jquery - dataTables .row( ) 不是服务器端处理行详细信息的函数

Question

 function format ( d ) {
return 'Full name: <br>'+
       'Salary:<br>'+
       'The child row can contain any data you wish, including links, images, inner     tables etc.';
  }

 $(function() {
 var dtable = $('#table_echipamente').dataTable( {
    "processing": true,
    "serverSide": true,
    "ajax": {
        "url": "inc/table_echipamente.php",
        "type": "POST"
    },
    "columns": [
        {
            "class":          "details-control",
            "orderable":      false,
            "data":           null,
            "defaultContent": ""
        },
        { "data": "beneficiar" },
        { "data": "distribuit" },
        { "data": "data_distribuit" },
        { "data": "denumire" },
        { "data": "nr_inventar_nou" },
        { "data": "nr_inventar_vechi" },
        { "data": "gestionar" },
        { "data": "observatii" },
        { "data": "optiuni" }
    ],
    "order": [[1, 'asc']]

  });

  // Array to track the ids of the details displayed rows
  var detailRows = [];

  $('#table_echipamente tbody').on( 'click', 'tr td:first-child', function () {
    var tr = $(this).closest('tr');
    var row = dtable.row( tr );
    var idx = $.inArray( tr.attr('id'), detailRows );

    if ( row.child.isShown() ) {
        tr.removeClass( 'details' );
        row.child.hide();

        // Remove from the 'open' array
        detailRows.splice( idx, 1 );
    } else {
        tr.addClass( 'details' );
        row.child( format( row.data() ) ).show();

        // Add to the 'open' array
        if ( idx === -1 ) {
            detailRows.push( tr.attr('id') );
        }
    }
   } );

  // On each draw, loop over the `detailRows` array and show any child rows
  dtable.on( 'draw', function () {
    $.each( detailRows, function ( i, id ) {
        $('#'+id+' td:first-child').trigger( 'click' );
      } );
  } );



  } );

当我点击打开按钮时出现以下错误:

TypeError: dtable.rows is not a function

我做错了什么？我按照官方网站上的示例进行操作。我找不到类似的问题，所以很奇怪。

Answer 1

我遇到了同样的问题，只是通过堆栈溢出的答案解决了它。使用 DataTable() 而不是 dataTable()。方法 row().child() 不适用于 dataTable()。所以你将拥有:

var dtable = $('#table_echipamente').DataTable( {})

而不是

var dtable = $('#table_echipamente').dataTable( {})