Laravel Schema Default Value Function - mysql

I want the default value for my Server model's sid to be the uniqid() function to be run every time. Something like this, for example
$table->string('sid')->default(uniqid);
How can I achieve this result?

You can use an event listener to set the sid attribute on your model. You can do this using an event listener, model observer, or just a closure function inside your model's boot function.
// app\Models\YourModel.php
/**
* Define model event callbacks.
*
* #return void
*/
public static function boot()
{
parent::boot();
static::creating(function ($model) {
$model->sid = uniqid();
});
}

Unfortunately no, MySQL requires constants for the default. So, the default value must be a constant, it cannot be a function or an expression.
Only way is to allow the fields nullable and add uniqid() while creating records on the database like this:
$table->string('sid')->nullable();
Now, When you add records set sid value to uinqid() value
If you have model named Table then
$record = new Table();
$record-> -----
--------
--------
$record->sid = uniqid();
$record->save();
This is the way you can achieve.
UPDATE
You can set default value in model as well like this:
protected $attributes = array(
'sid' => uniqid(),
);
Hope you understand.

i think this is the easiest way
$table->unique('sid')->index()

Related

Symfony 4 - update JSON user roles

I have entity User with field roles:
/**
* #ORM\Column(name="roles", type="json")
*/
private $roles = [];
public function getRoles(): array
{
return $roles = $this->roles;
}
public function setRoles($roles): self
{
$this->roles[] = $roles;
return $this;
}
I want to add functionality to update user role from ROLE_ADMIN to ROLE_USER. I tried this in my controller but instead of replacing ROLE_ADMIN with ROLE_USER it inerts this: "ROLE_ADMIN""ROLE_USER". This is my controller:
public function updateuser(Request $request, $id) {
$entityManager = $this->getDoctrine()->getManager();
$usr = $this->getDoctrine()->getRepository(User::class)->find($id);
$usr->setRoles("ROLE_USER");
$entityManager->persist($usr);
$entityManager->flush();
First of all its best practice that every users has at least a default role like ROLE_USER. If you want to give users extra roles then you add them with beside ROLE_USER, like for example ROLE_ADMIN.
Now take a close look how arrays work in PHP. Let's take the code of you setter function setRoles.
When you write the value assignment like this $this->roles[] = $roles, then a value is added to the array . Thats why you in you code you have both roles inside you array now. The already existing ROLE_ADMIN and after the function call you added ROLE_USER.
When you write the value assignment like this $this->roles = $roles, then the whole array is overwritten with the new value.
Conclusion:
Thats how you code should look like if you want a simple solution:
public function setRoles(array $roles): self
{
$this->roles = $roles;
return $this;
}
Then you can call it like this:
$user->setRoles(['ROLE_USER']);
The setRoles function only accepts array.
So your code should change accordingly:
$usr->setRoles(["ROLE_USER"]);
Furthermore, if you want to store it as json, you can use json_encode:
$usr->setRoles(json_encode(["ROLE_USER"]));

Laravel 6 - Updating the model timestamps

After some actions inside of my Laravel 6 app, I'm trying to update my model object $last (class CurrentConditions). As you can see below, I'm trying to update update_time property to current timestamp. Except this property, I have another timestamp property inside the CurrentConditions model: external_update_time and I want this property to stay the same after updating and saving $last object. Problem is that external_update_time property is updated to current time after saving the model.
/** #var CurrentConditions $last */
$last = $this->getLastUpdatedCurrentConditions($cityID);
$last->update_time = Carbon::now("UTC");
$last->save();
Here is my model:
class CurrentConditions extends Model
{
protected $table = 'current_conditions';
...
public $timestamps = false;
protected $dates = ['update_time', 'external_update_time'];
...
}
And here is my migration code:
Schema::create("current_conditions", function (Blueprint $table) {
$table->bigIncrements("current_conditions_id");
...
$table->timestamp("external_update_time");
$table->timestamp("update_time")->useCurrent();
});
So why is external_update_time updating as well? Thanks for any help.
Please check that external_update_time is not set on Update to set CurrentTimeStamp
in the schema table design
This is how MySQL 5.7 works - first timestamp will be updated with current timestamp. You should modify order of columns in your migration (assuming it's not deployed yet to live server) to achieve what you want:
$table->timestamp("update_time")->useCurrent();
$table->timestamp("external_update_time");

Accessing to the `latest` record stored in DB table right after recording it

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.

Yii2 relation based on attribute values instead of keys

I have 2 tables in the db (mysql), and between the 2 there is no classic relationship through keys or ids. The only way I could define relationship would be through attribute values. E.g. table wheel and car and certain wheels would match certain cars because of the size only. Can it be defined on DB level, and/or in yii2, and if yes, how?
In the relations I can add an onCondition(), but you have to define an attribute (???), too:
public function getWheels() {
return $this->hasMany(\app\models\Wheel::className(), ['???' => '???'])->onCondition(['<', 'wheelsize', $this->wheelsize]);
}
I could use a fake attribute and set it in all records like to 1, but it seems a little bit odd for me.
I find nothing on the web regarding this or maybe I'm just searching the wrong way, or maybe I'm trying something that's totally bad practice. Can you please point me to the right direction?
Hypothetically you can set an empty array as a link, but for security reasons (I think) the condition "0 = 1" is automatically added in the select.
I faced your own problem several times and the best solution I could find was to use ActiveQuery explicitly (similar to what happens for hasOne and hasMany):
public function getWheels() {
return new ActiveQuery(\app\models\Wheel::className(), [
'where' => 'my condition' // <--- inserte here your condition as string or array
'multiple' => true // true=hasMany, false=hasOne
// you can also add other configuration params (select, on condition, order by, ...
]);
}
This way you can get both the array and the ActiveQuery to add other conditions:
var_dump($model->wheels); // array of wheels objects
var_dump($model->getWheels()); // yii\db\ActiveQuery object
$model->getWheels()->andWhere(...); // customize active query
I don't think that you could achieve this through relation.
But there is a way to work around the limitation.
<?php
namespace app\models;
class Car extend \yii\db\ActiveRecord
{
/**
* #var \app\models\Wheel
*/
private $_wheels;
/**
* #return \app\models\Wheel[]
*/
public function getWheels()
{
if (!$this->_wheels) {
$this->_wheels = Wheel::find()
->where(['<', 'wheelsize', $this->wheelsize])
//->andWhere() customize your where here
->all();
}
return $this->_wheels;
}
}
Then you could access the wheels attribute just as relation does.
<?php
$car = Car::find(1);
$car->wheels;
Beware that this way does not support Eager Loading

Yii model: Dynamic table relations

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;