Getting data from datatable then passing it to php with ajax

Question

I am using jQuery DataTables, I can get data from selected row using this code

var str = $.map(table.rows('.selected').data(), function (item) {
                    return item[5]+" "+item[0]
                });

where item[5] is id, item[0] is a string.

I want to split return string for passing id and string. the error founded in ajax code specifically in

data : {}

where is the problem in this code.

<script>

$(document).ready(function() {
    var table = $('#liveSearch').DataTable();
    $('#liveSearch tbody').off('click', 'tr').on( 'click', 'tr', function () {
        if ( $(this).hasClass('selected') ) {
            $(this).removeClass('selected');
        }
        else {
            table.$('tr.selected').removeClass('selected');
            $(this).addClass('selected');
        }
    } );

        $('.example3-1').on('click', function () {
            if ((table.rows('.selected').data().length) > 0) {
                var str = $.map(table.rows('.selected').data(), function (item) {
                    return item[5]+" "+item[0]
                });
                console.log(str);
            $.confirm({
                confirmButtonClass: 'btn-info',
                cancelButtonClass: 'btn-danger',
                confirm: function () {
                    $.ajax({
                        type: 'post',
                        url: 'delete.php',
                        data: {
                            str1 : str.substr(0,str.indexOf(' ')),
                            str2 : str.substr(str.indexOf(' ')+1)
                        },
                        success: function( data ) {
                            console.log( data );
                        }
                    });
                    table.row('.selected').remove().draw(false);

                }
            });
        }
        });
} );

Have you tried using the javascript debugger in the browser? — RiggsFolly
– RiggsFolly, Commented Aug 15, 2015 at 23:40
No Use the javascript debugger its a great tool and lives in all browsers. Try F12 — RiggsFolly
– RiggsFolly, Commented Aug 15, 2015 at 23:47

Gyrocode.com · Accepted Answer · 2015-08-16 16:37:21Z

1

CAUSE

You're only allowing only one row selected. But you're using $.map which returns Array not a string as you expect. There is no sense in using $.map for just one row.

SOLUTION

Use the following code instead to get data for selected row and produce the string needed.

var rowdata = table.row('.selected').data();
var str = rowdata[5] + " " + rowdata[0];

It could be simplified further:

var rowdata = table.row('.selected').data();

// ... skipped ...

$.ajax({

   // ... skipped ...

   data: {
      str1: rowdata[5],
      str2: rowdata[0]
   }

   // ... skipped ...

});

NOTES

The solution would be different if you allow multiple row selection.

answered Aug 16, 2015 at 16:37

Gyrocode.com

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