CODEWITHSUNDEEP
phphtmllaravelviewlaravel-blade
Sahand
Sahand

Reputation: 8370

Laravel: Undefined variable in blade view

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.

Upvotes: 5

Views: 27215

Answers (4)

ram pratap singh
ram pratap singh

Reputation: 53

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

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

Upvotes: 1

Abdullah Al Shakib
Abdullah Al Shakib

Reputation: 2044

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.

Upvotes: 1

Joe
Joe

Reputation: 4738

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

Upvotes: 9

Rehan
Rehan

Reputation: 4013

Try passing data like this.

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

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

Upvotes: 2

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