Laravel: Undefined variable in blade view

Question

This is AdminController.php:

<?php

namespace App\Http\Controllers;

use Response;

use Illuminate\Support\Facades\DB;
use App\Caption;
use App\Image;
use App\Http\Controllers\Controller;

class AdminController extends Controller
{
    public function admin() {
        $images = Image::paginate();

        return view('admin',[ '$images' => $images]);
    }
}

And this is admin.blade.php:

@extends('template')

@section('title')
    Admin Page
@endsection

@section('header')
@endsection

@section('main')
    @if (Auth::check())
        @foreach ($images as $image)
            <p value='{{$image->id}}'>{{$image->content}}</p>
            <form action="image/{{$image->id}}/delete" method="post">
                <button type="submit">Delete caption</button>
            </form> 
            <form action="image/{{$image->id}}/approve" method="post">
                <button type="submit">Accept image</button>
            </form>
        @endforeach
    @else
        <p>Login first</p>
    @endif
@endsection

@section('footer')
@endsection

Why do I get the following error?

ErrorException in 408148d5409ae6eebf768dc5721bd0d1d9af48af.php line 9: Undefined variable: images (View: /Users/sahandz/Documents/School/Singapore/CS3226/backend/resources/views/admin.blade.php)

$imagesis clearly defined in my controller.

is Image::paginate() a scope you wrote? because normally you would do something like this: $images = Image::all()->paginate(15); — Christophvh
– Christophvh, Commented Mar 30, 2017 at 7:49
the correction code should be: return view('admin',[ 'images' => $images]); — Thanh Nguyen
– Thanh Nguyen, Commented Mar 30, 2017 at 9:06

Joe · Accepted Answer · 2017-03-30 07:47:15Z

9

You have used $images as your variable name when passing the data to the view. This results in blade creating a variable called $$images.

return view('admin',[ 'images' => $images]);

will result in the view creating a variable called $images

answered Mar 30, 2017 at 7:47

Joe

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1 Comment

user10678889 Over a year ago

great, I wasted my 4 hours and finally reached you

Rehan · Accepted Answer · 2017-03-30 07:46:30Z

2

Try passing data like this.

$images = Image::paginate();
return view("admin")->with('images',$images);

Basically,you need not use $ in the variable name.

answered Mar 30, 2017 at 7:46

Rehan

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Comments

Abdullah Al Shakib · Accepted Answer · 2017-03-30 07:49:35Z

1

Your code should be like this

public function admin() {
    $images = Image::paginate();

    return view('admin',[ 'images' => $images]);

}

Why you use [ '$images' => $images]. that is the problem.

answered Mar 30, 2017 at 7:49

Abdullah Al Shakib

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Comments

ram pratap singh · Accepted Answer · 2017-03-30 09:51:20Z

1

You may use this method also to pass your data to view

 return view('admin',compact('images'));

answered Mar 30, 2017 at 9:51

ram pratap singh

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