How to pass select option values from php code to Jquery?

Question

When clicking add row button, new row will add to the specific table. So I need to add a select option with php option values.

How to pass this php values to jQuery?

Jquery function I need to show select option inside the rowData.push('');

  $('.dt-add').each(function () {
        var whichtable = $(this).parents('form').attr('data-id');
        $(this).on('click', function(evt){
            //Create some data and insert it
            var rowData = [];
            var table = $('#teammembertable' + whichtable).DataTable();
//          rowData.push('');

            rowData.push('<select class="form-control addstafftype" id="addstafftype" name="addstafftype"><option value="">Select</option><option value="Leader">Leader</option><option value="Technician">Technician</option></select');
            rowData.push('<button type="button" data-id='+ whichtable +' class="btn-xs dt-delete dt-deletes"><i style="font-size:10px" class="fa">&#xf00d;</i></button>');
            table.row.add(rowData).draw( false );
        });
    });

PHP CODE

$dataadd_team_memb = array(
        'team_id' => $id,
        'Staff_id' => $this->input->post('getaddstaffname'),
        'Staff_type' => $this->input->post('getaddstafftype'),
        'status' => "active"
    );


    $insert_id = 0;
    if ($this->db->insert("team_members", $data)) {
        $insert_id = $this->db->insert_id();
    }

Put the select elements inside a form and submit that form to a PHP page. — Rory McCrossan
– Rory McCrossan, Commented Jul 31, 2019 at 7:50

Piyush Shukla · Accepted Answer · 2019-07-31 08:17:45Z

2

$('.dt-add').each(function () {
    var whichtable = $(this).parents('form').attr('data-id');
    $(this).on('click', function(evt){
        var rowData = [];
        var table = $('#teammembertable' + whichtable).DataTable();
        rowData.push('<select class="form-control addstafftype" id="addstafftype" name="addstafftype">'+
            '<option value="">Select</option>'+
            '<?php foreach($selectallstaff as $staffname){ ?>'+
                '<option value="<?php $staffname["Staff_id"]; ?>"><?php $staffname["Staff_name"]; ?></option>'+
            '<?php } ?>'+
            '</select');
        rowData.push('<button type="button" data-id='+ whichtable +' class="btn-xs dt-delete dt-deletes"><i style="font-size:10px" class="fa">&#xf00d;</i></button>');
        table.row.add(rowData).draw( false );
    });
});

answered Jul 31, 2019 at 8:17

Piyush Shukla

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4 Comments

sethu Over a year ago

its working now when i am giving <?= $staffname["Staff_id"]; ?>

Aksen P Over a year ago

remember also about <?= $staffname["Staff_name"]; ?>

Piyush Shukla Over a year ago

Ya, when you use the latest version on PHP.

Piyush Shukla Over a year ago

for($i=0;$i < count($addstaffname);$i++){ $addstaffname_final = $addstaffname[$i]; }

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