为什么赢得ajax让我打印php回音值？

Question

我是ajax的新手，我在尝试从我的php上传页面打印回显值时遇到了一些麻烦。我的目标是获得回声，这是结果，并在我的跨度div中打印。

这是我的JavaScript。

     $(document).ready(function (e) {
        $('#myForm').on('submit',(function(e) {
            e.preventDefault();
            var formData = new FormData(this); 
             $.ajax({
                type:'POST',
                url: $(this).attr('action'),
                data:formData,
                cache:false,
                contentType: false,
                processData: false,
                success:function(data){
                    console.log("success");
                    console.log(data);
                    $("#myForm")[0].reset();
                    $("#result").append(data);
                },
                error: function(data){
                    console.log("error");
                    console.log(data);
                }
            });

        }));

    }); 

这是我的HTML

<form action="postphoto" id="myForm" method="POST" enctype="multipart/form-data">
                      <input type="file" onchange="readURL(this);" name="profilepic" id="ImageBrowse"/>
                      <br>
                      <img id="blah" src="#" alt="" height="300" width="400" />
                      <br>
                      <br>
                      <textarea class="form-control" name="description" rows="1" cols="40" placeholder="Description"></textarea>  
                      <br>
                      <textarea class="form-control" name="tagone" rows="1" cols="10" placeholder="Tag"></textarea>  
                      <textarea class="form-control" name="tagtwo" rows="1" cols="10" placeholder="Tag"></textarea>  
                      <textarea class="form-control" name="tagthree" rows="1" cols="10" placeholder="Tag"></textarea>  
                      <br>
                      <textarea class="form-control" name="content" rows="5" cols="40" placeholder="Inspiration behind this.."></textarea> <br>
                      <textarea class="form-control" name="url" rows="1" cols="10" placeholder="Url"></textarea> 
                      <em>Separate words in the url by a dash. For example, fashion-fashion.</em><br>
                      <span id="result"></span>
                      <br><input type="submit" class="form-control" id="sub" name="uploadpic" value="Upload">
                      </form>

这是php       

       $chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
       $rand_dir_name = substr(str_shuffle($chars), 0, 15);
       $rand_dir_name = mysqli_real_escape_string($db_conx,$rand_dir_name);

       mkdir("userdata/user_photos/$rand_dir_name");

       if (file_exists("userdata/user_photos/$rand_dir_name/".@$_FILES["profilepic"]["name"]))
       {
        echo @$_FILES["profilepic"]["name"]." Already exists";
       }
       else
       {
         $query = "SELECT * FROM photos WHERE link='$url'";
         $query = $db_conx->query($query);
         $checks = mysqli_num_rows($query);

         if ($checks == 0) {

            move_uploaded_file(@$_FILES["profilepic"]["tmp_name"],"userdata/user_photos/$rand_dir_name/".$_FILES["profilepic"]["name"]);
            //echo "Uploaded and stored in: userdata/profile_pics/$rand_dir_name/".@$_FILES["profilepic"]["name"];
            $profile_pic_name = @$_FILES["profilepic"]["name"];
            $profile_pic_name = mysqli_real_escape_string($db_conx,$profile_pic_name);
            $img_id_before_md5 = "$rand_dir_name/$profile_pic_name";
            $img_id = md5($img_id_before_md5);
            $profile_pic_query = "INSERT INTO photos VALUES ('','test','$user','$date','$description','http://website.com/userdata/user_photos/$rand_dir_name/$profile_pic_name','no','$url','$tag1','$tag2','$tag3','$content')";
            $profile_pic_query1 = $db_conx->query($profile_pic_query);
            echo "Your photo was uploaded!";
         }
         else
         {
          echo "Link has already been taken.";
         }
       }
    }
    }
    ?>

问题仍然存在。为什么我的ajax不会在我的结果范围内打印回声。如果你能帮助我，我将不胜感激。