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)
{
$this->attributes['code'] = "Qwedf";
}
Anytime you are saving a record code will be assigned the Qwedf value
Related
In Laravel After recording last row to a DB table, can I safely access same recorded data right after recording it by calling latest() queries? Because transactions by other users may occur at the same time, and it may not really be the last record anymore?
Edit:
For example:
Public function StoreNotif($data){
auth()->user()->Notif()->create(store $data here..)
}
Public function SendNotif(){
$data="123";
$this->StoreNotif($data)
event(new Notification(stored Notif instance?));
}
No, you cannot rely on the database to return the record from your current script.
The ->latest() method will always sort the records with the most recent created_at date first.
https://laravel.com/docs/6.x/queries#ordering-grouping-limit-and-offset
But you haven't provided any code or explanation as to why this is a concern. If you just created a new record, why do you need to query it again? You should already have access to an instance of the model.
EDIT: I've made a few edits to demonstrate how you would pass the model from a controller to an event as referenced in the comments. Please post your code if you want more specific help.
SomeController.php
function store()
{
$model = Model::create([
'some_data' => 1
]);
// fire an event with the newly created model
event(new SomeEvent($model));
dd($model);
}
------------------------
Model {
// ...
attributes: [
'id' => 101,
'some_data' => 1
'created_at' => '2019-10-06 12:48:01',
'updated_at' => '2019-10-06 12:48:01',
]
// ...
}
SomeEvent.php
<?php
namespace App\Events;
use App\Model;
use Illuminate\Queue\SerializesModels;
class SomeEvent
{
use SerializesModels;
public $model;
public function __construct(Model $model)
{
$this->model = $model;
// ...
}
}
EDIT: Per your newly added code, you just need to pass the new model back to the original method. You could do something like this.
Public function StoreNotif($data)
{
// add a return statement
return auth()->user()->Notif()->create(store $data here..);
}
Public function SendNotif()
{
$data="123";
// store the returned data to a variable
$model = $this->StoreNotif($data);
// call the event with the model instance
event(new Notification(model));
}
I'm not sure what 'latest' is but I do know that MySQL uses SELECT LAST_INSERT_ID as the query to get the 'per-connection' id of the last inserted item. Under the covers it's using mysql_insert_id so if you are in a language that supports it, you could use that too.
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.
I am trying to update my rows. The figure provided is the Users table structure. When I register using Laravel Authentication. Only the yellow marked rows are used and others remain empty. I have a created another page called profile and want to add data to the remaining empty rows using a profile page. But I am stuck and can't figure out how to do it. Any suggestions?
You can create Profile controller for interaction with User table from profile page, like this:
use App\User;
class ProfileController extends Controller {
public function function_name(Request $request)
{
...
$user = Auth::User(); //get authorized user, or you can get user by ID
$user->city = 'City';
$user->country = 'Country ';
$user->save();
...
}
In your user model, dont forget to add thoes fields to fillable variable;
for exemple
protected $fillable = [
'name', 'email', 'password', 'company_name','street_name','whatever'
];
this way the ->create($data); method will work just fine
or you can just add those fields to registration form and update your registration method
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 = $model->linked;
I'm writing a simple app. I need to block user from a page if their credit is < 0.
I have a table "User_profiles" with a "credit" row.
How can I set up a model in conjunction with the controller to send the user to another page if the value of "credit" is 0?
This should be straightforward, but I'm new at the select->where stuff...
It has to be the row of the current user too--I don't know how to traverse arrays very well yet.
Thanks!
Well, the easiest solution would be to just load a different view...
As for the model, it would look like this:
class UserModel extends Model {
public function getUserCredit($id) {
$this->load->database();
//effectively generates: SELECT `credit` FROM `User_profiles` WHERE `id`=$id
$query = this->db->select('credit')->where('id',$id)->get('User_profiles');
//row() executes the query for a single result, returns the credit property
return $query->row()->credit;
}
}
Then in the controller:
class Users extends Controller {
//....
public function credit() {
$this->load->model('userModel','users');
// assuming the session library has been loaded
$user_id = $this->session->userdata('id');
$credit = $this->users->getUserCredit($user_id);
if ($credit == '0') $this->load->view('users/no_credit');
else $this->load->view('users/credit');
}
}
That's untested, but it should at least help you get the idea.
When you request the page /users/credit/1, CI will call the Users::credit(1) action.
It then loads UserModel as $this->users
You call $this->users->getUserCredit(1), which translates to UserModel::getUserCredit(1), to store as $credit
The model loads the database.
You tell the db to select('credit') (select the credit column), where('id',1) (where the id = 1), then get('User_profiles') (get matching rows from the User_profiles table). That returns a query, which you store as $query for readability.
getUserCredit returns the credit property of the single-row result of the query
If $credit == 0, you load the view views/users/no_credit.php
Otherwise, you load the view views/users/credit.php (it's conventional to name the views after the actions they represent and put them in a folder corresponding to the controller)