Javascript 文件上载无法使用AJAX

我在fileupload中使用ajax请参见，而使用fileupload值并不是将一个pageupload.php传递到另一个pagefile_upload_submit.php。我不知道如何使用ajax获取文件值

upload.php

你可以试试这个

<input id="sortpicture" type="file" name="sortpic" /><br>
    <input type='button' class="jos" value='upload' name='upload'>


<script src="jquery-2.2.1.js"></script>
<script>
$(document).ready(function(){

  var file_data = $('#sortpicture').prop('files')[0];   
    var form_data = new FormData();                  
    form_data.append('file', file_data);
    alert(form_data);   

 /*for ajax upload*/    
    $.ajax({
                url: 'upload.php', // point to server-side PHP script 
                dataType: 'text',  // what to expect back from the PHP script, if anything
                cache: false,
                contentType: false,
                processData: false,
                data: form_data,                         
                type: 'post',
                success: function(php_script_response){
                    alert(php_script_response); // display response from the PHP script, if any
                }


      }); 


});
</script>

和upload.php

---

希望它有帮助

看到这个答案它会有帮助这可能会帮助你：好吧，你的ajax在哪里？你不能使用。序列化以使用ajax发送文件。你需要使用新的FormData我在运行html页面时尝试了你的代码我得到了这样的[object FormData]是的，它只是一个警报，你可以对它进行评论。但您的文件将上载到img文件夹

**file_upload_submit.php**

<?php
$fstname=$_POST['fname'];// i got answer here

$filename  = basename($_FILES['file']['name']);// i am not getting filename here
$extension = pathinfo($filename, PATHINFO_EXTENSION);
$new_name= md5($filename.time()).'.'.$extension;
if (move_uploaded_file($_FILES['file']['tmp_name'], "upload/".$new_name))  {
echo "success";
}else{
echo "Error";
}   
?>

<input id="sortpicture" type="file" name="sortpic" /><br>
    <input type='button' class="jos" value='upload' name='upload'>


<script src="jquery-2.2.1.js"></script>
<script>
$(document).ready(function(){

  var file_data = $('#sortpicture').prop('files')[0];   
    var form_data = new FormData();                  
    form_data.append('file', file_data);
    alert(form_data);   

 /*for ajax upload*/    
    $.ajax({
                url: 'upload.php', // point to server-side PHP script 
                dataType: 'text',  // what to expect back from the PHP script, if anything
                cache: false,
                contentType: false,
                processData: false,
                data: form_data,                         
                type: 'post',
                success: function(php_script_response){
                    alert(php_script_response); // display response from the PHP script, if any
                }


      }); 


});
</script>

<?php


    if ( 0 < $_FILES['file']['error'] ) {
        echo 'Error: ' . $_FILES['file']['error'] . '<br>';
    }
    else {
        move_uploaded_file($_FILES['file']['tmp_name'], 'img/' . $_FILES['file']['name']);
    }





?>