javascript - Post files and input throught ajax formData -

i need post both files , input value throught 1 ajax call here code using actually, upload working problem when submit form 2 rows(see pic) image here
because posting 2 ajax calls same php page, if can post 1 ajajx call both values. html

<form  method="post" id="form_upload" enctype="multipart/form-data">   <input type="file" name="user_upload" id="mydrop" />   <input type="hidden" id="userid" value="1" />  </form> 

ajax

var form_upload = $("#form_upload").serialize();                      $.ajax({                     data: form_upload,                      type: 'post',                      url: 'upload.php',                      success : function(data){                         alert(data);                      }                 });                     var file_data = $('#mydrop').prop('files')[0];                     var form_data = new formdata();                                       form_data.append('file', file_data);                     $.ajax({                                 url: 'upload.php',                                  datatype: 'text',                                 cache: false,                                 contenttype: false,                                 processdata: false,                                 data: form_data,                                                          type: 'post',                      }); 

php

<?php     if(isset($_post)){      $dbhost = "localhost";     $dbuser = "root";     $dbpass = "";     $dbname = "494815";      try {          $dbcon = new pdo("mysql:host={$dbhost};dbname={$dbname}",$dbuser,$dbpass);         $dbcon->setattribute(pdo::attr_errmode, pdo::errmode_exception);         $targetdir = "users_upload/";         $temp = explode(".", $_files["file"]["name"]);         $newfilename = round(microtime(true)) . '.' . end($temp);         $targetfile = $targetdir.$newfilename;         if move_uploaded_file($_files['file']['tmp_name'],$targetfile);             $user_id = $_post['userid'];             $stmt = $dbcon->prepare("insert users_upload (user_id, file_name, file_date) values('$user_id', '".$newfilename."','".date("y-m-d h:i:s")."')");             $stmt->execute();     }     catch(pdoexception $e)         {             echo $e->getmessage();         }      $conn = null;     } ?>