使用codeigniter上传视频

Question

你好我试图上传一个视频到我的文件夹，我已经搜索了问题已经有了答案，并尝试了所有这些，似乎没有一个工作。 我想要在同一输入中上传图像和视频

这是我视图中的代码

$('#upload_image_button').on('click', function(){
          var imageFile = $('input[name=upload_image_input]');
          var imageToUpload = imageFile[0].files[0];
          var imageName = $('#upload_image_name').val();
          $('#Upload_Image').val(imageName);


           if(imageToUpload) {

                      //Provide the form data that would be sent to server through ajax
                      var formData = new FormData();
                      var uploadURI = '<?php echo base_url('uploadImage'); ?>';

                     formData.append('upload_image_input', imageToUpload);

                      //Upload the file using ajax
                      $.ajax({
                          type: 'POST'
                          , url: uploadURI
                          , data: formData
                          , processData: false
                          , contentType: false
                          , success: function() {
                            var image = '<?php echo base_url("assets/images/'+ imageName +'"); ?>';
                              //$('#profile_pic').attr('src', image);
                              Materialize.toast('Image Attached', 4000);
                          }
                          , error: function(errorw){
                              Materialize.toast('Something went wrong. Please try again', 4000);
                          }
                      });
                  }
                  else{
                      Materialize.toast('Field Empty', 4000);
                  }

    });

这是我控制器的代码

public function uploadImage(){
    $config['upload_path'] = 'assets/files';
    $config['allowed_types'] = 'mp4|3gp|jpg|png|jpeg';
    $config['max_size']     =0;
    $config['max_width']    =0;
    $config['max_height']   =0;

    $this->upload->initialize($config);

    if ($this->upload->do_upload('upload_image_input')) {
            //File upload
            $uploadData = $this->upload->data(); 
            $filename = $uploadData['file_name'];
         // $this->loginmodel->upload_model($filename,$account_id);
        }   
}