Laravel 4 Eloquent Column Alias - php

What i am trying to achieve is in my database i have a table named channel
i am using laravel's eloquent class to access these the properties from the table
The problem that i am facing is that
the table name is column and the column name is channel
so when accessing that property looks like this.
User::find(1)->channel->channel
How can i modify this to say
User::find(1)->channel->name
We cannot change the table name in the database.
Options i have thought of:
1)Create views for tables that need columns changed. Too messy...
2)Use column alias.... laravel documentation...sigh.. no clue how?
3)Use a property set with the create_function that would call this->channel
but i am pretty sure it won't work because laravel is using dynamic properties. and when it's fill out in the array im pretty sure it changes it to the name of the column.
I could in my belongs_to/hasOne/hasMany function change the property to the alias of the name i want to use so that later on i can change it. i dunno how well that would work..
any thoughts?
much appreciated

You could probably do it easily with Accessors / Mutators.
class Channel extends Eloquent {
public function getNameAttribute()
{
return $this->attributes['channel'];
}
public function setNameAttribute($value)
{
$this->attributes['channel'] = $value;
}
}
Reference
Laravel Accessors & Mutators

Related

Does a one-to-one relationship in Laravel always need first()?

I have a one-to-one relationship between User and UserSettings models,
But (after $user = auth()->user()) when I try $user->settings()->something it throws an Undefined property error.
It's gone when I use $user->settings()->first()->something...
My question is, is this how it's supposed to work? or am I doing something wrong?
You cannot directly run $user->settings()->something.
Because when you call $user->settings(), it just return Illuminate\Database\Eloquent\Relations\HasOne object.
So it is not the model's object, you need to take the model's object and call its attribute like this.
$user->settings()->first()->something;
Dynamic Properties
Since you have one-to-one relationship between User and UserSettings.
If you have a one-to-one relationship in your User model:
public function settings()
{
return $this->hasOne('App\Models\UserSettings', 'user_id', 'id');
}
According to Laravel doc
Once the relationship is defined, we may retrieve the related record using Eloquent's dynamic properties. Dynamic properties allow you to access relationship methods as if they were properties defined on the model:
Eloquent will automatically load the relationship for you, and is even smart enough to know whether to call the get (for one-to-many relationships) or first (for one-to-one relationships) method. It will then be accessible via a dynamic property by the same name as the relation.
So you can use eloquent's dynamic properties like this:
$user->settings->something; // settings is the dynamic property of $user.
This code will give you a result of collection.
$user->settings;
So calling 'something' is not available or it will return you of null, unless you get the specific index of it.
$user->settings()->something
while this one works because you used first() to get the first data of collection and accessed the properties of it .
$user->settings()->first()->something
The first method returns the first element in the collection that passes a given truth test
see docs here laravel docs
If you want to get the user settings itself simply do this:
$user->settings
Then you can get the fields of the settings doing this:
$user->settings->something
When you do this $user->settings() you can chain query after that. E.g.
$user->settings()->where('something', 'hello')->first()
That's why the output of $user->settings and $user->settings()->first() are the same.
Auth only gives you user info;
Try the following code:
$user = User::find(auth()->user()->id);//and then
$user->settings->something;

Laravel Virtual Columns fail to save

I added a couple of virtual columns to my database tables using Laravels virtualAs column modifier:
$table->decimal('grand_total')->virtualAs( '(total_value + (total_value*tax_rate))');
Basically it keeps a mysql virtual column that automatically calculates the grand total based on the total and tax rate stored in another column.
However, Laravel does not seem to play nice with virtual columns at all. When saving a record, it attempts to INSERT or UPDATE the virtual column, which is obviously not allowed in mySQL. I could not find a way to configure in the Eloquent model which fields are actually written to the database on an update or insert.
I've tried adding the field to the models $hidden, and $appends but nothing seems to work.
Looking at the Laravel Source code for an insert (https://github.com/laravel/framework/blob/5.6/src/Illuminate/Database/Eloquent/Model.php#L733), it seems to just insert whatever attributes are in $this->attributes. When the record is read from the database the grand_total field is read from the table and set as an attribute and then it is tried to be written again once the record is saved.
Is there any way to get this Laravel to stop trying to save columns that are virtual?
Here's a quick trait I wrote to solve your problem that will filter fields residing in the $virtualFields property before saving. It requires a select (refresh) after the save to get the new value for the virtual field. If you don't need to query this virtual field, I'd highly recommend you look into a mutator instead.
trait HasVirtualFields
{
public function save(array $options = [])
{
if (isset($this->virtualFields)) {
$this->attributes = array_diff_key($this->attributes, array_flip($this->virtualFields));
}
$return = parent::save($options);
$this->refresh(); // Refresh the model for the new virtual column values
return $return;
}
}
class YourModel
{
use HasVirtualFields;
protected $virtualFields = ['grand_total'];
}

SQLSTATE[42S02]: Base table or view not found laravel

So, going into the problem straight away. someone told me that we dont need to make a pivot table if we only want to have ids of the table. laravel can itself handle this situation. I dont know how this works. I have a table community and another table idea. relation is like this;
One community can contain many ideas and an idea can be found in many
communities.
Relation in idea Model:
public function community() {
return $this->belongsToMany('App\Community')->withTimestamps();
}
Relation in community Model:
public function idea() {
return $this->belongsToMany('App\idea');
}
Now i want to fetch all the records related to a single community to show on its page Let's say the community is Arts.
Here is Controller function:
public function showCommunities($id) {
$community = Community::findOrFail($id)->community()->get();
return view('publicPages.ideas_in_community', compact('community'));
}
When i attach ->community()->get() to the Community::findOrFail($id) Then it throws the error
SQLSTATE[42S02]: Base table or view not found laravel
Any help would be appreciated.
Edit:
Logically, this piece of code Community::findOrFail($id)->community()->get() should be like this Community::findOrFail($id)->idea()->get(). Now it is true but it has little issue. it throws an error
Fatal error: Class 'App\idea' not found
The way you define the many-to-many relation looks ok - I'd just call them communities() and ideas(), as they'll return a collection of objects, not a single object.
Make sure you use correct class names - I can see you refering to your model classes using different case - see App\Community and App\idea.
In order to find related models, Eloquent will look for matching rows in the pivot table - in your case it should be named community_idea and have 3 fields: community_id, idea_id and autoincrement primary key id.
With that in place, you should be able to get all ideas linked to given community with:
$ideas = Community::findOrFail($communityId)->ideas;
If you need communities linked to given idea, just do:
$communities = Idea::findOrFail($ideaId)->communities;
You can read more about how to use many-to-many relationships here: https://laravel.com/docs/5.1/eloquent-relationships#many-to-many
someone told me that we dont need to make a pivot table if we only want to have ids of the table
The above is not true (unless I've just misunderstood).
For a many-to-many (belongsToMany) their must be the two related table and then an intermediate (pivot) table. The intermediate table will contain the primary key for table 1 and the primary key for table 2.
In laravel, the convention for naming tables is plural for your main tables i.e. Community = 'communities' and Idea = 'ideas'. The pivot table name will be derived from the alphabetical order of the related model names i.e.
community_idea.
Now, if you don't want/can't to follow these conventions that's absolutely fine. For more information you can refer to the documentation: https://laravel.com/docs/5.2/eloquent-relationships#many-to-many
Once you're happy that you have the necessary tables with the necessary fields you can access the relationship by:
$ideas = $community->ideas()->get();
//or
$ideas = $community->ideas;
So you controller would look something like:
public function showCommunities($id)
{
$community = Community::findOrFail($id);
//The below isn't necessary as you're passing the Model to a view
// but it's good for self documentation
$community->load('ideas');
return view('publicPages.ideas_in_community', compact('community'));
}
Alternatively, you could add the ideas to the array of data passed to the view to be a bit more verbose:
public function showCommunities($id)
{
$community = Community::findOrFail($id);
$ideas = $community->ideas
return view('publicPages.ideas_in_community', compact('community', 'ideas));
}
Hope this helps!
UPDATE
I would imagine the reason that you're receiving the App\idea not found is because the model names don't match. It's good practice (and in certain environments essential) to Capitalise you class names so make sure of the following:
Your class name is Idea and it's file is called Idea.php
The class has it's namespace declared i.e. namespace App;
If you've added a new class and it's not being found you might need to run composer dump-autoload from the command line to update the autoloader.

Upgrading to Yii2, ORM not functional

I'm upgrading a big project form Yii1 to Yii2. I'm having some problems regarding to ORM.
I have several relation declared in the following fashion(basically a copy-paste from the guidebook):
class Order extends \yii\db\ActiveRecord {
/* other code */
public function getAffiliate()
{
return $this->hasOne(Affiliate::className(), ['id_affiliate' => 'affiliate_id']);
}
Whenever I try to echo or w/e $order->affiliate->name; I get the following error:
yii\base\ErrorException: Trying to get property of non-object
I've got no experience with Yii1 what so ever. Something weird about this project is the database. All tables start with yii_tablename and id's are: id_tablename. Was that normal for Yii1 and could this be causing the issue above?
Edit: When I execute the function like so: $order->getAffilate() it returns an ActiveQuery WITHOUT the data from the affiliate.
When I execute the following:
$order->getBillingAddress()->one();
I get a weird error:
Getting unknown property: app\models\Order::billing
return $this->hasOne(Affiliate::className(), ['id_affiliate' => 'affiliate_id']);
It's mean that when you call $order->affiliate yii2 will find in Affiliate table on id_affiliate field current Order affiliate_id value and selected one value.
Check that you have right field names and database have right data.
When you call $order->affiliate you will get Affiliate object. But if you call $order->getAffiliate() you will get ActiveQuery object.
I found a solution. One which I don't really like though, but it does the job. Was reading this thread: link.
Kartik V
The problem is clearly in uniqueness in naming your relation and your model attribute. In your User model, you have an attribute named role and you also have a relation getter named getRole.
So I changed the name of the getter like so:
public function getOrderAffiliate()
{
return $this->hasOne(Affiliate::className(), ['id_affiliate' => 'affiliate_id']);
}
And that fixed the issue. Never had this issue before and wonder why this happened though.

Cakephp and database with uncomon structure

how can I access any table from database in my model?
For example, I have Indexcontroller and code inside it:
$results = $this->Index->query("SELECT COUNT(*) FROM my_own_table");
Error: Database table indices for model Index was not found.
So, as I understand, I can access only table with naming related to model/controller name. But what to do if I can't modify the table naming and I want to access it's data?
You're not limited to using a model that's directly associated with your controller (this is just default behaviour); you can use any model.
To achieve what you want, create a new model for this table, eg. MyOwnTable, and in your controller, you can add this property to the class:
public $uses = array('Index', 'MyOwnTable');
Now you can access MyOwnTable using CakePHP's built in ActiveRecord functionality:
$results = $this->MyOwnTable->find('count');
If you have other tables you want to access, simply create models for those and add them to the $uses property. (You can also use $this->loadModel('Model') inside the action if you prefer).
If you have a table name that isn't very readable (eg. my_tb_own_1_x or some such), you can call the model class something human readable (eg. MyTable), and add the $useTable property to the model:
public $useTable = 'my_tb_own_1_x';
/* and change the default primary key if you have an unusual one */
public $primaryKey = 'my_tb_own_1_x_idx_pk';
See the CakePHP manual for more info on how to change default model and controller behaviour:
1.3 - Model Attributes
2.0 - Model Attributes
1.3 - Controller Attributes
2.0 - Controller Attributes
Nope. You can access different tables. However, CakePHP stumbles over the fact that the table that is associated by default to the Index model doesn't exist.
In other words, the model Index expects a table 'indices' to exist (and an error is thrown when it doesn't). You can do one of two things:
Create a table indices
Add the following to your Index model: var $useTable = false;
If you have any use for an indices table I'd go with option 1. If you're not going to use the indices table, go with option 2.
If you go with either step 1 or 2, your example should start working.

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