从视图中将CakePHP交叉发布到不同的控制器

Question

I have a 'jobs' controller which allows users to assign tasks/jobs to each other. Each of these jobs has a 'client' which is a foreign key.

Within my 'add' view for the jobs controller, I've added a checkbox and another field to allow the user to add a client from within the 'add' job form. Basically, I need an ajax request that sends the entered client name, to the 'clients' controller, who will then add the new client to the database, and in turn return a new list of clients, for me to update my drop-down list.

I'm not sure where to begin with this, as I've not used AJAX from a PHP environment before and I'm not sure how I should structure it, or how I should begin processing it.

Answer 1

I've hammed a solution out. It's not very elegant, but it works.

At the bottom of my View:

echo $this->Html->scriptBlock(
            'function sendAjaxRequest(){
            $.ajax({
               type: "POST",
                url: "' . $this->Html->url(array('controller' => 'clients', 'action' => 'add_external')) . '",
                data: "name="+$(\'#JobNewClientName\').val(),
                success: function(msg){
                    //parse and sort the entries alphabetically
                    msg = $.parseJSON(msg);
                    var values = [];
                    for(var i in msg) {
                        values.push({ key: i, value: msg[i] });
                    }
                    values.sort(function(a, b) {return a.value.toLowerCase().localeCompare(b.value.toLowerCase())});
                    var str = values.map(function (kvp) { return kvp.value; }).join(\'\n\');
                    //remake the selectbox options
                    var options = \'<option value=""></option>\';
                    $.each(values, function(i){
                        options += \'<option value="\'+values[i].key+\'">\'+values[i].value+\'</option>\';
                    });        
                    $(\'#JobClientId\').html(options);
                }
            });
     }');

Controller:

function add_external() {
        $this->autoRender = false;        
        $this->Client->create();
        $this->Client->set('name', $_POST['name']);
        $this->Client->save();
        echo json_encode($this->Client->find('list', array('order' => array('Client.name' => 'ASC'))));     
    }

Essentially, the add_external() function takes a posted name from the ajax and adds it to the database. Then it json_encode s all clients.

This JSON object is then parsed and sorted in the javascript success statement, this sorting retains the order as defined by the controller, ie alphabetically by value (not key).

Next, I am able to create a brand new options variable containing the same markup that the select box requires. Then all I need to do is switch them out.

If anybody can think of a more elegant solution, I'd be happy to award them the acceptance.