Laravel ajax模特展示帖子

Question

I have a foreach() from database table, I want to show a popup/model to show its extra information.

I am showing just title and description and on click i want to open up a popup and show its extra information.

@foreach($myProjects as $project)
    <div class="col-sm-4 col-md-4 notes notes--blue">
        <a href="#edit-note" data-toggle="modal" style="background-color: #f9f9f9;border-bottom: 5px solid #42A5F5">
            <div class="notes__title">{{$project->title}}</div>
            <div class="notes__body">{{$project->description}}</div>
        </a>
        <div class="notes__actions" data-demo-action="delete-listing">
             <i class="zmdi zmdi-delete"></i>
        </div>
        <div class="notes__actions1" data-demo-action="delete-listing">
              <i class="zmdi zmdi-edit"></i>
        </div>
        <div class="notes__actions2" data-demo-action="delete-listing">
              <i class="zmdi zmdi-eye"></i>
        </div>
    </div>
@endforeach

I am completely blank, Should i fetch post id to a hidden html tag and on model button click an ajax call will fetch the record info based on the id ?

Answer 1

I would add a data-id attribute to one of the elements, possibly the wrapper, then add something like

$(document.body).on('click', '.clickable_element', function(e){
  if ($(this).data('id')) {
        $.ajax({
           url : 'your detail url', 
           data: { id: parseInt( $(this).data('id'), 10 ), 
           success : function(response){
               // open popup and add response into it.
           }
        })
  }

});

Update

I just noticed you already have bootstrap modal there.

you can add your data-id to data-toggle element then in javascript

$('[data-toggle=modal]').on('shown.bs.modal' , function(){
  // do your ajax stuff
  // add response in `.modal-body`
})