Laravel Compact variable blade file display to syntax error

Question

Here is my Controller code:

 public function method(Request $request){
   $typeType = $request->type; //this variable show 'Booking'

   return view('home.profile',compact('type'));
 }

Here is blade file :

 @extends('layouts.dashboard')
 @section('page_heading',{{$taskType}})
 @section('section')
    //here some code
 @stop

if I use this blade code. then I have got this problem:

Parse error: syntax error, unexpected '<' (View:resources/view/home/profile.blade.php)

if I use this code on blade

 @extends('layouts.dashboard')
 @section('page_heading','{{$taskType}}')
 @section('section')
    //here some code
 @stop

then blade file display it:

<?php echo e($taskType); ?>

I want to display it on blade file:

 Booking

How can I solve this problem?

You named the variable $task, not $taskType

Derek Pollard
– Derek Pollard

2018-06-01 05:21:14 +00:00
Commented Jun 1, 2018 at 5:21 — Derek Pollard
– Derek Pollard, Commented Jun 1, 2018 at 5:21
yes, I'm changing but also show error.

Shohidur Rahman
– Shohidur Rahman

2018-06-01 05:43:56 +00:00
Commented Jun 1, 2018 at 5:43 — Shohidur Rahman
– Shohidur Rahman, Commented Jun 1, 2018 at 5:43

Hamed Kamrava · Accepted Answer · 2018-06-01 06:12:10Z

2

Probably you don't need {{ }} inside a Blade Directive.

So change that to:

 @section('page_heading', $taskType)

answered Jun 1, 2018 at 6:12

Hamed Kamrava

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1 Comment

Hamed Kamrava Over a year ago

If it was your answer, please mark it as best answer

Rob · Accepted Answer · 2018-06-01 05:47:32Z

0

public function method(Request $request){
   $typeType = $request->type; //this variable show 'Booking'

   return view('home.profile')->with(['taskType' => $typeType]);
 }

answered Jun 1, 2018 at 5:47

Rob

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1 Comment

Rob Over a year ago

Can you please go ahead and let me know by printing what error are you getting? I mean print_r($typeType) in controller and refresh the page.

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