如何解决jQuery扩展了Datatable试图从Laravel变量获取数据
我正试图通过从控制器传递来使用compact()函数进行查看的变量将数据传递到jquery Datatable,但是我面临的问题是: DataTables警告:表格ID = DataTables_Table_0-Ajax错误。有关此错误的更多信息,请参见http://datatables.net/tn/7
控制器
public function show(Expense $expense)
{
$expenseTypes = ExpenseType::all()->sortBy('label');
$charges = json_encode($expense->charges());
/*$charges = $expense->charges()->get();*/
return view('expenses.show',compact('expense','expenseTypes','charges'));
}
}
HTML
<table class="table m-b-0 table-hover expand-table dataTable table-custom">
<thead>
<tr>
<th></th>
<th style="width: 5%">id</th>
<th>date</th>
<th>description</th>
<th>details</th>
<th style="width: 25%">observation</th>
<th>sum</th>
</tr>
</thead>
<tbody>
</tbody>
<tr class="bg-light">
<td colspan="4"></td>
<th class="text-nowrap" scope="row">{{ __('all.total sum')}}</th>
<td colspan="1"><b>{{ $expense->charges()->sum('sum') }} <small>MAD</small></b></td>
</tr>
</table>
jQuery
<script>
/* Formatting function for row details - modify as you need */
function format ( d ) {
// `d` is the original data object for the row
return '<table cellpadding="5" cellspacing="0" border="0" style="padding-left:50px;">'+
'<tr>'+
'<td>description:</td>'+
'<td>'+d.description+'</td>'+
'</tr>'+
'<tr>'+
'<td>details:</td>'+
'<td>'+d.details+'</td>'+
'</tr>'+
'<tr>'+
'<td>observation:</td>'+
'<td>'+d.observation+'</td>'+
'</tr>'+
'</table>';
}
$(document).ready(function() {
var table = $('.expand-table').DataTable( {
serverSide: false,"ajax": {
"url" : "{{$charges}}","type" : "POST"
},"columns": [
{
"className": 'details-control',"orderable": false,"data": null,"defaultContent": ''
},{ "data": "id" },{ "data": "date" },{ "data": "description" },{ "data": "details" },{ "data": "observation" },{ "data": "sum" }
],"order": [[1,'asc']]
} );
// Add event listener for opening and closing details
$('.expand-table tbody').on('click','td.details-control',function () {
var tr = $(this).closest('tr');
var row = table.row( tr );
if ( row.child.isShown() ) {
// This row is already open - close it
row.child.hide();
tr.removeClass('shown');
}
else {
// Open this row
row.child( format(row.data()) ).show();
tr.addClass('shown');
}
} );
} );
</script>
解决方法
您正试图将null
作为键传递给第一列数据,请使用id
或传递的JSON数据中存在的任何其他有效键。当标题中传递的json OR列计数中的键不可用且数据不匹配时,会出现此错误。
"columns": [
{
"className": 'details-control',"orderable": false,"data": 'id',/// instead of 'null use id
"defaultContent": ''
},{ "data": "id" },{ "data": "date" },{ "data": "description" },{ "data": "details" },{ "data": "observation" },{ "data": "sum" }
],
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