I have a view created using Bake that has the following:
<fieldset>
<legend><?php echo __('Edit Device'); ?></legend>
<?php
echo $this->Form->input('DeviceID');
echo $this->Form->input('DeviceTypeID');
echo $this->Form->input('UserID');
echo $this->Form->input('Type');
echo $this->Form->input('KeyPadID');
echo $this->Form->input('Version');
echo $this->Form->input('Description');
echo $this->Form->input('UpdateID');
?>
</fieldset>
Which saves to the table:
CREATE TABLE `device` (
`DeviceID` VARCHAR(255) NOT NULL ,
`DeviceTypeID` INT(11) NOT NULL ,
`UserID` INT(10) NOT NULL ,
`Type` VARCHAR(10) NULL DEFAULT NULL ,
`KeyPadID` INT(10) NULL DEFAULT NULL ,
`Version` VARCHAR(255) NULL DEFAULT NULL ,
`Description` TINYBLOB NULL ,
`UpdateID` INT(11) NULL DEFAULT NULL ,
PRIMARY KEY (`DeviceID`),
INDEX `FK_USER` (`UserID`),
INDEX `FK_devices_updates` (`UpdateID`),
INDEX `FK_device_devicetype` (`DeviceTypeID`),
CONSTRAINT `FK_device_devicetype` FOREIGN KEY (`DeviceTypeID`) REFERENCES `devicetype` (`DeviceTypeID`),
CONSTRAINT `FK_devices_updates` FOREIGN KEY (`UpdateID`) REFERENCES `update` (`ID`) ON UPDATE CASCADE ON DELETE CASCADE,
CONSTRAINT `FK_USER` FOREIGN KEY (`UserID`) REFERENCES `user` (`UserID`) ON UPDATE CASCADE ON DELETE CASCADE
)
My problem is that when the form is displayed, it shows DeviceTypeID and UserID as well as UpdateID as the foreign key value instead of a drop down with the caption being the text and the value being the ID column. How would I go about setting a field from the foreign table to be the display field and the id as being the value?
Update 11-02-2013
First of all I strongly suggest to convert your primary and foreign keys accordingly
so that they meet the CakePHP naming conventions.
This means that:
DeviceID should be id.
DeviceTypeID should be device_type_id
UserID should be user_id
Also all primary keys in your tables should be named as id.
This way you will never have to worry about anything, concerning your models etc.
After that, all your tables must be in plural form. This means that device table should be devices, so you should rename it also.
I assume that you also have the following tables: devices_types and users.
At this point, I should notice that I would prefer to have a table named devicetype. I avoid underscored names, because it's very easy to make mistakes using the correct form for each object, class etc. So I don't have to worry whether I should use the CamelCase or not.
Anyway
Your Device model should be something like that:
<?php
/** Device.php **/
class Device extends AppModel {
public $name = 'Device';
public $belongsTo = array(
'DeviceType' => array(
'className' => 'DeviceType',
'foreignKey' => 'device_type_id'
/** Specify other keys that meet your needs **/
),
'User' => array(
'className' => 'User',
'foreignKey' => 'user_id'
)
);
};
?>
Also your DeviceType model should be similar to
<?php
/** DeviceType.php **/
class DeviceType extends AppModel {
public $name = 'DeviceType';
};
In your edit() method, you should query your DeviceType in something like this:
...
$devicetypes = $this->Device->DeviceType->find('list', array('id', 'caption'));
$this->set(compact('devicetypes'));
...
This way in your view the respective form element sets the <select> menu correctly.
PS: You should follow the CakePHP conventions about model-naming etc... Mine was just an example.
Related
I am using a drop down box with a foreign key relationship. I have got the drop down filling in the correct values but the only problem is when I add a user there is a foreign key constraint. But I can make users if I just use the normal input box and type an id that exists in the other table.
For example when I enter the id with this in my add.ctp, it works:
echo $this->Form->input('location');
but when I use this it doesn't
echo $this->Form->input('location_id', array('type' => 'select', 'options' => $CompanyLocations));
This is my add function in my UsersController
public function add()
{
$user = $this->Users->newEntity();
if ($this->request->is('post')) {
$user = $this->Users->patchEntity($user, $this->request->data);
if ($this->Users->save($user)) {
$this->Flash->success(__('The user has been saved.'));
return $this->redirect(['action' => 'index']);
}
$this->Flash->error(__('The user could not be saved. Please, try again.'));
}
$CompanyLocations= $this->Users->CompanyLocations->find('list');
$this->set(compact('CompanyLocations'));
$this->set(compact('user'));
$this->set('_serialize', ['user']);
This is in my UsersTable
$this->belongsTo('CompanyLocations');
and my CompanyLocationsTable
public function initialize(array $config)
{
parent::initialize($config);
$this->table('company_locations');
$this->displayField('location_name');
$this->primaryKey('location_id');
$this->belongsTo('Locations', [
'foreignKey' => 'location_id',
'joinType' => 'INNER'
]);
}
and my MySQL code
CREATE TABLE IF NOT EXISTS southpac_team.company_locations (
location_id INT NOT NULL AUTO_INCREMENT,
location_name VARCHAR(45) NULL,
PRIMARY KEY (location_id))
ENGINE = InnoDB;
DROP TABLE IF EXISTS southpac_team.users ;
CREATE TABLE IF NOT EXISTS southpac_team.users (
id INT NOT NULL AUTO_INCREMENT,
username VARCHAR(20) NOT NULL,
password VARCHAR(255) NOT NULL,
name VARCHAR(255) NOT NULL,
department INT NULL,
mobile VARCHAR(255) NULL,
email VARCHAR(255) NULL,
extension INT NULL,
lame_number INT NULL,
spa_auth_number VARCHAR(15) NULL,
creation_date DATE NULL,
picture VARCHAR(255) NULL,
employed TINYINT(1) NOT NULL,
location INT NOT NULL,
PRIMARY KEY (id),
INDEX get location_idx (location ASC),
CONSTRAINT get location
FOREIGN KEY (location)
REFERENCES southpac_team.company_locations(location_id)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;
Naming conventions
You are not following the naming conventions, by default the foreign key name for a belongsTo association is the singular underscored variant of the association alias, postfixed with _id so in the case of CompanyLocations that would be company_location_id, not just location.
echo $this->Form->input('company_location_id');
Also the variable holding the list should use camel casing, then you don't even need to specify it via the options argument:
$companyLocations= $this->Users->CompanyLocations->find('list');
$this->set(compact('companyLocations'));
Change the association defaults
If you are working with a legacy database that you cannot modify, then you need to configure CakePHP accordingly, ie specify the custom foreign key via the options argument of Table::belongsTo().
$this->belongsTo('CompanyLocations', [
'foreignKey' => 'location'
]);
Bake gets confused
The belongsTo association in CompanyLocationsTable looks fishy too, unless you really have a LocationsTable that should be associated with CompanyLocationsTable via:
company_locations.location_id > locations.primary_key
I guess you've created the model via bake, which treated location_id as a foreign key since it matches the default foreign key naming scheme for a belongsTo association.
See also
Cookbook > CakePHP at a Glance > Conventions > Model and Database Conventions
Cookbook > Database Access & ORM > Associations - Linking Tables Together > BelongsTo Associations
Cookbook > Views > Helpers > Form > Creating Form Controls
Cookbook > Views > Helpers > Form > Creating Inputs for Associated Data
I'm now to CakePHP (a few days) and I'm having issues with inserting records with Foreign Keys.
Basic idea, Warehouses.
I have a warehouses table that holds all the available warehouses and I have a warehouse_types table to hold the different types of warehouses. The warehouse_types table already has some data in it.
I have a simple add view (based off the blog example in CakePHP's Cookbook) and I have it displaying the names of the different types from the database.
When I save the page I get a Database error.
Error: SQLSTATE[23000]: Integrity constraint violation: 1452 Cannot add or update a child row: a foreign key constraint fails (`kvs`.`warehouses`, CONSTRAINT `warehouses_ibfk_1` FOREIGN KEY (`warehouse_type_id`) REFERENCES `warehouse_types` (`id`))
The SQL that is being generated is:
SQL Query: INSERT INTO `kvs`.`warehouses` (`name`, `location`, `modified`, `created`) VALUES ('Name Text', 'Location Text', '2014-04-02 12:07:13', '2014-04-02 12:07:13')
If I output the information in the request->data object, it shows that it is getting a value for the warehouseType_id.
array(
'Warehouse' => array(
'name' => 'Name Text',
'location' => 'Location Text',
'warehouseType_id' => '1'
)
)
Any help on this would be awesome!
My database tables are as follows:
CREATE TABLE `warehouses` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`name` varchar(50) DEFAULT NULL,
`location` varchar(255) DEFAULT NULL,
`warehouse_type_id` int(10) unsigned NOT NULL,
`created` datetime DEFAULT NULL,
`modified` datetime DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `warehouse_type_id` (`warehouse_type_id`),
CONSTRAINT `warehouses_ibfk_1` FOREIGN KEY (`warehouse_type_id`) REFERENCES `warehouse_types` (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=10 DEFAULT CHARSET=latin1;
CREATE TABLE `warehouse_types` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`name` varchar(50) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1;
My models are as follows:
class Warehouse extends AppModel{
public $belongsTo = 'WarehouseType';
}
class WarehouseType extends AppModel{
}
My view (add.ctp) is as follows:
<h1>Add Warehouse</h1>
<?php
echo $this->Form->create('Warehouse');
echo $this->Form->input('name');
echo $this->Form->input('location');
echo $this->Form->input('warehouseType_id');
echo $this->Form->end('Create Warehouse');
?>
Finally, the add method in my WarehousesController:
public function add(){
//Check Request to see if it is a Post
if($this->request->is('post')){
$this->Warehouse->create(); //Make a new Warehouse object
debug($this->request->data);
if($this->Warehouse->save($this->request->data)){ //Try and save Warehouse
$this->Session->setFlash(__('Warehouse '.$this->request->data['Warehouse']['name'].' created'));
return $this->redirect(array('action' => 'index')); //Redirect back to Warehouse list
}
$this->Session->setFlash(__('Unable to add Warehouse :('));
}
$this->set('warehouseTypes', $this->Warehouse->WarehouseType->find('list'));
}
I've read over this answer here that seems the most relevant, but I'm lost: CakePHP Can't insert record with foreign key error
Also reviewed their documentation on saving: http://book.cakephp.org/2.0/en/models/saving-your-data.html#saving-related-model-data-hasone-hasmany-belongsto
Your column name is different than what is being used in CakePHP. Change CakePHP to use warehouse_type_id. (Your CakePHP code is using warehouseType_id)
Change this- on your view file-
Add Warehouse
<?php
echo $this->Form->create('Warehouse');
echo $this->Form->input('name');
echo $this->Form->input('location');
echo $this->Form->input('warehouseTypes');
echo $this->Form->end('Create Warehouse');
?>
I have a requirement where I need to build the relations between more than 3 tables.
I have 4 tables namely, Message, Flat, Person, Mapping tables.
Now, below tables have the following fields:
Message:
`Id` int(11) NOT NULL AUTO_INCREMENT,
`Mapid` int(11) DEFAULT NULL,
PRIMARY KEY (`Id`),
KEY `FK41715B218022FC0` (`MapId`)
Mapping
`Id` int(11) NOT NULL AUTO_INCREMENT,
`FlatId` int(11) DEFAULT NULL,
PRIMARY KEY (`Id`),
KEY `FKE2B3C68A24F94F50` (`FlatId`),
Flat
`Id` int(11) NOT NULL AUTO_INCREMENT,
`PersonId` int(11) DEFAULT NULL,
PRIMARY KEY (`Id`),
KEY `FK2FFF79122B94A6` (`PersonId`),
Person
`Id` int(11) NOT NULL AUTO_INCREMENT,
`Name` varchar(255) DEFAULT NULL,
`FlatId` int(11) DEFAULT NULL,
`Phone` varchar(255) DEFAULT NULL,
PRIMARY KEY (`Id`),
KEY `FKC4E39B55AF5432C` (`FlatId`),
Now, I have to build relations in such a way that in the Cgridview(admin.php) of Message, i should display PersonId of flat table and Name and Phone of Person table along with the columns of Message table.
I have defined relations like this in model class of message(message.php)
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(
'mapping' => array(self::BELONGS_TO, 'Mapping', 'MapId'),
'flat'=>array(self::HAS_ONE,'Flat',array('FlatId'=>'Id'),'through'=>'mapping'),
'person'=>array(self::HAS_ONE,'Person',array('PersonId'=>'Id'),'through'=>'flat'),
);
}
Can anyone explain me the step by step procedure to display the columns of person table in message gridview.
With assuming you can create dataProvider for the CGridView:
<?php
$this->widget('zii.widgets.grid.CGridView',array(
'id'=>'message-grid',
'dataProvider'=>$yourDataProvider //such as $model->search();
'filter'=>$model,
'columns'=>array(
'Id',
'Mapid',
'person.name',
'person.FlatId',
'person.Phone',
));
?>
You can pass a DataProvider to the view and use it in CgridView widget or use a 'search()' action from $model.
You can personalize the relation columns like this:
<?php
$this->widget('zii.widgets.grid.CGridView',array(
'id'=>'messagePerson-grid',
'dataProvider'=>$model->search(),
'filter'=>$model,
'columns'=>array(
'Id',
'Mapid',
array(
'header'=>'Person Name', // Personalize column name
'value'=>'$data->flat->person->Name',
'htmlOptions'=>array('style'=>'width:10%;'), // Personalize html attributes
),
'flat.person.FlatId', // Or directly with default relation name.
'flat.person.Phone',
));
?>
I'm using Datamapper ORM for the first time together with the array and htmlforms extensions for CRUD. So far both extensions worked well for all my models but today I'm facing a weird problem for one model.
When I try to save an object I get this SQL error:
A Database Error Occurred
Error Number: 1054
Unknown Column 'orderitems.orderitem_id' in where clause
SELECT COUNT(*) AS `numrows` FROM (`orderitems`) WHERE `orderitems`.`order_id` = 2 AND `orderitems`.`orderitem_id` NOT IN (8, 9)
Filename: libraries/datamapper.php
Line Number: 2526
I think the correct SQL should be ...AND orderitems.id NOT IN (...) becasue the Orderitem model is referencing itself and not another related model. This error is triggered inside the count() method which is triggered by the save() method which is triggered by the from_array() method of the array DM extension.
Here is my controller code
//Get order
$order = new Order(2);
//Fields to render in the form
$fields = array('orderitem');
//Save changes from $_POST
if($post = $this->input->post())
{
$order->trans_begin();
if ( ! $order->from_array($post, $fields, TRUE) OR $order->trans_status() === FALSE)
$order->trans_rollback();
else
$order->trans_commit();
}
//Load view
$data = array(
'order' => $order,
'fields'=> $fields,
);
$this->load->view('order/edit', $data);
Here is my view code
echo $order->render_form($fields);
And here are the implied models
class Order extends DataMapper {
public $has_one = array('orderstatus', 'place', 'user', 'paymenttype');
public $has_many = array('orderitem');
...
/*SQL:
CREATE TABLE IF NOT EXISTS orders (
id mediumint(4) unsigned NOT NULL auto_increment PRIMARY KEY,
user_id mediumint(1) unsigned, FOREIGN KEY (user_id) REFERENCES users(id) ON UPDATE CASCADE ON DELETE RESTRICT,
orderstatus_id tinyint(1) unsigned, FOREIGN KEY (orderstatus_id) REFERENCES orderstatuses(id) ON UPDATE CASCADE ON DELETE RESTRICT,
place_id tinyint(1) unsigned, FOREIGN KEY (place_id) REFERENCES places(id) ON UPDATE CASCADE ON DELETE SET NULL,
paymenttype_id tinyint(1) unsigned, FOREIGN KEY (paymenttype_id) REFERENCES paymenttypes(id) ON UPDATE CASCADE ON DELETE SET NULL,
total float NOT NULL DEFAULT 0
) ENGINE = InnoDB;
*/
class Orderitem extends DataMapper {
public $has_one = array('order', 'orderitemstatus');
public $has_many = array('orderitemextra');
...
/*SQL:
CREATE TABLE IF NOT EXISTS orderitems (
id int(1) unsigned NOT NULL auto_increment PRIMARY KEY,
order_id mediumint(1) unsigned, FOREIGN KEY (order_id) REFERENCES orders(id) ON UPDATE CASCADE ON DELETE CASCADE,
orderitemstatus_id tinyint(1) unsigned, FOREIGN KEY (orderitemstatus_id) REFERENCES orderitemstatuses(id) ON UPDATE CASCADE ON DELETE RESTRICT,
name varchar(128) NOT NULL,
price float NOT NULL,
) ENGINE = InnoDB;
*/
As I said I had no problems with other models so I don't know if it's a DM problem or a problem with my code or any of my models definitions.
I'll appreciate any kind of help.
Thanks!
I am going to attempt to keep this as simple as possible, but the use case is outside the original intention of Zend_Db I fear. It concerns a set of tables I have for tagging pages (or anything else eg. documents) in a CMS.
I have three tables:
Pages (pages)
Tags (tags)
TagLink (tags_link) which is a many-to-many linking table between Pages and Tags
Pages is a simple table (I have removed the inconsequential columns from the code below):
CREATE TABLE `pages` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`name` varchar(255) NOT NULL DEFAULT '',
PRIMARY KEY (`id`),
FULLTEXT KEY `search` (`name`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8
Tags is quite simple as well although there is a self-referential column (parent_tag_id):
CREATE TABLE `tags` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`tag` varchar(255) NOT NULL,
`parent_tag_id` int(11) NOT NULL DEFAULT '0',
`updated` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
KEY `GetByParentTagId` (`parent_tag_id`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8
TagLink is again fairly simple:
CREATE TABLE `tags_link` (
`tag_id` int(11) NOT NULL,
`module_type` varchar(50) NOT NULL,
`foreign_key` int(11) NOT NULL,
UNIQUE KEY `Unique` (`tag_id`,`module_type`,`foreign_key`),
KEY `Search` (`module_type`,`foreign_key`),
KEY `AllByTagId` (`tag_id`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8
The complicating factor is that TagLink is able to link against any other table in the database and not just Pages. So if for example I had a documents upload section then that could also be tagged. To facilitate this way of working there is effectively a multi-column key.
To make this clearer I will demonstrate a couple of insert queries that might be run when tags are added to a table (eg. Pages):
INSERT INTO `tags_link`
SET `tag_id` = '1',
`module_type` = 'Pages',
`foreign_key` = '2'
INSERT INTO `tags_link`
SET `tag_id` = '1',
`module_type` = 'Documents',
`foreign_key` = '3'
So as you can see the module_type column is simply an arbitrary string that describes where the foreign key can be found. This is not the name of the table however as anything with an ID can have tags linked to it even if it is not necessarily in the MySQL database.
Now to the Zend_Db_Table $_referenceMap in PageTable:
protected $_referenceMap = array(
'TagLink' => array(
'columns' => 'id',
'refTableClass' => 'Models_Tag_TagLinkTable',
'refColumns' => 'foreign_key'
),
);
But this does not take into account my arbitrary module_type column and will return any TagLink with the same foreign key. Obviously this is bad because you get TagLinks for documents mixed in with those for pages for instance.
So my question is how can I take into account this additional column when setting up this reference? The aim is to avoid having a TagLink class for each module_type as I have now.
I would imagine something like the following could explain my requirements although obviously this is not how it would be done:
protected $_referenceMap = array(
'TagLink' => array(
'columns' => 'id',
'refTableClass' => 'Models_Tag_TagLinkTable',
'refColumns' => 'foreign_key',
'where' => 'module_type = "Pages"'
),
);
My current implementation overrides the _fetch method in the Documents_TagLinkTable in the following way:
protected function _fetch(Zend_Db_Table_Select $select) {
$select->where("module_type = 'Documents_Secondary_Tags' OR module_type = 'Documents_Primary_Tags' OR module_type = 'Documents'");
return parent::_fetch($select);
}
As you can see there maybe more than one set of tags added to any object as well.
Example 3 in "Fetching Dependent Rowsets" in the Zend Framework reference demonstrates a technique you could use:
http://framework.zend.com/manual/en/zend.db.table.relationships.html
Whilst it doesnt show a "where" clause being included in the select, it should work.
Duncan