I\'m trying to get first cell (td
) for each row and getting it but only for current page. If I navigate to next page then the checkbox checked on the previous p
jQuery DataTables removes non-visible rows from DOM for performance reasons. When form is submitted, only data for visible checkboxes is sent to the server.
You need to turn elements that are checked and don't exist in DOM into
upon form submission.
var table = $('#example').DataTable({
// ... skipped ...
});
$('form').on('submit', function(e){
var $form = $(this);
// Iterate over all checkboxes in the table
table.$('input[type="checkbox"]').each(function(){
// If checkbox doesn't exist in DOM
if(!$.contains(document, this)){
// If checkbox is checked
if(this.checked){
// Create a hidden element
$form.append(
$('')
.attr('type', 'hidden')
.attr('name', this.name)
.val(this.value)
);
}
}
});
});
var table = $('#example').DataTable({
// ... skipped ...
});
$('#btn-submit').on('click', function(e){
e.preventDefault();
var data = table.$('input[type="checkbox"]').serializeArray();
// Include extra data if necessary
// data.push({'name': 'extra_param', 'value': 'extra_value'});
$.ajax({
url: '/path/to/your/script.php',
data: data
}).done(function(response){
console.log('Response', response);
});
});
See jQuery DataTables: How to submit all pages form data for more details and demonstration.
value
attribute assigned with unique value.id
attribute check
for multiple elements, this attribute is supposed to be unique.paging
, info
, etc. options for jQuery DataTables, these are enabled by default.
.