Question

我在Laravel 5.3项目中使用数据表插件我可以显示我的所有数据，但无法实现编辑和删除如何实现。

<script>
$.noConflict();
jQuery(document).ready(function() {
    $('#myTable').DataTable({
        processing : true,
        serverSide : true,
        ajax       : '{{URL::asset('data')}}',
        columns    :[
            {data:'id',name:'id'},
            {data:'prizebond_number',name:'prizebond_number'},
            {data:'prizebond_location',name:'prizebond_location'},

            {
                mRender: function (data, type, row) {
                    var linkEdit='<a href="{{url('user_prizebonds.edit')}}" class="btn btn-warning btn-sm glyphicon glyphicon-edit" data-id="' + row['id'] + '">EDIT</a>';
                    var linkDelete='<a href="{{url('user_prizebonds.destroy')}}" class="editor_remove btn btn-danger btn-sm glyphicon glyphicon-trash" data-id="' + row["data:id"] + '">DELETE</a>';
                    return  linkEdit + "  " + linkDelete;
                }
            }

        ]
    });
} );

// mycontroller code

public function data(){
    $current_user_id=Auth::id();
    $user_prizebonds=UserPrizebond::where('user_id', "$current_user_id")->get();
   return Datatables::of($user_prizebonds)->make(true);

}

Answer 1

写入您的控制器功能。像这样

return Datatables::of($user_prizebonds)->addColumn('action', function ($user_prizebonds) {
    return = '<button type="button" class="btn btn-default">Default</button>';
})->make(true);

和js代码

{data: 'action', name: 'action', orderable: false, searchable: false}

如何在laravel 5.3中使用jquery数据表实现CRUD

1 个答案: