How can I return a view from an AJAX call in Laravel 5?

Question

I'm trying to get an html table to return on an ajax call.

route:

Route::post('job/userjobs', 'JobController@userjobs');  

ajax on calling page:

function getUserJobs(userid) {
    $_token = "{{ csrf_token() }}";
    var userid = userid;
    $.ajax({
        headers: { 'X-CSRF-Token' : $('meta[name=_token]').attr('content') },
        url: "{{ url('/job/userjobs') }}",
        type: 'POST',
        cache: false,
        data: { 'userid': userid, '_token': $_token }, //see the $_token
        datatype: 'html',
        beforeSend: function() {
            //something before send
        },
        success: function(data) {
            console.log('success');
            console.log(data);
            //success
            //var data = $.parseJSON(data);
            if(data.success == true) {
              //user_jobs div defined on page
              $('#user_jobs').html(data.html);
            } else {
              $('#user_jobs').html(data.html + '{{ $user->username }}');
            }
        },
        error: function(xhr,textStatus,thrownError) {
            alert(xhr + "\n" + textStatus + "\n" + thrownError);
        }
    });
}



//on page load
getUserJobs("{{ $user->id }}");

controller:

public function userjobs() {
    $input = Request::all();
    if(Request::isMethod('post') && Request::ajax()) {
        if($input['userid']) {
            $userjobs = Userjob::select('select * from user_jobs where user_id = ?', array($input['userid']));
            if(! $userjobs) {
                return response()->json( array('success' => false, 'html'=>'No Jobs assigned to ') );
            }
            $returnHTML = view('job.userjobs')->with('userjobs', $userjobs);
            return response()->json( array('success' => true, 'html'=>$returnHTML) );

        }
    }   
}

view:

@section('content')
<table class="table table-striped">
    <tbody>
@foreach ($userjobs as $userjob)
        <tr>
            <td><strong>{{ $userjob->title }}</strong><br />
            {{ $userjob->description }}
            </td>
        </tr>
@endforeach
</table>
@stop

Im not getting anything in the json.html data. nothing. If in the controller I say:

return response()->json( array('success' => true, 'html'=>'<span>html here</html>') );

This works just fine.

How can I return a view from an ajax call in Laravel 5.

Answer 1

The view() function just creates an instance of the View class. Not just an HTML string. For that you should call render() :

$returnHTML = view('job.userjobs')->with('userjobs', $userjobs)->render();
return response()->json(array('success' => true, 'html'=>$returnHTML));

Answer 2

if your ajax is correct and you are getting results from your DB

 $returnHTML = view('job.userjobs',[' userjobs'=> $userjobs])->render();// or method that you prefere to return data + RENDER is the key here
            return response()->json( array('success' => true, 'html'=>$returnHTML) );

Answer 3

在查看文件名之前使用字符串函数，如

return (String) view('Company.allUserAjax');

Answer 4

$returnHTML = view('job.userjobs')->with('userjobs', $userjobs)->renderSections()['content'];

So this question is old and the most upvoted answer did not solve the problem for the asker and for me neither. I ran into the same problem and it took me two days to find the solution. Everywhere I came looking for answers the said use ->render() . But nothing returned. I have a partial view which I wanted to include. I had not extended my partial view or given it a section name. Since this is not necessary when including it inside a blade file with the include directive.

So the solution is, enclose your html inside a section and instead of render() , use renderSections()['sectionname'] . Just using render() will not work.

I hope this will safe somebody some time and frustration!

Answer 5

Don't return your view as JSON, just return the view from your controller For example:

$view = view("<your url>",compact('data'))->render();
return $view;

This will work for sure.

Answer 6

If you use AJAX in Laravel then when you want to display view with AJAX response don't use return command instead use echo command.

For example, don't use:

return view('ajax', ['ajax_response'=> $response]);

use:

echo view('ajax', ['ajax_response'=> $response]);

Other example, don't use:

$view = view("job.userjobs",compact($userjobs))->render();
return response()->json(['html'=>$view]);

use:

$view = view("job.userjobs",compact($userjobs))->render();
echo response()->json(['html'=>$view]);

Answer 7

Ok - I found by trial and error that you don't include the blade template commands in the view file for this to work.

I had:

@section('content')
<table class="table table-striped">
    <tbody>
@foreach ($userjobs as $userjob)
        <tr>
            <td><strong>{{ $userjob->title }}</strong><br />
            {{ $userjob->description }}
            </td>
        </tr>
@endforeach
</table>
@stop

and now its working with:

<table class="table table-striped">
    <tbody>
    <?php foreach ($userjobs as $userjob){?>
        <tr>
            <td><strong><?php echo $userjob->title; ?></strong><br />
            <?php echo $userjob->description; ?>
            </td>
        </tr>
    <?php } ?>
</table>

Altho I searched - I never found documentation stating this.

Answer 8

删除这个 if 条件，看看它是否有效

if(Request::isMethod('post') && Request::ajax()) {

Answer 9

idk if you are still looking for answer but i ran into same issue as you, it kept returning blank. the key to success was to remove @section @endsection part in your partial view

Answer 10

This helped me:

echo view('ajaxView')->with(['value' => 'some value']);

I have used echo view('ajaxView') instead return view('ajaxView') .

Answer 11

I got it working without these : @section & @stop

<table class="table table-striped">
    <tbody>
@foreach ($userjobs as $userjob)
        <tr>
            <td><strong>{{ $userjob->title }}</strong><br />
            {{ $userjob->description }}
            </td>
        </tr>
@endforeach
</table>

either

$returnHTML = view('job.userjobs',[' userjobs'=> $userjobs])->render();// or method that you prefere to return data + RENDER is the key here

or this

$returnHTML = view('job.userjobs',compact($userjobs))->render();// or method that you prefere to return data + RENDER is the key here`

both worked

Answer 12

$view = view("job.userjobs",compact($userjobs))->render();
return response()->json(['html'=>$view]);