I'm getting an error "Object doesn't support property or method 'column'. How do I fix this?

kmh158

I'm running the same code on IE11 using sharepoint 2013. I used the same example of coding that you have provided. I couldn't find anything to fix this issue.

kmh158

$('#example thead tr').clone(true).appendTo( '#example thead' );
$('#example thead tr:eq(1) th').each( function (i) {
var title = $(this).text();
$(this).html( '<input type="text" placeholder="Search '+title+'" />' );

    $( 'input', this ).on( 'keyup change', function () {
        if ( table.column(i).search() !== this.value ) {
            table
                .column(i)
                .search( this.value )
                .draw();
        }
    } );

This what I'm using

allan

I don't see where you are defining your table variable. Please could you post a link to a test case showing the issue per the forum rules?

My guess is that you are using $().dataTable() rather than $().DataTable() as noted in the top FAQ.

Allan

kthorngren

My guess is the problem is with table.column(i) in line 2. What does the variable table contain? In your code it is expected to be the Datatables API, ie, var table = $('#myTable').DataTable();. Seems like it must be something else.

Hard to say without seeing it. Please post a link to your page or a test case so we can take a look.
https://datatables.net/manual/tech-notes/10#How-to-provide-a-test-case

Kevin

kmh158

this what I have for "table"

var $tabel = $('#mytable');
var $parent = $table.children('tbody');
var table = $table.dataTable();

kmh158

var $tabel = $('#mytable');

var $parent = $table.children('tbody');

var table = $table.dataTable();

$('#example thead tr').clone(true).appendTo( '#example thead' );

$('#example thead tr:eq(1) th').each( function (i)
{
var title = $(this).text();
$(this).html( '<input type="text" placeholder="Search '+title+'" />' );

$( 'input', this ).on( 'keyup change', function () {
if ( table.column(i).search() !== this.value ) {
table
.column(i)
.search( this.value )
.draw();
}
} );
});

This the how the code looks like that I have

kthorngren

var table = $table.dataTable(); needs to be var table = $table.DataTable(); to use the API. Note the D in `DataTable. Please see this FAQ for more details.

Kevin

kmh158

I changed it to DataTable(), but it's still giving me the same error

colin

Hi @kmh158 ,

We're happy to take a look, but as per the forum rules, please link to a test case - a test case that replicates the issue will ensure you'll get a quick and accurate response. Information on how to create a test case (if you aren't able to link to the page you are working on) is available here.

Cheers,

Colin

kthorngren

var $tabel = $('#mytable');

var $parent = $table.children('tbody');

var table = $table.dataTable();

Looks like a typo. In line 1 you have $label and in line 5 you have $table. If this doesn't help then as Colin mentions we will need to see it.

Kevin