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我一直在尝试找到一个简单的教程，该教程显示了如何为我的文件上传添加进度百分比的基础知识，我已经构建了我的文件上传部分，所以我不想要两者都附带的插件，我想成为能够自己编写进度条，但我需要一些关于如何做到这一点的帮助。我想了解它是如何工作的，我不只是想要一些可以为我完成这一切的插件。

任何帮助将不胜感激，谢谢！

我只是对如何获取文件上传的百分比感兴趣，而不是对进度条本身感兴趣。我希望能够有一个准确的百分比。

4

1 回答 1

17

看这里：

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

试试这个：-

这是我的 html 代码

<!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 2012-08-26T16:57:08.700 回答