Activerecord-association: create a new object (find a class)

Question

I have a model with a relationship, and I want to instantiate a new relationship type object.

Example: a person has a company, and I have a person-object: now I want to create a company-object.

The companyobject class is defined in relation, so I don’t think I need to “know” this class, but should I ask the personal object to provide me with a new instance of the Company type? But I dont know how.

This - I think - is the same question as the New Model Object through association , but I use PHPActiveRecord , not ruby.

<h / "> The reason for this: I have an abstract superclass person , and two children have their own attitude to the type of corporate object. I need to be able to create an instance of the correct class in an abstract person.

The workaround is to get it directly from the static $has_one :

 $class = $this::$has_one[1]['class_name']; $company = new $class; 

a hard-coded number, of course, can be eliminated by looking for the association name in the array, but it's still pretty ugly.

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If someone knows how this is implemented in Ruby, and how is the implementation of phpactiverecord different, can I get some ideas from there?

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Some tests showed that although “searching for my class name in an array” looks strange, it has no effect on performance and is quite functional in use.

Answer 1

You can also use build_association() in relationship classes.
The simplest way to use it is through Model __call, i.e. If your relationship is like $person->company , you can create an instance of the company with
$company = $person->build_company()

Please note that this will NOT make a “connection” between your objects $person->company ( $person->company will not be established).
Alternatively, instead of build_company() you can use create_company() , which will save the new entry and associate it with $ person

Answer 2

I am currently using the solution below. This is the actual solution, instead of the $has_one[1] hack I mentioned in the question. If there is a method in phpactiverecord I will feel very silly exposing msyelf. But please prove me stupid, so I do not need to use this solution: D

I am stupid. The functionality below is implemented by calling create_associationname, as @Bogdan_D answers

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Two features added. You should probably add them to the \ActiveRecord\Model class. In my case, there is a class between our classes and this model that contains additional functions like this, so I posted it there.

These are 2 functions:

• public function findClassByAssociation($associationName)
  • Called with the name of the association you are looking for.
  • Checks three static vars ( has_many , belongs_to and has_one ) for association
  • calls findClassFromArray if an association is found.
  • from person / company: $person->findClassByAssociation('company');
• private function findClassFromArray($associationName,$associationArray)
  • Just a working function that tries to match a name.

A source:

 /** * Find the classname of an explicitly defined * association (has_one, has_many, belongs_to). * Unsure if this works for standard associations * without specific mention of the class_name, but I suppose it doesn't! * @todo Check if works without an explicitly set 'class_name', if not: is this even possible (namespacing?) * @todo Support for 'through' associations. * @param String $associationName the association you want to find the class for * @return mixed String|false if an association is found, return the class name (with namespace!), else return false * @see findClassFromArray */ public function findClassByAssociation($associationName){ //$class = $this::$has_one[1]['class_name']; $that = get_called_class(); if(isset($that::$has_many)){ $cl = $this->findClassFromArray($associationName,$that::$has_many); if($cl){return $cl;} } if(isset($that::$belongs_to)){ $cl = $this->findClassFromArray($associationName,$that::$belongs_to); if($cl){return $cl;} } if(isset($that::$has_one)){ $cl = $this->findClassFromArray($associationName,$that::$has_one); if($cl){return $cl;} } return false; } /** * Find a class in a php-activerecord "association-array". It probably should have a specifically defined class name! * @todo check if works without explicitly set 'class_name', and if not find it like standard * @param String $associationName * @param Array[] $associationArray phpactiverecord array with associations (like has_many) * @return mixed String|false if an association is found, return the class name, else return false * @see findClassFromArray */ private function findClassFromArray($associationName,$associationArray){ if(is_array($associationArray)){ foreach($associationArray as $association){ if($association['0'] === $associationName){ return $association['class_name']; } } } return false; } 

Answer 3

In PHPActiveRecord, you have access to an array of relationships. The relationship must have a name that YOU NEED TO KNOW THE LINK / ASSOCIATION NAME WHAT YOU WANT . This should not be the name of the class, but the class name of the model you are talking about should be explicitly referenced. Just a basic example without checking errors or detailing db relationships such as table reference or foreign key column name:

 class Person extends ActiveRecord\Model { static $belongs_to = array( array('company', 'class_name' => 'SomeCompanyClass') ); //general function get a classname from a relationship public static function getClassNameFromRelationship($relationshipName) foreach(self::$belongs_to as $relationship){ //the first element in all relationships is it name if($relationship[0] == $relationshipName){ $className = null; if(isset($relationship['class_name'])){ $className = $relationship['class_name']; }else{ // if no classname specified explicitly, // assume the clasename is the relationship name // with first letter capitalized $className = ucfirst($relationship); } return $className } } return null; } } 

To with this function, if you have a person object and want the object defined by the company relationship to use:

 $className = $person::getClassNameFromRelationship('company'); $company = new $className();