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如何查找已上傳了多少數據？

2012-09-05 55 views

回答

1

試試這個： -

演示網址： -

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

你可以從這個網址下載jQuery的文件，並添加HTML標籤

http://ajax.googleapis.com/ajax/libs/jquery/1.7/jquery.js

http://malsup.github.com/jquery.form.js

試試這個：

這是我的HTML標記：

<!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代碼：

<?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!"; 
} 
?>

來源

2012-09-05 16:00:57

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