我正在尝试制作一个基于ajax的进度条。但是，我不知道如何计算上传了多少数据，我想将其显示为上传数据的百分比。感谢
您可以使用APC或PEAR Package Upload进度。
http://pecl.php.net/package/uploadprogress
我已经有一段时间没有这样做了，我记得Webkit有一个问题，必须使用iframe。可能想调查一下。
试试这个：-
演示网址：--
http://jquery.malsup.com/form/progress.html
你可以从这个url下载jQuery文件并添加html标签
http://ajax.googleapis.com/ajax/libs/jquery/1.7/jquery.js
http://malsup.github.com/jquery.form.js
试试这个：
这是我的html标记：
<!doctype html>
<head>
<title>File Upload Progress Demo #1</title>
<style>
body { padding: 30px }
form { display: block; margin: 20px auto; background: #eee; border-radius: 10px; padding: 15px }
.progress { position:relative; width:400px; border: 1px solid #ddd; padding: 1px; border-radius: 3px; }
.bar { background-color: #B4F5B4; width:0%; height:20px; border-radius: 3px; }
.percent { position:absolute; display:inline-block; top:3px; left:48%; }
</style>
</head>
<body>
<h1>File Upload Progress Demo #1</h1>
<code><input type="file" name="myfile"></code>
<form action="upload.php" method="post" enctype="multipart/form-data">
<input type="file" name="uploadedfile"><br>
<input type="submit" value="Upload File to Server">
</form>
<div class="progress">
<div class="bar"></div >
<div class="percent">0%</div >
</div>
<div id="status"></div>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7/jquery.js"></script>
<script src="http://malsup.github.com/jquery.form.js"></script>
<script>
(function() {
var bar = $('.bar');
var percent = $('.percent');
var status = $('#status');
$('form').ajaxForm({
beforeSend: function() {
status.empty();
var percentVal = '0%';
bar.width(percentVal)
percent.html(percentVal);
},
uploadProgress: function(event, position, total, percentComplete) {
var percentVal = percentComplete + '%';
bar.width(percentVal)
percent.html(percentVal);
},
complete: function(xhr) {
bar.width("100%");
percent.html("100%");
status.html(xhr.responseText);
}
});
})();
</script>
</body>
</html>
我的php代码：
<?php
$target_path = "uploads/";
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
echo "The file ". basename( $_FILES['uploadedfile']['name']).
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
?>