我想使用ajax和php实现文件上传。我有一个表单输入标签。我想要输入标签的 onchange 事件，文件将被上传到服务器，我将在 javascript 的变量中获取文件的路径！所以，我想保持在同一页面上并上传文件，在 javascript 变量中获取文件路径。
任何伪代码、示例或教程将不胜感激。
问问题
5185 次
3 回答
5
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<head>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><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 2012-08-02T12:07:18.447 回答
1
这是一种方法，也是我做到的。使用 XHR。在我们说话的时候，我已经启动并运行了
通过 AJAX 和 jQuery 使用 HTML5 文件上传
http://dev.w3.org/2006/webapi/FileAPI/#FileReader-interface
$(':file').change(function(){
var file = this.files[0];
name = file.name;
size = file.size;
type = file.type;
if(file.name.length < 1) {
}
else if(file.size > 100000) {
alert("File is to big");
}
else if(file.type != 'image/png' && file.type != 'image/jpg' && !file.type != 'image/gif' && file.type != 'image/jpeg' ) {
alert("File doesnt match png, jpg or gif");
}
else {
$(':submit').click(function(){
var formData = new FormData($('*formId*')[0]);
$.ajax({
url: 'script', //server script to process data
type: 'POST',
xhr: function() { // custom xhr
myXhr = $.ajaxSettings.xhr();
if(myXhr.upload){ // if upload property exists
myXhr.upload.addEventListener('progress', progressHandlingFunction, false); // progressbar
}
return myXhr;
},
//Ajax events
success: completeHandler = function(data) {
/*
* workaround for crome browser // delete the fakepath
*/
if(navigator.userAgent.indexOf('Chrome')) {
var catchFile = $(":file").val().replace(/C:\\fakepath\\/i, '');
}
else {
var catchFile = $(":file").val();
}
var writeFile = $(":file");
writeFile.html(writer(catchFile));
$("*setIdOfImageInHiddenInput*").val(data.logo_id);
},
error: errorHandler = function() {
alert("Något gick fel");
},
// Form data
data: formData,
//Options to tell JQuery not to process data or worry about content-type
cache: false,
contentType: false,
processData: false
}, 'json');
});
}
});
于 2012-08-02T11:25:46.170 回答
0
于 2012-08-02T11:20:46.477 回答