CODEWITHSUNDEEP
phpjquerycakephpcakephp-2.0
FernandoPaiva
FernandoPaiva

Reputation: 4460

Creating a link with CakePHP and JQuery?

I'm trying create a $this->Html->link inside a JQuery script. This link goes redirect to an action in my controller the problem is that I can not do this.

How could I do this ?

Trying this.

$(document).ready(function() {
    $('#dataTables-example').DataTable({
        "processing": true,
        "serverSide": true,
        "ajax":{
            url: "<?php echo $this->Html->url("/Empresas/indexAjax.json");?>",
            dataSrc:""
        },
        "columns": [
            {"data": "Empresa.id"},
            {"data": "Empresa.nomeFantasia"},
            {"data": "Empresa.cnpj"},
            {"data": "Empresa.telefone1"},
            {"data": "Empresa.telefone2"},
            {"data": "Empresa.celular"},
            {"data": "Empresa.aberto"},
            {"data":null,
                "bSortable": false,
                "render": function(obj) {
                    return "<?php echo $this->Html->link('<i class="glyphicon glyphicon-eye-open"></i>',
                                                        array('action' => 'view', obj["Empresa"].id), 
                                                        array('title'=>'view', 'escape' => false)); ?>";

                    //return '<a href=/Project/Empresas/view/' + o["Empresa"].id + '>' + 'View' + '</a>';
                }
            }
        ]
    });   
});

Upvotes: 0

Views: 159

Answers (1)

user3082321
user3082321

Reputation: 654

You have typos in this. You're trying to get property of $obj["Empresa"] with wrong operator. You can do what you want with the code below.

<?php $link = $this->Html->link('<i class="glyphicon glyphicon-eye-open"></i>',
                                ['action' => 'view', $obj["Empresa"]->id], 
                                ['title'=>'view', 'escapeTitle' => false)]); 
?>

return "<? echo $link; ?>";

Upvotes: 2

Related Questions