Pagination with selected check boxes. Checkboxes w

Question

I'm using the jquery datatables plugin.

I have just a straight html table layout for it.

<table  class="display" id="contactsTable">  
    <thead> 
        <tr>  
            <th>Id</th> 
            <th>Email</th> 
            <th>Name</th> 
            <th>Phone</th>  
            <th>City</th>    
            <th>State</th>    
            <th>Arrival</th>  
            <th>Departure</th> 
            <th>Inserted</th> 
            <th>Check</th> 
        </tr> 
    </thead> 
    <tbody>    
                    <tr>             
            <td>301</td> 
            <td>email address</td>    
            <td>Test</td> 
            <td></td> 
            <td></td> 
            <td></td> 
            <td>July 14 2011</td> 
            <td>July 23 2011</td> 
            <td>April 12 2011 07:05</td> 
            <td><input type="checkbox" name="selected[]" value="301" class="chkbox"/></td> 
        </tr> 
                    <tr>             
            <td>300</td> 
            <td>email</td>    
            <td>Test</td>  
<td></td> 
            <td></td> 
            <td></td>               
            <td>September 02 2011</td> 
            <td>September 10 2011</td> 
            <td>April 11 2011 12:01</td> 
            <td><input type="checkbox" name="selected[]" value="300" class="chkbox"/></td> 
        </tr> 

Here is my code for submit (just temp).

<input id="submitButton" type="submit" value="Submit" onclick="test()" />

And my javascript to map the checkboxes to an array.

function test() {
        var values = $('input:checkbox:checked.chkbox').map(function () {
            return this.value;
        }).get();
        console.log(values);
    }

Here is the datatables code

 $(document).ready(function() {  

    var selected;
    var selectedOutput = '#selectedOutput';
    var template = 'selectedTemplate';
    var submitButton = '#submitButton'

    var  dTable = $('#contactsTable').dataTable({
        "bJQueryUI": true,
        "sPaginationType": "full_numbers",
        "bPaginate": true,
        "bLengthChange": true,
        "bFilter": true,
        "bSort": false,
        "bInfo": true,
        "bAutoWidth": false,
        "bProcessing": true,
        "aoColumns": [
            {"bVisible": false }, //keep the id invisble    
            null,
            null,     
            null,
            null,
            null,
            null,
            null,
            null,
            null
        ]
    });     

The problem is, I can't do a checkbox selection across a page. The table has mutiple pages to it. If I click submit, it only submits the array of checkboxes for the current page I am on.

I hope this is clear enough. I'm not sure what is happening. Thanks for any help.

Answer 1

i think you might have a look at this, it might solve your problem.

This example might be of help to you as it seems to be exactly your case (this one uses fnGetNodes(), while in the other link it is explained how to use fnGetHiddenNodes() if the first function doesn't work ): http://datatables.net/examples/api/form.html

Answer 2

CAUSE

jQuery DataTables removes non-visible rows from DOM for performance reasons, that is why when you submit the form only visible checkboxes get submitted.

SOLUTION

You may need to turn those <input type="checkbox"> that are checked and don't exist in DOM into <input type="hidden"> upon form submission.

For example, to submit form with values of all checkboxes:

var table = $('#example').DataTable();

$("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(
               $('<input>')
                  .attr('type', 'hidden')
                  .attr('name', this.name)
                  .val(this.value)
            );
         }
      } 
   });          
});

If you're submitting the form via Ajax, it's even simpler.

For example, to submit form via Ajax with values of all checkboxes:

var table = $('#example').DataTable();

$("#btn-submit").on('click', function(e){
   e.preventDefault();

   $.ajax({
      url: "/path/to/your/script.php",
      data: table.$('input[type="checkbox"]').serialize();
   }).done(function(data){
      console.log("Response", data);
   });
});

DEMO

See jQuery DataTables: How to submit all pages form data for more information and demonstration.

Answer 3

var oSettings = oTable.fnSettings();
oSettings._iDisplayLength = CHECK_BOXES_COUNT_OR_ROW_COUNT;
oTable.fnDraw();

This should solve the issue for you