When I try to update a user, let's say his account balance
$this->User->id = $validUserId;
$this->User->saveField('balance', 100);
I get this error
SQL Error: 1054: Unknown column 'User.group_id' in 'field list'
with this automatically generated query druring the save process
SELECT `User`.`group_id` FROM `users` AS `User` WHERE `User`.`id` = *validUserId* LIMIT 1
The user belongsTo a UserGroup and therefore the user has user_group_id attribute, is there any way to tell cake, that the attribute is related to a UserGroup?
Thanks in advance,
EL
Maybe, if you are just saving a field, you don't need to do validations. In that case just do
$this->User->saveField('balance', 100, false);
http://api.cakephp.org/class/model#method-ModelsaveField
Or if there is a callback like beforeSave or something, I think you can do something like:
$this->save($data, array('validate'=>false, 'callbacks'=>false));
Hope this helps
Most certainly, the models should look like this:
class UserGroup extends AppModel {
var $belongsTo = array(
'UserGroup' => array(
'className' => 'UserGroup',
'foreignKey' => 'user_group_id'
)
);
}
class User extends AppModel {
var $hasMany = array(
'User' => array(
'className' => 'User',
'foreignKey' => 'user_group_id'
)
);
}
Read about associating CakePHP models in the book
Related
I just cant get data from 2 tables in a HABTM relationship and I cant get the answer for these cakephp docs (again). Sorry for asking the question on something that should be explained in the docs.
This should be a simple but I keep getting undefined index and the relationship I set up seems in accordance with the docs. I have another post on a more complicated matter but this issue needs to be isolated .
I dont know how to get related data from the lessons table and the students table via this lessons-students HABTM table.
I want all the lessons a student does. I want the name of the student from the student table and lesson details from the lesson table which has lesson_id as the common link in this lessons-students table. Sounds simple but I cant do it.
public $hasAndBelongsToMany = array(
'Student' => array(
'className' => 'Student',
'joinTable' => 'lessons_students',
'foreignKey' => 'lesson_id',
'associationForeignKey' => 'student_id',
'unique' => 'keepExisting',
)
);
class LessonsController extends AppController {
....
$this->Lesson->recursive = -1;
$options['joins'] = array(
array('table' => 'lessons_students',
'alias' => 'LessonsStudent',
'type' => 'LEFT',
'conditions' => array(
'Lesson.id = LessonsStudent.lesson_id',
)
),
array('table' => 'students',
'alias' => 'Student',
'type' => 'LEFT',
'conditions' => array(
'LessonsStudent.student_id=Student.id',
)
),
);
$options['conditions'] = array('Student.id' => 2);
// $Item->find('all', $options);
$student=$this->set( 'student',$this->Lesson->find('all', $options));
In the view I get the error undefined index Student
<?php
// debug($student);
foreach ($student as $item2):
echo '<td>'. $item2['Lesson']['id'].'</td>';
echo '<td>'. $item2['Student']['id'].'</td>';
http://stackoverflow.com/questions/17250215/cant-get-all-user-data-from-habtm-in-cakephp
http://book.cakephp.org/2.0/en/models/associations-linking-models-together.html#hasandbelongstomany-habtm
If you want to use model association you need to remove $this->Lesson->recursive = -1; because it disable any associations defined in model. And than you can remove $options['joins'].
If you have defined many association in your Lesson model and don't want to fetch all of it use unbindModel() or use Containable behavior
$this->Lesson->Behaviors->load('Containable');
$this->Lesson->contain('Student');
If you need to do find many times in many actions it's better to enable Containable in model itself
class Lesson extends AppModel {
public $actsAs = array('Containable');
}
so you can avoid $this->Lesson->Behaviors->load('Containable'); line;
I'm working on an already existing cakephp 1.3 project and I needed to add a new table to the database. I have this in my controller:
$conditions = array('ShootingPlacement.person_id' => $id, 'Email.person_id' => $id, 'Email.shooting_placement_id' => 'ShootingPlacement.id');
$shootingPlacements = $this->ShootingPlacement->find('all', compact('conditions'));
And it's giving me this error:
Warning (512): SQL Error: 1054: Unknown column 'Email.person_id' in 'where clause' [CORE/cake/libs/model/datasources/dbo_source.php, line 684]
And ths is the query it's trying to create:
SELECT `ShootingPlacement`.`id`, ... FROM `shooting_placements` AS `ShootingPlacement`
LEFT JOIN `people` AS `Person` ON (`ShootingPlacement`.`person_id` = `Person`.`id`)
LEFT JOIN `shootings` AS `Shooting` ON (`ShootingPlacement`.`shooting_id` = `Shooting`.`id`)
WHERE `ShootingPlacement`.`person_id` = 123688 AND `Email`.`person_id` = 123688 AND `Email`.`shooting_placement_id` = 'ShootingPlacement.id'
ORDER BY `lastname` ASC
Obviously my controller code is wrong, but I'm not sure how to relate the Email table to the ShootingPlacement one. I think my models are correct. So far if I have this:
$conditions = array('ShootingPlacement.person_id' => $id);
$shootingPlacements = $this->ShootingPlacement->find('all', compact('conditions'));
It will retrieve the rows from Shooting, ShootingPlacement and Person, I want Email to be there too. Email has 2 foreign keys: one from ShootinPlacement and one from Person.
These are the models, the only one I created is Email, the rest where working correctly.
class Email extends AppModel
{
var $name = 'Email';
var $belongsTo = array
(
'Person' => array
(
'className' => 'Person',
'foreignKey' => 'person_id'
),
'ShootingPlacement' => array
(
'className' => 'ShootingPlacement',
'foreignKey' => 'shooting_placement_id'
)
);
}
class ShootingPlacement extends AppModel
{
var $name = 'ShootingPlacement';
var $belongsTo = array
(
'Person' => array
(
'className' => 'Person',
'foreignKey' => 'person_id',
'order' => 'lastname ASC'
),
'Shooting' => array
(
'className' => 'Shooting',
'foreignKey' => 'shooting_id'
)
);
}
class Person extends AppModel
{
var $name = 'Person';
var $belongsTo = array
(
'PersonOrigin' => array
(
'className' => 'PersonOrigin',
'foreignKey' => 'person_origin_id'
)
);
var $hasMany = array
(
'ShootingPlacement' => array
(
'className' => 'ShootingPlacement',
'foreignKey' => 'person_id',
'dependent' => false
)
);
}
class Shooting extends AppModel
{
var $name = 'Shooting';
var $belongsTo = array
(
'ShootingLocation' => array
(
'className' => 'ShootingLocation',
'foreignKey' => 'shooting_location_id'
),
'Emission' => array
(
'className' => 'Emission',
'foreignKey' => 'emission_id'
)
);
}
What I need on the view is to loop through the ShootingPlacement variable and I need it to contain the Email table data for that specific id of ShootingPlacement and Person (As you see in the query, Person and ShootingPlacement are in a relationship already, I only need there to be Email too)
You should be very careful with the relationship you're after. From a quick glance at some of these answers, they seem to suggest you simply add a join to the Email model into your Person model and rely on the conditions of your find to ensure your query doesn't ransack your server's memory.
I'm going to assume that first of all, you want this Email relationship to be implicit in all your queries on Person, otherwise you could simply specify the join on each query you wanted it for. In this case, you definitely want to link it using model relationships.
Your code shows that Shooting and ShootingPlacement (presume this is a model to model mapping relationship) both belong to two models. Incidentally, Shooting belongsTo Emission - which we haven't seen here yet. I assume this isn't applicable to the current scenario.
Now, let's assume off the bad that because your Email table has foreign keys, it will be a hasOne relationship, rather than a hasMany - so that's what you need to link it by. I'm going to link it to the ShootingPlacement model because this is the model you are querying, so it should be the central point at which models are joined around it. Structure wise, because everything seems to originate from your Person model, I would have to suggest you query that model instead. But the way it's set up so far will allow you to query from nearly anywhere and still retrieve mostly the same results bar a few model names and table aliases.
Purely because your foreign key between Email and ShootingPlacement has a different name, and CakePHP 1.3 doesn't handle this very well, I'm also going to suggest you don't use a foreign key, instead putting it into the relationship as conditions.
class ShootingPlacement extends AppModel
{
var $name = 'ShootingPlacement';
var $actsAs = array('Containable');
var $hasOne = array(
'Email' => array(
'className' => 'Email',
'foreignKey' => false,
'conditions' => array(
'Email.shooting_placement_id = ShootingPlacement.id',
'Email.person_id = ShootingPlacement.person_id'
)
)
);
var $belongsTo = array (
'Person' => array (
'className' => 'Person',
'foreignKey' => 'person_id',
'order' => 'lastname ASC'
),
'Shooting' => array (
'className' => 'Shooting',
'foreignKey' => 'shooting_id'
)
);
}
I've also added the containable behaviour in there. This allows you to control from each query which associated models you'd like to return with your primary model results. It will default to all, but can be handy when you only want something specific and/or for memory reasons (these kinds of queries can destroy your server memory pretty quickly if you don't limit them or specify only the field names you want to return).
Now when you create your Email model, I wouldn't suggest complicating this mess of entangled models any further by linking it back to ShootingPlacement again. As you've said, it also has a foreign key to the Person model. So you might want to do exactly the same thing as above for your Person model (changing the conditions to reflect the Person foreign key of course). This way your model is a little more flexible; it will still join to ShootingPlacement and Person, and will also allow you to query it seperately if required without the other associated models.
Documentation
CakePHP 1.3 Model Associations
CakePHP 1.3 Containable Behaviour
See also
This article on Stack
In your model add containable behavior
class Email extends AppModel {
var $name = 'Email';
var $actsAs = array('Containable');
var $belongsTo = array
(
'Person' => array
(
'className' => 'Person',
'foreignKey' => 'person_id'
),
'ShootingPlacement' => array
(
'className' => 'ShootingPlacement',
'foreignKey' => 'shooting_placement_id'
)
);
}
Just write the below code in your controller.
$this->ShootingPlacement->recursive = 2;
$this->ShootingPlacement->contain = array(
'Shooting',
'Person' => array(
'Email'
)
);
$conditions = array(
'ShootingPlacement.person_id' => $id,
'Email.shooting_placement_id' => 'ShootingPlacement.id'
);
$shootingPlacements = $this->ShootingPlacement->find('all', compact('conditions'));
Hope this helps you.
Add a $hasOne relation to Person model with Email like below
var $hasOne = array(
'Email' => array(
'className' => 'Email',
'foreignKey' => 'person_id' // Column defined for person ids in Email table
)
);
Then add
$this->ShootingPlacement->recursive = 2;
OR
you can simply use joins in cakephp to join email model. Refer cakephp joining tables
You need to link your model ShootingPlacement with "Email" with which you call it.
class ShootingPlacement extends AppModel
var $name = 'Shooting';
var $hasMany= array
(
'Email' => array
(
'className' => 'Email',
'foreignKey' => 'yourfk'
),
);
}
And uses it s very powerful ContainableBehavior !
exemple :
$contain=array('Email'=>array('fields'=>array('id','...')));
$conditions=array('ShootingPlacement.id'=>$yourId);
$this->ShootingPlacement->attachBehaviros('Containable');
$this->ShootingPlacement->find('all',$conditions);// your will retrieve yoru SHootingItem + Emails linked
This would provide the required join:
$conditions = array('ShootingPlacement.person_id' => $id, 'Email.person_id' => $id, 'Email.shooting_placement_id' => 'ShootingPlacement.id');
$joins = array(
array(
'table' => 'emails',
'alias' => 'Email',
'type' => 'LEFT',
'conditions' => array('Email.shooting_placement_id = ShootingPlacement.id')
)
);
$shootingPlacements = $this->ShootingPlacement->find('all',
array(
'conditions' => $conditions,
'joins' => $joins
)
);
community,
there is a problem with multiple tables which receive their entries of the same model "Post" on the same page. When clicking a paginator-number the app will change the page for both tables.
I tried it with dummy classes in the AppModel which should work but I get an error saying an internal error has occurred (error 500). I found out that there is a problem with the corresponding SQL-statement with a not found row.
In the AppModel:
class PostHood extends Post {
public $useTable = 'posts';
};
That code uses the table "cake_posts" of "class Post extends AppModel". In the controller I tried to get the PostHood like following:
$this->paginate['PostHood'] = array(
'conditions' => array('OR' => array(stuff)),
'limit' => 5
);
$this->set('postsHood', $this->paginate('PostHood'));
In the view there is a foreach-loop using the $postsHood as $post.
Maybe you have an idea, thanks in advance :)
EDIT 1:
I got some error notices after changing the code. May be you have an idea what to do.
Change of the AppModel:
class PostHood extends AppModel {
var $name = 'Post';
public $useTable = 'posts';
};
The Controller:
$this->loadModel('PostHood');
$this->paginate = array(
'conditions' => array('OR' =>
array(
array('AND' => array(
array('PostHood.ZIPCODE LIKE' => $userArea . '%'),
array('PostHood.ALTDATE >' => date("Y-m-d")),
array('PostHood.AGENT' => '0'),
array('PostHood.OWNER <>' => $this->UserAuth->getUserId()),
array('PostHood.PARENTID' => '0'),
array('PostHood.ACCEPTED' => '0')
)), [MORE AND ARRAYS]
$this->set('postsHood', $this->paginate('PostHood'));
The corresponding error in the view:
Error: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'Post.DATE' in 'order clause'
SQL Query: SELECT PostHood.id, PostHood.B/S, PostHood.H, PostHood.CITY,
PostHood.MARKET, PostHood.DATE, PostHood.ALTDATE, PostHood.TIME, PostHood.INCOME, PostHood.ZIPCODE, PostHood.ALIAS, PostHood.VEHICLE, PostHood.DELIVERYAREA, PostHood.SPACE, PostHood.STREET, PostHood.HOUSENUMBER, PostHood.NAME, PostHood.CART, PostHood.TEL, PostHood.OWNER, PostHood.created, PostHood.modified, PostHood.AGENT, PostHood.EXTRADATA, PostHood.PARENTID, PostHood.BRATED, PostHood.SRATED, PostHood.REQUESTED, PostHood.ACCEPTED FROM usr_web126986_4.cake_posts AS PostHood WHERE 1 = 1 ORDER BY Post.DATE asc LIMIT 5
Obviously Cake tries to fetch data with "PostHood" but the table "posts" which I actually want to use is listening to "Post.field". How can I fix that? Thanks :)
You should try something like this:
$this->paginate = array(
'conditions' => array('OR' => array(stuff)),
'limit' => 5
);
$this->set('postsHood', $this->paginate('PostHood'));
I am just curious as to why you want to structure your model extending another model.
Got it working with the great plugin DataTables. That overrides the cake-pagination and uses JSON to display the results.
In order to implement it into cake you need
1.) a cake component for interpreting sql-statements: https://github.com/cnizzdotcom/cakephp-datatable
2.) the plugin: http://www.datatables.net/index
3.) implementation:
$this->paginate = array(
'fields' => array('Model.field1', 'Model.field2'),
'conditions' => array( stuff )
);
$response = $this->DataTable->getResponse();
$this->set('response', $response);
$this->set('_serialize','response');
$encode = json_encode($response);
$this->set('dataTablesData', $encode);
Then you can grab the data in the view.
When I was working on my current project, I ran into a rather complex issue. I'll point out my problem much more detailed right now:
There are three Models: User, UsersExtendedField and UsersExtended.
UsersExtendedField contains custom fields that can be managed manually. All custom fields can be shown in the Users view as well, and be filled out of course. The values are then stored in UsersExtended, which has got two foreignKeys: user_id and field_id.
The relations look like this: User hasMany UsersExtendedField, UsersExtendedField hasMany UsersExtended, UsersExtended belongsTo User, UsersExtendedField.
The problem: When accessing the Users view, a form with user information input is shown. Any UsersExtendedFields are available as well, and since these hasMany UsersExtended, they've got plenty of UsersExtended values. But I want to reduce those to only the value(s) that belong to the User, whose view is shown at the moment. Here are my (desired) relations:
Croogo::hookBehavior('User', 'Crooboard.ExtendedUser', array(
'relationship' => array(
'hasMany' => array(
'UsersExtendedField' => array(
'className' => 'Crooboard.UsersExtendedField',
'foreignKey' => '',
'conditions' => array('status' => 1)
),
),
),
));
class UsersExtendedField extends AppModel {
var $name = 'UsersExtendedField';
var $displayField = 'fieldname';
var $hasMany = array(
'UsersExtended' => array(
'className' => 'Crooboard.UsersExtended',
'foreignKey' => 'field_id',
'conditions' => array(
'UsersExtended.user_id = User.id'
)
),
);
}
This is not the full code, these are the important parts. The problem starts right where I wrote 'UsersExtended.user_id = User.id'. Obviously, this won't work. But I do not have any idea how to access the User.id here. I also could not imagine a HABTM structure to solve this task. Do you have any idea how to get the semantics of this 'UsersExtended.user_id = User.id' to work?
Thank your very much for taking the time to read through this and helping me!
It sounds like you need to set up your HABTM relationship properly.
You already have the join table, UsersExtended, which contains your foreign keys.
Remove all previous relationships and set up HABTM in each of your User and UserExtendedField models.
The relationship code in your User model would look like this:
var $hasAndBelongsToMany = array(
'UsersExtended' => array(
'className' => 'UsersExtended',
'joinTable' => 'UsersExtended', //assuming this is the
//name of that model's table
'foreignKey' => 'user_id',
'associationForeignKey' => 'field_id'
)
);
For more information check out the page in the cakephp book
In addition, this blog post helped me grasp the relationship concepts when I was learning cakephp.
I have two table
1. Airline -id(primary), name
2. Form - id(primary), operator, other unwanted fields
I want to relate Airline.name to Form.operator. Is it possible since Form.operator is not primary key, if yes give me the query.
Can some one also guide me as how will the cakephp model relation be in this case
I would advise you to not use the name Form as is it used elsewhere in the system, however try this (or something similar) and read http://book.cakephp.org/view/1039/Associations-Linking-Models-Together
In app/models/airline.php:
<?php
class Airline extends AppModel
{
var $name = 'Airline';
var $hasOne = array(
'Form' => array(
'className' => 'Form',
'foreignKey' => 'operator')
);
// other stuff
// ... //
?>
In app/models/form.php:
<?php
class Form extends AppModel
{
var $name = 'Form';
var $belongsTo = array(
'Airline' => array(
'className' => 'Airline',
'foreignKey' => 'operator')
)
;
// other stuff
// ... //
?>
var $hasOne = array(
'airline' => array(
'className' => 'airline',
'foreignKey' => false,
'conditions' => array(
'`form`.`yourfield` = `airline`.`yourfield`'
)
)
}
This should work. just replace your fields
in order to make the relations, as Leo suggested, work, you have to follow the cake conventions. In order to save you some headaches later on, I would therefore suggest the nicely written and short material here and here. You will learn e.g. that a good foreign key for which cakephp can do some lifting for you is named operator_id, instead of simply operator (if operator is not yet a foreign key, it could be that you have a database design issue).
Lifting here refers to automatically recognizing relations once defined in e.g. a $belongsTo.
select * from `airline`, `form` where `airline.id`=`form.operator`