Question

我将laravel中的项目存储在子目录中，并且我正在使用数据表，但是，当我想恢复信息时，我得到了典型的错误＆＃34;无效的json＆＃34; ，我的项目网址如下

http://www.loupper.com/loupper-b/public/

但我得到的答案是这个

http://www.loupper.com/subcategory/data 绘制= 1＆安培;柱％5B0％5D％5Bdata％5D =对号＆安培;柱％5B0％5D％5Bname％5D = ID＆安培;柱％5B0％5D％ 5Bsearchable％5D =假安培;柱％5B0％5D％5Borderable％5D =假安培;柱％5B0％5D％5Bsearch％5D％5Bvalue％5D =安培;柱％5B0％5D％5Bsearch％5D％5Bregex％5D =假安培;列％5B1％5D％5Bdata％5D = subcategoria＆安培;柱％5B1％5D％5Bname％5D = subcategoria＆安培;柱％5B1％5D％5Bsearchable％5D =真安培;柱％5B1％5D％5Borderable％5D =真安培;柱％ 5B1％5D％5Bsearch％5D％5Bvalue％5D =安培;柱％5B1％5D％5Bsearch％5D％5Bregex％5D =假安培;柱％5B2％5D％5Bdata％5D = categoria＆安培;柱％5B2％5D％5Bname％ 5D = categoria＆安培;柱％5B2％5D％5Bsearchable％5D =真安培;柱％5B2％5D％5Borderable％5D =真安培;柱％5B2％5D％5Bsearch％5D％5Bvalue％5D =安培;柱％5B2％5D％ 5Bsearch％5D％5Bregex％5D =假安培;柱％5B3％5D％5Bdata％5D =动作＆安培;柱％5B3％5D％5Bname％5D =动作＆安培;柱％5B3％5D％5Bsearchable％5D =假安培;柱％5B3％图5D％5Borderable％5D =假安培;柱％5B3％5D％5Bsearch％5D％5Bvalue％5D =安培;柱％5B3％5D％5Bsearc H％5D％5Bregex％5D =假安培;为了％5B0％5D％5Bcolumn％5D = 1＆安培;为了％5B0％5D％5Bdir％5D =降序＆安培;开始= 0＆安培;长度= 10安培;搜索％5Bvalue％5D =安培;搜索％5Bregex％5D =假安培; _ = 1505922295346

看来，路线还没有占据子目录＆＃34; loupper-b / public /＆＃34; ，我怎么能将子目录添加到网址？

这是我的代码，jquery

$(function () {
$.noConflict();

user_tbl = $('#categoria-table').DataTable({
    processing: true,
    serverSide: true,
    order: [[1, 'desc']],
    ajax: {
        url: '/category/data',
        data: function (d) {
          d.filter = $('select[name=filter]').val();

        },

    },
    columns: [
        {data: 'checkmark', name:'id', orderable: false, searchable: false},
        {data: 'nombre', name: 'nombre'},
        {data: 'action', name: 'action', orderable: false, searchable: false, class:"text-center"}
    ]
});

$('#search-form').on('submit', function(e) {
    user_tbl.draw();
    e.preventDefault();
    return false;
     });

   });

laravel

public function indexDatatable()
{
    $categoria=DB::table("categoria")->select("id","nombre")->get();

    return Datatables::of($categoria)
    ->addColumn('action', function ($categoria) {

                return '<button onClick=deleteCategoria('.$categoria->id.'); type="button" class="btn btn-xs btn-danger" data-id="'.$categoria->id.'"><i class="glyphicon glyphicon-trash"></i></button><a href="/categoria/edit/'.$categoria->id.'" class="btn btn-xs btn-primary f-left"><i class="fa fa-pencil"></i></a>';

    })
   ->editColumn('checkmark', function($categoria) {
        return '<input type="checkbox" id="chk" name="chk[]" class="case" value="'.$categoria->id.'">';
     })

    ->rawColumns(['checkmark','action'])
    ->make(true);

}

Answer 1

您可以使用url('/category/data')让它运行。

user_tbl = $('#categoria-table').DataTable({
processing: true,
serverSide: true,
order: [[1, 'desc']],
ajax: {
    url: '{{url('/category/data')}}',
    data: function (d) {
      d.filter = $('select[name=filter]').val();

    },

},
columns: [
    {data: 'checkmark', name:'id', orderable: false, searchable: false},
    {data: 'nombre', name: 'nombre'},
    {data: 'action', name: 'action', orderable: false, searchable: false, class:"text-center"}
]
});

Answer 2

固定。我只需删除第一个斜杠：

ajax: {
  url: 'category/data',
}

如何更改数据表的响应？，laravel

2 个答案: