how to show popup using ajax output - codeigniter

Question

i want to show a popup message about to login the user. here iam using ajax function.
if the success: part is true i want to show the popup message. pls help me thanks in adv.
<button data-toggle="modal" data-target="#squarespaceModal" class="btn green btn-success" onclick="going('<?echo $user_id;?>','<?echo $event_id;?>')">Going</button>
<div class="modal fade" id="squarespaceModal" tabindex="-1" role="dialog" aria-labelledby="modalLabel" aria-hidden="true">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal"><span aria-hidden="true">×</span><span class="sr-only">Close</span></button>
<h3 class="modal-title" id="lineModalLabel">My Modal</h3>
</div>
<div class="modal-body">
<!-- content goes here -->
<form>
<div class="form-group">
<label for="exampleInputEmail1">Email address</label>
<input type="email" class="form-control" id="exampleInputEmail1" placeholder="Enter email">
</div>
<div class="form-group">
<label for="exampleInputPassword1">Password</label>
<input type="password" class="form-control" id="exampleInputPassword1" placeholder="Password">
</div>
<div class="form-group">
<label for="exampleInputFile">File input</label>
<input type="file" id="exampleInputFile">
<p class="help-block">Example block-level help text here.</p>
</div>
<div class="checkbox">
<label>
<input type="checkbox"> Check me out
</label>
</div>
<button type="submit" class="btn btn-default">Submit</button>
</form>
</div>
<div class="modal-footer">
<div class="btn-group btn-group-justified" role="group" aria-label="group button">
<div class="btn-group" role="group">
<button type="button" class="btn btn-default" data-dismiss="modal" role="button">Close</button>
</div>
<div class="btn-group btn-delete hidden" role="group">
<button type="button" id="delImage" class="btn btn-default btn-hover-red" data-dismiss="modal" role="button">Delete</button>
</div>
<div class="btn-group" role="group">
<button type="button" id="saveImage" class="btn btn-default btn-hover-green" data-action="save" role="button">Save</button>
</div>
</div>
</div>
</div>
</div>
</div>
function going()
function going(user_id,event_id)
{
$.ajax({
url: "<?echo base_url()?>events/event_going",
type: 'post', // HTTP METHOD
data:
{user_id:user_id,event_id:event_id },
success: function(data)
{
if(data='true')
{
//pls check & add any suggsns
$('#squarespaceModal').dialog(data);
}
// document.getElementById('mysubmit').value="active";
}
});
}
controller
public function event_going()
{
$user_id=$this->input->post('user_id');
$event_id=$this->input->post('event_id');
$temp=$this->session->userdata('user');
$going=$this->EM->is_going($event_id,$user_id);
$going1=$this->EM->is_going1($event_id,$user_id);
if($temp=="")
{
echo "Please log in";
}
else if($going==$user_id)
{
echo "Already confirmed as going..";
}
else if($going1==$user_id)
{
echo "Are you sure you want to change... Press ok to continue";
$this->EM->event_updation($event_id,$user_id);
}
else
{
$data=array('event_id'=>$event_id,'ev_going'=>$user_id);
$this->EM->eventgoing($data);
}
}

Answer 1

What you can do is, instead of showing a new popup, you can get the content of popup from ajax. What i mean is, on click of a button, call ajax function, get the content and have a div inside the modal-body, for example
<div class="dynamic-data">
</div>
and in ajax success function do like this
success: function(data){
$(".dynamic-data").html(data);
}