What's the fastest way to implement Gravatar URLs in Laravel? I have a mandatory email address field, but I don't want to create a new column for Gravatars, and I'd prefer to use the native Auth::user() attributes.
Turns out, you can use a Laravel mutator to create attributes that don't exist in your model. Assuming you have a User model with a mandatory email column in the corresponding users table, just stick this in your User model:
public function getGravatarAttribute()
{
$hash = md5(strtolower(trim($this->attributes['email'])));
return "http://www.gravatar.com/avatar/$hash";
}
Now when you do this:
Auth::user()->gravatar
You'll get the gravatar.com URL you're expecting. Without creating a gravatar column, variable, method, or anything else.
Expanding on Wogan's answer a bit...
Another example using a Trait:
namespace App\Traits;
trait HasGravatar {
/**
* The attribute name containing the email address.
*
* #var string
*/
public $gravatarEmail = 'email';
/**
* Get the model's gravatar
*
* #return string
*/
public function getGravatarAttribute()
{
$hash = md5(strtolower(trim($this->attributes[$this->gravatarEmail])));
return "https://www.gravatar.com/avatar/$hash";
}
}
Now on a given model (i.e. User) where you want to support Gravatar, simply import the trait and use it:
use App\Traits\HasGravatar;
class User extends Model
{
use HasGravatar;
}
If the model doesn't have an email column/attribute, simply override the default by setting it in the constructor of your model like so:
public function __construct() {
// override the HasGravatar Trait's gravatarEmail property
$this->gravatarEmail = 'email_address';
}
Related
I'm using Laravel 9 and I have a request can contains :
Parameter called SEASON the value can be an array or null
so SEASON parameter can be an array and can be also null
Parameter called EXPIRY can be an array and can be also null
I have two classes one for the SEASON feature and the other class for EXPIRY both they extends from Repository. and both have a method called execute that return an array
abstract class Repository
{
abstract public function execute(): array;
}
class Expiry extends Repository
{
public function execute()
{
return ['The Request contain Expiry Parameter, and seasonal behaviours is done'];
}
}
class Season extends Repository
{
public function execute()
{
return ['The Request contain Season Parameter, and expiry behaviours is done'];
}
}
I would like to call execute method of Season class if my request contains SEASON, or call the execute method of expiry if my request contains Expiry. OR Call both of them and merge the execute return of execute in one array so I can have as result.
['The Request contain Expiry Parameter, and seasonal behaviours is done', 'The Request contain Expiry Parameter, and expiry behaviours is done']
That's what I tried inside my controller :
public function bootstrap($data)
{
$parseTopics = Helper::parseTopicsRequest();
$basicProgram = new BasicProgramRepository();
$seasonalProgram = new SeasonalProgramRepository($parseTopics['SEASONAL']);
$object = count($parseTopics['SEASONAL']) ? $seasonalProgram : $basicProgram;
// Polymorphism
return $object->execute();
}
Question 1 :
I'm not sure if I should use this way or something like to fix my need:
$employe = new Program(new BasicProgramRepository());
Expected Result :
The expected result depends on if I have season parameter and expiry. What I want to achieve is to use different behaviours ( execute method )
if you want to achieve Polymorphism method, it will be better creating repository or something only for managing that logic.
here is sample.
class SampleRepository
{
/**
* repository instance value
*
* #var string[] | null
*/
private $sampleArray; // maybe here is SEASON or EXPIRY or null
/**
* constructor
*
* #param string[] | null $sampleArray
*/
public function __construct($sampleArray)
{
$this->sampleArray = $sampleArray;
}
/**
* execute like class interface role
*
* #return array
*/
public function execute()
{
return (!$this->sampleArray) ? [] : $this->getResult();
}
/**
* get result
*
* #return array
*/
private function getResult()
{
// maybe pattern will be better to manage another class or trait.
$pattern = [
"SEASON" => new Season(),
"EXPIRY" => new Expiry()
];
return collect($this->sampleArray)->map(function($itemKey){
$requestClass = data_get($pattern,$itemKey);
if (!$requestClass){ // here is space you don't expect class or canIt find correct class
return ["something wrong"];
}
return $requestClass->execute();
})->flatten();
}
}
and you can call like this.
$sampleRepository = new SampleRepository($sampleValue); // expect string[] or null like ["SEASON"],["SEASON","EXPIRY"],null
$result = $sampleRepository->execute(); // [string] or [string,string] or []
this approach is only what your parameter is secified value.
if your return result is almost same both of Season class and Expiry class, it will be better to manage on trait. (that is $pattern on sample code)
try some.
I read comments,so following..
For example, it prefers to be only getting result of getResult().
so, some pattern and so many logics shouldn't be written on getResult();
If you use trait, this is sample.
first, you need to create managing behaviors class.
Behavior.php
<?php
namespace App\Repositories;
class Behavior
{
use Behavior\BehaviorTrait;
// if you need to add another pattern, you can add trait here.
}
and then, you need to create Behavior directory at same level place.
you move that directory, you create trait file like this.
<?php
namespace App\Repositories\Behavior;
trait BehaviorTrait
{
public static function findAccessibleClass(string $itemKey)
{
return data_get([
"SEASON" => new Season(),
"EXPIRY" => new Expiry()
],$itemKey);
}
}
findAccessibleClass() method has responsible of finding correct class.
then, you call this method like this.
private function getResult()
{
return collect($this->sampleArray)->map(function($itemKey){
$requestClass = Behavior::findAccessibleClass($itemKey); // fix here.
if (!$requestClass){ // here is space you don't expect class or canIt find correct class
return ["something wrong"];
}
return $requestClass->execute();
})->flatten();
}
if your code is so much in getResult(), you will be better to separate code for responsible.
To create Behavior trait, getResult don't need to have responsible of behavior logic. it will be easy testing or fixable in short.
hope well.
I'm new to OOP and MVC with PHP, and I'm currently learning by making my own custom framework from scratch, for testing purposes. I have set up my controllers, models and views and everything works fine.
My app has the following architecture :
It’s a small blog that follows the rules of the MVC pattern. To summarize, it works like this :
The called Controller will fetch the data using the right models
Models return objects of the class \Classes\{MyObject}
Controller call the right template to render the view, and passes it the data and objects to display
The problem
In some views, I need to display related data. For example, in the article view, I need to display the author's first name. In the database, an article contains only the author’s ID, not his first name : this is the same thing in my class \Classes\Article.
What I've tried
To display the author’s first name in my view, I've updated the model Find method to use a LEFT JOIN in the SQL query. Then, I've updated my \Classes\Article class to have a user_firstname property :
class Article
{
private $pk_id;
private $title;
private $excerpt;
private $content;
private $created_at;
private $fk_user_id;
private $updated_at;
private $user_firstname; // <-- I've added this property to retrieve author's firstname
// (...)
}
What I did works well, but my teacher tells me it’s not the right way to do it because the author’s firstname is not part of the definition of an article.
In this case, my teacher tells me to use a DTO (Data Transfert Object) between Article and User classes.
Questions
What is the right way to set up a DTO in this case?
Do I need to create a new ArticleUserDTO class in a new namespace ?
How to use it ?
I think I understood the problem : the Article class should only contain what defines an article. But I can’t understand the logic of setting up a DTO. I’ve done some research on it, I understand the usefulness of the DTO but I can’t set up into my app.
Setting up a DTO in my app was easy ! As mentioned by #daremachine, the diagram below helped me to understand what a DTO is for.
Diagram source : martinfowler.com
We can see DTOs as an object assembler, in which we place all the elements we need on the view side.
For example, in the post view of an article, I needed to display other items, such as the author and posted comments. So I have created a Post class that groups all these items.
Setting up a DTO
In my \Classes\ namespace, I've created a new Post class. First, we define the properties we will need. Then we add the getters and setters for each of them. Finally, we set up the constructor, which will call each of the classes we need in the view.
namespace Classes;
use DateTime;
class Post
{
private int $pk_id;
private string $title;
private string $excerpt;
private string $content;
private DateTime $created_at;
private DateTime $updated_at;
private int $author_id;
private string $author_firstname;
private array $comments;
public function __construct(Article $article, User $author, array $comments)
{
$this->setPkId($article->getId());
$this->setTitle($article->getTitle());
$this->setExcerpt($article->getExcerpt());
$this->setContent($article->getContent());
$this->setCreatedAt($article->getCreatedAt());
$this->setUpdatedAt($article->getUpdatedAt());
$this->setAuthorId($article->getAuthorId());
$this->setAuthorFirstname($author->getFirstname());
$this->setComments($comments);
}
/**
* #param int $pk_id
*/
public function setPkId(int $pk_id): void
{
$this->pk_id = $pk_id;
}
/**
* #return int
*/
public function getPkId(): int
{
return $this->pk_id;
}
// (etc)
}
We now need to update the ArticleController, which should no longer pass the Article, Comment and User objects, but only the new Post object.
namespace Controllers;
class ArticleController extends Controller
{
// (...)
/**
* Get an article and display it
*
* #return void
*/
public function show(): void
{
// (...)
// Find Article :
$article = $this->articleModel->find($article_id);
if (!$article) {
Http::error404();
}
// Find Comments :
$commentaires = $this->commentModel->findAllByArticle($article_id);
// Find User (author)
$user = $this->userModel->find($article->getAuthorId());
// Data Transfert Object instance :
$post = new Post($article, $user, $commentaires);
$pageTitle = $post->getTitle();
// Pass DTO to view :
Renderer::render('articles/show', compact('pageTitle', 'post'));
}
}
We just need to update our view to use the new Post object and it's done ! Thanks to #daremachine for his help :)
I made a library class that I am using for some common functions not provided by Laravel. It's been loaded into /config/app.php under the 'aliases' array, so that shouldn't be the problem.
When I call a method from my class ("InfoParse"), my conroller returns a blank page. I think this has to do with the fact that I'm calling a method from the library which uses Eloquent to interface with the database. I tried adding
use Illuminate\Database\Eloquent\Model;
to the top of the file, but that didn't help either.
Is there a specific way I should be setting up my class file so I can use either the DB:: class or Eloquent class?
Below is the function in question:
/**
* Check to see if this student is already recorded in our student table.
* If not, add the entry, then return true.
* #param int $cwid
* #return boolean
*/
public static function checkStudentTableRecords($cwid)
{
if(Student::where('cwid', '=', $cwid)->count() != 0)
{
return TRUE;
}
else
{ ##insert the student into our student table
$studentInfo = self::queryInfoFromCWID($cwid);
$studentEntry = new Student;
$studentEntry->cwid = $cwid;
$studentEntry->fName = $studentInfo['fName'];
$studentEntry->lName = $studentInfo['lName'];
$studentEntry->email = $studentInfo['email'];
$studentEntry->save();
return TRUE;
}
}
(note: the self::queryInfoFromCWID() function is calling a function defined earlier in the class)
After some investigation, it turns out I need to format my Eloquent Model calls like this:
if(\Student::where('cwid', '=', $cwid)->count() != 0)
...
$studentEntry = new \Student;
The backslash is necessary to avoid namespace collision within the Laravel4 application.
In controller A, I load a model not associated with this controller. I'm interested in managing the model name of controller B with a single variable, so I don't have to manually change many lines if the table/model B's name changes.
For example below is the controller A's code:
public $modelBName = 'ModelB';
public function controller_a_function() {
$this->loadModel($this->modelBName); // I use the variable here for model B
$this->ModelB->model_b_function(); // COMMENT #1
}
Question:
For the line commented "COMMENT #1," how do I use the variable name instead of explicitly written out word 'ModelB'? This line appears multiple times throughout the code, and I would like to use the variable $modelBName if possible. ModelB will likely not change, but if it does for some reason, it would be nice to just change one variable instead of editting multiple lines.
The simple answer; use this:
$this->{$this->modelBName}->find('all');
Note the curly brackets {} around the property name. more information can be found in the manual;
http://php.net/manual/en/language.variables.variable.php
A cleaner approach may be a sort of 'factory' method;
/**
* Load and return a model
*
* #var string $modelName
*
* #return Model
* #throws MissingModelException if the model class cannot be found.
*/
protected function model($modelName)
{
if (!isset($this->{$modelName})) {
$this->loadModel($modelName);
}
return $this->{$modelName};
}
Which can be used like this;
$result = $this->model($this->modelBName)->find('all');
debug($result);
And, if you don't want to specify the model, but want it to return a '$this->modelBName' automatically;
/**
* Load and return the model as specified in the 'modelBName' property
*
* #return Model
* #throws MissingModelException if the model class cannot be found.
*/
protected function modelB()
{
if (!isset($this->{$this->modelBName})) {
$this->loadModel($this->modelBName);
}
return $this->{$this->modelBName};
}
Which can be used like this:
$result = $this->modelB()->find('all');
debug($result);
I think you are confused between model name and table name. You can set a model to use a different database table by using the $useTable property, for example:
class User extends AppModel {
public $useTable = 'users_table'; // Database table used
}
class Product extends AppModel {
public function foo() {
$this->loadModel('User');
$this->User->find('all');
}
}
You should never need to change the name of the model, and if the name of the database table changes you can simply update the $useTable property in the model.
I have an model with a relation, and I want to instantiate a new object of the relations type.
Example: A person has a company, and I have a person-object: now I
want to create a company-object.
The class of the companyobject is defined in the relation, so I don't think I should need to 'know' that class, but I should be able to ask the person-object to provide me with a new instance of type company? But I don't know how.
This is -I think- the same question as New model object through an association , but I'm using PHPActiveRecord, and not the ruby one.
Reason behind this: I have an abstract superclass person, and two children have their own relation with a type of company object. I need to be able to instantiate the correct class in the abstract person.
A workaround is to get it directly from the static $has_one array:
$class = $this::$has_one[1]['class_name'];
$company = new $class;
the hardcoded number can of course be eliminated by searching for the association-name in the array, but that's still quite ugly.
If there is anyone who knows how this is implemented in Ruby, and how the phpactiverecord implementation differs, I might get some Ideas from there?
Some testing has revealed that although the "search my classname in an array" looks kinda weird, it does not have any impact on performance, and in use it is functional enough.
You can also use build_association() in the relationship classes.
Simplest way to use it is through the Model's __call, i.e. if your relation is something like $person->company, then you could instantiate the company with $company = $person->build_company()
Note that this will NOT also make the "connection" between your objects ($person->company will not be set).
Alternatively, instead of build_company(), you can use create_company(), which will save a new record and link it to $person
In PHPActiveRecord, you have access to the relations array. The relation should have a name an you NEED TO KNOW THE NAME OF THE RELATIONSHIP/ASSOCIATION YOU WANT. It doesn't need to be the classname, but the classname of the Model you're relating to should be explicitly indicated in the relation. Just a basic example without error checking or gritty relationship db details like linking table 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 $relationship){
//the first element in all relationships is it's name
if($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 an object defined by the 'company' relationship use:
$className = $person::getClassNameFromRelationship('company');
$company = new $className();
I'm currently using below solution. It's an actual solution, instead
of the $has_one[1] hack I mentioned in the question. If there is a
method in phpactiverecord I'm going to feel very silly exposing
msyelf. But please, prove me silly so I don't need to use this
solution :D
I am silly. Below functionality is implemented by the create_associationname call, as answered by #Bogdan_D
Two functions are added. You should probably add them in the \ActiveRecord\Model class. In my case there is a class between our classes and that model that contains extra functionality like this, so I put it there.
These are the 2 functions:
public function findClassByAssociation($associationName)
Called with the name of the association you are looking for.
Checks three static vars (has_many,belongs_to and has_one) for the association
calls findClassFromArray if an association is found.
from the person/company example: $person->findClassByAssociation('company');
private function findClassFromArray($associationName,$associationArray)
Just a worker-function that tries to match the name.
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;
}