Hi I followed a tutorial to implement a friend system. It all works find, but I need to post other columns to the row that just the id's. How would I expand that.
This is the method that is accessed when the add friend button is clicked
public function getAdd($id){
$user = User::where('id', $id)->first();
//After passing all checks. Add other account
Auth::user()->addFriend($user);
echo "Sent";
}
AddTenancy Method
public function addFriend(User $user){
$this->friendsOf()->attach($user->id);
}
I assume the relationship is many-to-many between users. And you need to add additional data to the pivot.
Here's how you'd do that:
public function addFriend(User $user){
$this->friendsOf()->attach($user->id, ['another_col' => 'some data']);
}
Replace 'another_col' and some data with your column and your data. You can also add more than 1 column into the array.
Related
I have 2 tables called jobs & job_records. Its relationship as below:
JobRecords Model
public function job()
{
return $this->belongsTo(Job::class);
}
Job Model:
public function jobRecord()
{
return $this->hasOne(JobRecord::class);
}
jobs table has 2 columns that I need to display alongside my job_records table view. It's total_pges & status.
In my JobRecords Controller, I have tried the following method. It throws me an error of Call to undefined relationship.
JobRecordController:
$job_records = JobRecord::whereStatus('In Progress')
->with('jobs', 'jobs.status', 'jobs.total_pges')
->get();
return DataTables::of($job_records)
I am still beginning with Laravel and PHP. I can sense that there is something wrong with the relationship. But I couldn't figure out what it is exactly. Can anyone help me out with this matter?
In your JobRecord model change the relation ship as
public function job()
{
return $this->hasOne('App\Models\Job','foreign_key','local_key');
}
Similarly, in Job model
public function job()
{
return $this->belongsTo('App\Models\JobRecord','foreign_key','local_key');
}
Replace foreign_key and local_key with appropriate values...
I deleted my previous answer. What are you trying to do exactly? You can't use "jobs" in the "with function" without to define "jobs" as function in the model.
If you change it to "job" (instead of "jobs), then it would work, but I don't know if you want this. With your query you saying that a record have many jobs? But your model doesn't define that.
I want to store in my database all the user actions done about an entity.
For example, for 1 entity, I want to store :
Created by (= author)
Updated by
Date of creation
Date of update
I want to store the history of the actions of a user, not the last ones. I thought I could create a table with these columns :
log_id
user_id
entity_id
action (= "create" or "update" or something else)
date
And then, I could easily get the last update of my entity and display the date and the user who did it.
Is there a Symfony bundle to do this ? Should I use Monolog ?
I will do this for many entities and I'm not sure if this is the correct way to do...
Is it possible to create only one logs table to store each log about each entity ? It bothers me to create 1 logs table per entity.
Since Doctrine is event based, it's easy:
Either use an extension, like Gedmo Loggable
Or hook into Doctrine's events and log, using Monolog, everything that happens in your app.
Personally I would prefer option 2 since I'm a control maniac, it's a little more complex though. Personally I would also use Monolog so I could abstract away the way how and where the log entries are stored.
When you decide how to approach this and you will need any assistance along the way, please ask another question.
Good luck.
I don't know if that would fit what you need, but you could easily add a Listener to the symfony kernel to log every controller used.
Something like this :
class UserLogListener {
protected $authChecker;
protected $tokenStorage;
protected $entityManager;
public function __construct(TokenStorageInterface $tokenStorage, AuthorizationChecker $authChecker, EntityManager $entityManager)
{
$this->authChecker = $authChecker;
$this->tokenStorage = $tokenStorage;
$this->entityManager = $entityManager;
}
public function onKernelRequest(GetResponseEvent $event)
{
if( $this->tokenStorage->getToken() != null){
$user = $this->tokenStorage->getToken()->getUser();
$currentDate = new \Datetime();
$action = $event->getRequest()->attributes->get('_controller');
$method = $event->getRequest()->getMethod();
$userIp = $event->getRequest()->getClientIp();
$userLogRepository = $this->entityManager->getRepository(UserLog::class);
if($user instanceof User){
$userLog = new UserLog();
$userLog->setUser($user);
$userLog->setIp($userIp);
$userLog->setAction($action);
$userLog->setMethode($method);
$userLog->setDate($currentDate);
if($event->getRequest()->request && $methode=='POST'){
$userLog->setData(json_encode($event->getRequest()->request->all()));
}else{
$userLog->setData($event->getRequest()->getPathInfo());
}
$this->entityManager->persist($userLog);
$this->entityManager->flush();
}
}
}
}
What it does is add to the database (with an entity called UserLog) information about every page called. So you can know which action is made by knowing which controller is called, and you can also log the request data so you can find out what modification/creation the user did.
I used many to many relationship in my user table to make the logged in user to follow another one, but I didn't figuerd it out by myself, I checked what others did and tried to do something similar and it worked. In my methods I have:
function follow(User $user) {
$this->followers()->attach($user->id);
}
function unfollow(User $user) {
$this->followers()->detach($user->id);
}
which allows me follow.
the tables are related with a function something like:
return $this->belongsToMany('App\User', 'followers', 'user_id', 'follower_id');
now I pass the $user value by the controller and the controller is pretty simple:
$userId = User::find($user);
$willfollow = Auth::user();
$willfollow->unfollow($userId);
I know the controller information is probably not needed but in case it is easy to check the relation within the controller I would prefer to do it that way, because I'm obviously don't have that much knowledge of method use.
I am using Laravel 5.4.
Since Laravel 5.3 you can use syncWithoutDetaching (the most efficient):
$this->followers()->syncWithoutDetaching([$user->id]);
Other ways:
$this->followers()->sync([$user->id], false);
Check for existing before save (efficient only if you have already loaded $this->followers):
function follow(User $user) {
if(!$this->followers->contains($user)) {
$this->followers()->attach($user->id);
}
}
I want to insert a field with value whenever new row is created for a modal.
Ex: Suppose this is my user.php modal
class User extends Authenticatable
{
protected $guarded = ['id'];
}
What i want is in my application anywhere when i insert a row in user table, then i want to insert an extra column code with its value in user table.
Ex: If i do below in my application
User::create(['name'=>'xyz', 'password' => 'kajsndjk']);
then it should insert an extra column code =>'Qwedf' also in my table.
In my application there are many places where i am creating the users, so i don't want to remember every time to insert code column.
Please suggest how can i achieve it.
Overriding the static create function on the User class is the only thing that will work in my opinion.
public static function create(array $attributes = [])
{
$object = parent::create($attributes);
$object->code = 'some text';
$object->save();
return $object;
}
I've tested and like I expected, oseintow's answer will not work, because it would work only if you directly modified code variable, which you obviously are not doing.
Add this mutator to your User model
public function setCodeAttribute($value)
{
$this->attributes['code'] = "Qwedf";
}
Anytime you are saving a record code will be assigned the Qwedf value
Table.linkedIndex is related to LinkedIndex.ID. The value of the field LinkedIndex.TableName is either Linked1 or Linked2 and defines which of these tables is related to a row in Table.
Now i want to make a dynamical link with Yii models so that i can easily get from a Table row to the corresponding Linked1 or Linked2 row:
Table.linkedID = [LinkedIndex.TableName].ID
Example
Table values:
LinkedIndex values:
Now I should get the row from Linked2 where ID=2:
$model = Table::model()->findByPk(0);
$row = $model->linked;
Model
In the model Table, I tried to make the relation to the table with the name of the value of linkedIndex.TableName:
public function relations()
{
return array(
'linkedIndex' => array(self::HAS_ONE, 'LinkedIndex', array('ID' => 'linkedIndex')),
'linked' => array(
self::HAS_ONE,
'linkedIndex.TableName',
array('ID' => 'linkedID'),
)
)
}
But then I get the error:
include(linkedIndex.TableName.php) [function.include]: failed to open stream: No such file or directory
Is there any way to make a dynamic relation Table.linkedID -> [LinkedIndex.TableName].ID with Yii Models?
Per the Yii docs here:
http://www.yiiframework.com/doc/api/1.1/CActiveRecord#relations-detail
I'd suggest using self::HAS_ONE instead (unless there can be multiple rows in LinkedIndex with the same ID - although from the looks of above, I doubt that's the case).
You can link tables together that have different keys by following the schema:
foreign_key => primary_key
In case you need to specify custom PK->FK association you can define it as array('fk'=>'pk'). For composite keys it will be array('fk_c1'=>'pk_с1','fk_c2'=>'pk_c2').
so in your case:
public function relations(){
return array(
'linkedIndex' => array(self::HAS_ONE, 'LinkedIndex', array('ID' => 'linkedIndex')),
);
}
where LinkedIndex is the class name for the LinkedIndex model (relative to your Table model - i.e. same folder. You could change that, of course) and array('ID' => 'linkedIndex') specifies the relationship as LinkedIndex.ID = Table.linkedIndex.
Edit
Looking at your updated example, I think you're misunderstanding how the relations function works. You're getting the error
include(linkedIndex.TableName.php) [function.include]: failed to open stream: No such file or directory
because you're trying to create another relation here:
'linked' => array(
self::BELONGS_TO,
'linkedIndex.TableName',
array('ID' => 'linkedID'),
)
This part: linkedIndex.TableName refers to a new model class linkedIndex.TableName, so Yii attempts to load that class' file linkedIndex.TableName.php and throws an error since it doesn't exist.
I think what you're looking for is to be able to access the value TableName within the table LinkedIndex, correct? If so, that's accessible from within the Table model via:
$this->linkedIndex->TableName
This is made possible by the relation we set up above. $this refers to the Table model, linkedIndex refers to the LinkedIndex relation we made above, and TableName is an attribute of that LinkedIndex model.
Edit 2
Per your comments, it looks like you're trying to make a more complex relationship. I'll be honest that this isn't really the way you should be using linking tables (ideally you should have a linking table between two tables, not a linking table that says which 3rd table to link to) but I'll try and answer your question as best as possible within Yii.
Ideally, this relationship should be made from within the LinkedIndex model, since that's where the relationship lies.
Since you're using the table name as the linking factor, you'll need to create a way to dynamically pass in the table you want to use after the record is found.
You can use the LinkedIndex model's afterFind function to create the secondary link after the model is created within Yii, and instantiate the new linked model there.
Something like this for your LinkedIndex model:
class LinkedIndex extends CActiveRecord{
public $linked;
public static function model($className = __CLASS__){
return parent::model($className);
}
public function tableName(){
return 'LinkedIndex';
}
public function afterFind(){
$this->linked = new Linked($this->TableName);
parent::afterFind();
}
//...etc.
}
The afterFind instantiates a new Linked model, and passes in the table name to use. That allows us to do something like this from within the Linked model:
class Linked extends CActiveRecord{
private $table_name;
public function __construct($table_name){
$this->table_name = $table_name;
}
public static function model($className = __CLASS__){
return parent::model($className);
}
public function tableName(){
return $this->table_name;
}
//...etc.
}
which is how we dynamically create a class with interchangeable table names. Of course, this fails of the classes need to have separate operations done per-method, but you could check what the table_name is and act accordingly (that's pretty janky, but would work).
All of this would result in being to access a property of the linked table via (from within the Table model):
$this->linkedIndex->linked->foo;
Because the value of LinkedIndex.TableName and Table.linkedID is needed to get the values, I moved the afterFind, suggested by M Sost, directly into the Table-Class and changed its content accordingly. No more need for a virtual model.
class Table extends CActiveRecord {
public $linked; // Needs to be public, to be accessible
// ...etc.
public function afterFind() {
$model = new $this->linkedIndex->TableName;
$this->linked = $model::model()->findByPk( $this->linkedID );
parent::afterFind();
}
// ...
}
Now I get the row from Linked2 where ID=2:
$model = Table::model()->findByPk(0);
$row = $model->linked;