duanfei1268 2018-04-30 15:39
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Json编码codeigniter中的错误

I tried to load a view in my codeigniter controller and also i pass a status to my ajax function. In the ajax function i wrote a sweetalert popup in the sucess function when my status is 1. now my actual problem is when i load a view in the controller the status is not passing to my ajax function.can anyone figureout what the problem is any help is appreciable.

controller

$query = $this->package_view_model->enquiry_history();
    if (isset($query))
      {
        $status = 1;
        $this->load->view('packages/enquiry');
      }
    echo json_encode (array("status"=>$status));

ajax function

function sendMail(package_url)
{   
    var formData = $("#enqform").serialize();
      jQuery.ajax({
          type: 'POST',
          url: '<?php echo base_url() ?>tour-package/send-mail',
          data: formData,
          dataType: 'json',
          success: function(data){
            if(data.status == 1){
                swal({
                       title:'Thankyou for your interest!!!',
                       text: "Our excecutives will contact you soon.",
                       html: true,
                       type: "success",
                       showCancelButton: false,
                       showConfirmButton:false
                   });
                window.setTimeout(function() {
                    window.location.href = '<?php echo base_url() ?>tour-packages';
                }, 1000000);
            }
          }
      }); 
      return false;
      e.preventDefault();
  }
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2条回答 默认 最新

  • download2014711 2018-04-30 16:05
    关注

    Hope this will help you :

    $query = $this->package_view_model->enquiry_history();
    $status = 1;
    if (! empty($query))
    { 
       $this->load->view('packages/enquiry');
    }
    echo json_encode(array("status" => $status));
    exit;
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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