PHP AJAX图片上传示例不上传


PHP AJAX Image Upload Example Not Uploading

我的代码在控制台上没有产生错误。当我点击上传按钮时,什么也没发生。按照我使用的教程中的说明,有一个帖子被发送到了自己身上,但图像没有上传到我的文件夹中,也没有显示在我的页面上。除非我知道我应该使用jquery(一旦我把它上传到文件夹,我就会进行转换),否则我的代码怎么了?

<?php
if (!empty($_FILES)) {
    $name = $_FILES['file']['name'];
    if ($_FILES['file']['error'] == 0) {move_uploaded_file($_FILES['file']
            ['tmp_name'], "post_images/" . $name))} 
}
?>

<script type="text/javascript">
var handleUpload = function (event) {
event.preventDefault();
event.stopPropagation();
var fileInput = document.getElementById('file');
var data = new FormData();  
data.append('file', fileInput.files[1]);    
var request = new XMLHttpRequest();
request.upload.addEventListener('progress', function(event){
    if (event.lengthComputable) 
    {
    var percent = event.loaded / event.total;
    var progress = document.getElementById('upload_progress');
    while(progress.hasChildNodes()) {
        progress.removeChild(progress.firstChild);          
    }   
progress.appendChild(document.createTextNode(Math.round(percent * 100) + 
    '%'));          
}
});
request.upload.addEventListener('load',function(event) {
document.getElementById('upload_progress').style.display = 'none';
});
request.upload.addEventListener('error', function(event) {
alert('Upload failed');
});
request.open('POST', 'upload.php');
request.setRequestHeader('Cache-Control', 'no-cache');
document.getElementById('upload_progress').style.display = 'block';
request.send(data); 
};

window.addEventListener('load', function(event) {
var submit = document.getElementById('submit');
submit.addEventListener('click', handleUpload);
});
</script>

<div id="uploaded">
<?php
if (!empty($name)) {
echo '<img src="post_images/' . $name . '" width="100" height="100" />';    
}
?>
</div>
<div id="upload_progress"></div>
<div>
<form action="" method="post" enctype="multipart/form-data">
<div>
    <input type="file" id="file" name="file" />
    <input type="submit" id="submit" value="upload" />
</div>
</form>
</div>

在非多文件输入元素中只有一个文件,fileInput.files[1]尝试使用第二个文件。

data.append('file', fileInput.files[0]);