How to return an array of object properties from a returned object of Laravel Eloquent

Question

I have say simple example of getting an object using this query:

return App\models\SuccessIndicatorYearWeight::find(1);

This returns:

{
    success_indicator_year_weight_id: "108",
    success_indicator_id_fk: "174",
    year_id_fk: "1",
    criteria_excel: "05-05-2015",
    criteria_very_good: "08-05-2015",
    criteria_good: "11-05-2015",
    criteria_fair: "15-05-2015",
    criteria_poor: "20-05-2015",
    created_by: "4",
    create_date: "2015-05-01 21:40:35",
    updated_by: "4",
    update_date: "2015-05-02 06:49:30",
}

Now I want to return only these properties from the object, criteria_excel, criteria_very_good, criteria_fair, criteria_poor.

I have tried with these:

return App\models\SuccessIndicatorYearWeight::find(1)->pluck('criteria_excel', 'criteria_very_good', 'criteria_fair', 'criteria_poor');

And

return App\models\SuccessIndicatorYearWeight::find(1)->select('criteria_excel', 'criteria_very_good', 'criteria_fair', 'criteria_poor')->get();

And was unsuccessful.

A simple workaround is this:

return App\models\SuccessIndicatorYearWeight::whereId(1)->select('criteria_excel', 'criteria_very_good', 'criteria_good', 'criteria_fair', 'criteria_poor')->get();

But, I was wondering if there are any other ways to return only selected properties of an object.

Akarun · Accepted Answer · 2015-05-02 08:35:02Z

1

If you use select(...)->get(), it's always return Object.

But you can add toArray to extract your criteria. Like this :

return App\models\SuccessIndicatorYearWeight::find(1)
    ->select('criteria_excel', 'criteria_very_good', 'criteria_fair', 'criteria_poor')
    ->get()
    ->toArray();

It will return an Array of Associative Array (with your keys).

answered May 2, 2015 at 8:35

Akarun

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panos · Accepted Answer · 2015-05-02 08:39:23Z

0

If you will always return only those fields when you return JSON (or array) you can use protected $visible = ['fields', 'you', 'need', '...']; in your model.

http://laravel.com/docs/5.0/eloquent#converting-to-arrays-or-json

answered May 2, 2015 at 8:39

panos

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vanadium23 · Accepted Answer · 2015-05-02 07:48:14Z

0

No, the latest method is aprecciated. Under Eloquent there is an SQL Builder to database. If you not specific select, that would produce following query:

SELECT * FROM indicator_year_weight;

That means to DB to return all fields from table and Eloquent will put them all to object. After adding select:

SELECT criteria_excel, criteria_very_good, criteria_good, criteria_fair, criteria_poor FROM indicator_year_weight;

That means to return only a specified fields, so Eloquent will put only it,

answered May 2, 2015 at 7:48

vanadium23

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borracciaBlu · Accepted Answer · 2015-05-02 13:27:12Z

0

Probably what you need is list as arguments it takes the column you want to extract.

In your case it will be :

$criterias = App\models\SuccessIndicatorYearWeight::whereId(1)->list(
    'criteria_excel', 
    'criteria_very_good', 
    'criteria_fair', 
    'criteria_poor'
);

Here you can find the documentation: http://laravel.com/docs/5.0/queries#selects

answered May 2, 2015 at 13:27

borracciaBlu

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