Laravel 5.6 inject blade into view via ajax call

Question

I need to inject some blade logic into my view via an ajax call. I am not sure if this is even possible or if there is an other way?

What I am trying to do:

axios.post('/buildingimage', formData, { headers: {'Content-Type': 'multipart/form-data' }})
        .then(response => {
            e.preventDefault();

            $('.errorMessages').hide();
            let building = response.data.building;

            let resultInfo = "";

            $('.successMessages').show();

            let newBuilding = "<tr>" +
                "                  <td>" + building.location + " </td>" +
                "                  <td>" + building.source + "</td>" +
                "                  <td>" + building.disc + "</td>" +
                "                  <td>" + building.result + "</td>" +
                "                  <td>" + resultInfo + "</td>" +
                "                  <td> Edit form </td>" +
                "                  @can('Delete buildings')" +
                "                  <td>" +
                "                      {!! Form::open(['class' => 'deleteBuildingForm', 'method' => 'DELETE', 'route' => ['buildingimage.destroy', " + building.id + "] ]) !!}" +
                "                              <button type='submit' class='btn btn-danger'>" +
                "                              <i class='far fa-trash-alt'></i>" +
                "                          </button>" +
                "                      {!! Form::close() !!}" +
                "                  </td>" +
                "                  @endcan" +
                "              </tr>";

            $(".buildingsTable > tbody:last-child").append(newBuilding);
        })
        .catch(error => {
            if (error.response) {
                $('.errorMessages').show();
                $('.successMessages').hide();

                $.each(error.response.data.errors, function(key, value) {
                    $('.errorMessagesList').append('<li>' + value + '</li>');
                });
            }
    });

As you can see the form that I try to render isn't actually a form in my view but results in just a string. I am not sure how I can fix this.

The tricky part is that is should be in td of a table.

you can't do it like this. handle blade logic in your backend then fetch result from ajax. — Amir
– Amir, Commented Nov 14, 2018 at 9:52

Yoram de Langen · Accepted Answer · 2018-11-14 09:56:41Z

1

I assume you're having a JSON response? In that case, you can achieve it by rendering the view directly with the render() method on the View Factory class.

return response()->json([
    'data' => [
        'building' => $buildingCollection,
        'view' => view('path.to.view')->render(),
    ],
]);

Your javascript tr and td should be unnecessary at this point.

Your table rendering logic can be on the server-side instead of the frontend. My personal preference would be rendering it on the frontend.

edited Nov 14, 2018 at 9:56

answered Nov 14, 2018 at 9:53

Yoram de Langen

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6 Comments

Michael Over a year ago

What do I put in my view then?

Yoram de Langen Over a year ago

Everything that you do within the let newBuilding javascript part. This is pretty basic blade stuff. You can check it here: laravel.com/docs/5.7/blade

Michael Over a year ago

I am a bit confused, so I make a new view with the <tr>...</tr> that I create via javascript and I render this?

Michael Over a year ago

What do I do in my ajax call then?

Yoram de Langen Over a year ago

Within your javascript you only have to put the response.data.data.view content within your HTML. Something like; $(".buildingsTable > tbody:last-child").html(response.data.data.view)

|

PHP Geek · Accepted Answer · 2018-11-14 10:25:29Z

0

Laravel provides render() method to render any of the view from view directory. 
if you want to render any of the view then you can write it as 
view('auth.viewname',$dataarray)->render()

where $dataarray contains the data to be carried to the that view. Here is the example - 
in the controller (on ajax method)
write it like 
return response()->json(view('viewname)->render()));
and then use the response in the ajax to append any html block

answered Nov 14, 2018 at 10:25

PHP Geek

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