this->Form->input('group_id');
should fix things.
If I'm not wrong, the find('list') is looking for id and name, so you have to specify the fields in you case because you are missing the name field
I hope that helps
Schema:
CITY
ID (int)
Name (string)
Status (int)
CITY_STATUS
ID (int)
Name (string)
When I display a city (in the View view), I want to display the related CITY_STATUS.Name value, instead of the CITY.Status value
When I add or update a city, I want to display a drop down of all CITY_STATUS.Names in the drop down
How do I do this in Yii?
Guess I'll answer it myself.
Question 1
Relations are easier if you setup a foreign key in your database first. To do this you need to use MySQL (not SQLite) with the InnoDB engine (not MyISAM), and the field in question needs an index on it. Then, Gii will setup the relations function for you automatically. Otherwise, you'll have to do it manually in the relations() function of the model in question.
To use a related value in a View:
In protected/views/[model name]/view.php, in the CDetailView attributes array, change
'Status'
to
array('label'=>'Status', 'value'=>$model->RelationName->Name)
where RelationName is the name of the relation
To use a related value in an Index view, change protected/views/[model name]/_view.php (note the underscore), for example in this case you would change
$data->Status
to
$data->RelationName->Name
To use a related value in an Admin view, in the CGridView widget call, in the columns array, replace say
'Status'
with
array('name'=>'Status', 'header'=>'Status', 'value'=>'$data->RelationName->Name')
(note the use of the variable $data, and not say $model or $dataProvider). Still trying to figure out how to sort and filter...
Question 2
To use a drop-down menu, in protected/views/[model name]/_form.php:
change
<?php echo $form->textField($model,'Status'); ?>
to
<?php echo $form->dropDownList($model,'Status', CHtml::listData(Status::model()->findAll(), 'ID', 'Name')); ?>
el chief, you are a gentleman and a scholar.
Hi This is either a very specific or very generic quetion - I'm not sure, and I'm new to the Zend framework / oo generally. Please be patient if this is a stupid Q...
Anyway, I want to create a model which does something like:
Read all the itmes from a table 'gifts' into a row set
for each row in the table, read from a second table which shows how many have been bought, the append this as another "field" in the returned row
return the row set, with the number bought included.
Most of the simple Zend examples seem to only use one table in a model, but my reading seems to suggest that I should do most of the work there, rather than in the controller. If this is too generic a question, any example of a model that works with 2 tables and returns an array would be great!
thanks for your help in advance!
I assume second tables is something like "gift_order" or something.
In this case you need to specify tables relationships beetween "gift" and and "gift_order" via foreign keys and describe it in table class.
It will look like this
class GiftOrder extends Zend_Db_Table_Abstract
{
/** Table name */
protected $_name = 'gif_order';
protected $_referenceMap = array(
"Fileset" =>array(
"columns" => array("gifId"),
"refTableClass" => "Gift",
"refColumns" => array("id")
));
........................
You need to specify foreigh key constraint while create table with SQL
ALTER TABLE `gift_order`
ADD CONSTRAINT `order_to_gift` FOREIGN KEY (`giftId`) REFERENCES `gift` (`id`) ON DELETE CASCADE;
If this is something you looking for you could find more on this at this link link http://framework.zend.com/manual/en/zend.db.table.relationships.html
With such solution you will be able to loop gifts and get their orders without any complex SQL's
$rowSetGifts = $this->findGifts();
while($rowSetGifts->next()){
$gift = $rowSetGifts->current();
$orders = $gift->findGiftOrder();//This is magick methods, this is the same $gift->findDependentRowset('GiftOrder');
//Now you can do something with orders - count($orders), loop them or edit
}
I would recommend creating a function in your gifts model class that returns what you want. It would probably look something like:
public function getGiftWithAdditionalField($giftId) {
$select = $this->getAdapter()->select()
->from(array('g' => 'gifts'))
->joinLeft(array('table2' => 't2'), 'g.gift_id = t2.gift_id', array('field' => 'field'))
->where('g.gift_id = ?', $giftId);
return $this->getAdapter->fetchAll($select);
}
You can check out the Zend Framework Docs on Joins for more info.