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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。谢谢

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2 回答 2

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您可以使用 APC 或 PEAR 包上传进度。

http://pecl.php.net/package/uploadprogress

有一段时间没有这样做了，我记得 Webkit 存在问题并且必须使用 iframe。可能想调查一下。

于 2012-09-05T15:41:23.900 回答
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尝试这个：-

演示网址：--

http://jquery.malsup.com/form/progress.html

您可以从此 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

尝试这个：

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>&lt;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!";
}
?>
于 2012-09-05T16:00:57.573 回答