Laravel + CodeSleeve stapler bundle - php

Question

I am using the Codesleeve Stapler and i have a small problem.
I did exactly what the last example described on this page:
https://github.com/CodeSleeve/stapler
The difference is that my new model is called Pictures instead of ProfilePictures
and my model is not User but Trip.
The <img src="<?= asset($picture->photo->url('thumbnail')) ?>">
on the view shows the last picture that has been uploaded.
I want to show the Picture that belongs to each Trip. How could i perform this?
Thanks.

Answer 1

So, you have two models: 'Trip', and 'Picture'. In your Trip model, you need to define a 'hasMany' relationship to the Picture model:
public function pictures(){
return $this->hasMany('Picture');
}
Then, in your Picture model, you define the Stapler attachment:
// Be sure and use the stapler trait, this will not work if you don't:
use Codesleeve\Stapler\Stapler;
// In your model's constructor function, define your attachment:
public function __construct(array $attributes = array()) {
// Pictures have an attached file (we'll call it image, but you can name it whatever you like).
$this->hasAttachedFile('image', [
'styles' => [
'thumbnail' => '100x100#',
'foo' => '75x75',
'bar' => '50x50'
]
]);
parent::__construct($attributes);
}
Now that you've defined the attachment on the Picture model, every time you access a Picture object, you'll also have access to the file attachment. Assuming you had a Trip record, you could do this:
<?php foreach ($trip->pictures as $picture): ?>
<img src="<?= asset($picture->image->url('thumbnail')) ?>">
<?php endforeach ?>
You could access the original image like this:
<img src="<?= asset($picture->image->url()) ?>">
// or
<img src="<?= asset($picture->image->url('original')) ?>">
In fact, you can access any of the style you defined:
<img src="<?= asset($picture->image->url('foo')) ?>">
<img src="<?= asset($picture->image->url('bar')) ?>">
Hope this helps.