我想用jQuery异步上传文件。
$（文档）.ready（函数（）{$（“#uploadbutton”）.click（函数（）{var filename=$（“#file”）.val（）；$.ajax美元({类型：“POST”，url:“addFile.do”，enctype:'多部分/表单数据'，数据：{文件：文件名},成功：函数（）{alert（“上传的数据：”）；}});});});<script src=“https://cdnjs.cloudflare.com/ajax/libs/jquery/2.2.0/jquery.min.js“></script><span>文件</span><input type=“file”id=“file”name=“file”size=“10”/><input id=“uploadbutton”type=“button”value=“Upload”/>
我只得到文件名，而不是上传文件。我可以做什么来解决这个问题？
当前回答
我一直在使用下面的脚本来上传图像，这很好。
HTML
<input id="file" type="file" name="file"/>
<div id="response"></div>
JavaScript
jQuery('document').ready(function(){
var input = document.getElementById("file");
var formdata = false;
if (window.FormData) {
formdata = new FormData();
}
input.addEventListener("change", function (evt) {
var i = 0, len = this.files.length, img, reader, file;
for ( ; i < len; i++ ) {
file = this.files[i];
if (!!file.type.match(/image.*/)) {
if ( window.FileReader ) {
reader = new FileReader();
reader.onloadend = function (e) {
//showUploadedItem(e.target.result, file.fileName);
};
reader.readAsDataURL(file);
}
if (formdata) {
formdata.append("image", file);
formdata.append("extra",'extra-data');
}
if (formdata) {
jQuery('div#response').html('<br /><img src="ajax-loader.gif"/>');
jQuery.ajax({
url: "upload.php",
type: "POST",
data: formdata,
processData: false,
contentType: false,
success: function (res) {
jQuery('div#response').html("Successfully uploaded");
}
});
}
}
else
{
alert('Not a vaild image!');
}
}
}, false);
});
解释
我使用response div显示上传动画和上传完成后的响应。
最好的部分是，当您使用此脚本时，可以随文件发送额外的数据，如id等。我在脚本中提到了额外的数据。
在PHP级别，这将作为正常的文件上传工作。额外数据可以作为$_POST数据检索。
这里你没有使用插件之类的东西。您可以根据需要更改代码。你不是盲目地在这里编码。这是任何jQuery文件上传的核心功能。实际上是Javascript。
其他回答
这个AJAX文件上传jQuery插件在某处上传文件，并传递对回调的响应。
它不依赖于特定的HTML，只需给它一个＜input-type=“file”＞它不要求服务器以任何特定方式响应使用多少文件或文件在页面上的位置无关紧要
--尽可能少地使用--
$('#one-specific-file').ajaxfileupload({
'action': '/upload.php'
});
--或者--
$('input[type="file"]').ajaxfileupload({
'action': '/upload.php',
'params': {
'extra': 'info'
},
'onComplete': function(response) {
console.log('custom handler for file:');
alert(JSON.stringify(response));
},
'onStart': function() {
if(weWantedTo) return false; // cancels upload
},
'onCancel': function() {
console.log('no file selected');
}
});
您可以使用
$(function() {
$("#file_upload_1").uploadify({
height : 30,
swf : '/uploadify/uploadify.swf',
uploader : '/uploadify/uploadify.php',
width : 120
});
});
Demo
对于PHP，请查找https://developer.hyvor.com/php/image-upload-ajax-php-mysql
HTML
<html>
<head>
<title>Image Upload with AJAX, PHP and MYSQL</title>
</head>
<body>
<form onsubmit="submitForm(event);">
<input type="file" name="image" id="image-selecter" accept="image/*">
<input type="submit" name="submit" value="Upload Image">
</form>
<div id="uploading-text" style="display:none;">Uploading...</div>
<img id="preview">
</body>
</html>
JAVASCRIPT语言
var previewImage = document.getElementById("preview"),
uploadingText = document.getElementById("uploading-text");
function submitForm(event) {
// prevent default form submission
event.preventDefault();
uploadImage();
}
function uploadImage() {
var imageSelecter = document.getElementById("image-selecter"),
file = imageSelecter.files[0];
if (!file)
return alert("Please select a file");
// clear the previous image
previewImage.removeAttribute("src");
// show uploading text
uploadingText.style.display = "block";
// create form data and append the file
var formData = new FormData();
formData.append("image", file);
// do the ajax part
var ajax = new XMLHttpRequest();
ajax.onreadystatechange = function() {
if (this.readyState === 4 && this.status === 200) {
var json = JSON.parse(this.responseText);
if (!json || json.status !== true)
return uploadError(json.error);
showImage(json.url);
}
}
ajax.open("POST", "upload.php", true);
ajax.send(formData); // send the form data
}
PHP
<?php
$host = 'localhost';
$user = 'user';
$password = 'password';
$database = 'database';
$mysqli = new mysqli($host, $user, $password, $database);
try {
if (empty($_FILES['image'])) {
throw new Exception('Image file is missing');
}
$image = $_FILES['image'];
// check INI error
if ($image['error'] !== 0) {
if ($image['error'] === 1)
throw new Exception('Max upload size exceeded');
throw new Exception('Image uploading error: INI Error');
}
// check if the file exists
if (!file_exists($image['tmp_name']))
throw new Exception('Image file is missing in the server');
$maxFileSize = 2 * 10e6; // in bytes
if ($image['size'] > $maxFileSize)
throw new Exception('Max size limit exceeded');
// check if uploaded file is an image
$imageData = getimagesize($image['tmp_name']);
if (!$imageData)
throw new Exception('Invalid image');
$mimeType = $imageData['mime'];
// validate mime type
$allowedMimeTypes = ['image/jpeg', 'image/png', 'image/gif'];
if (!in_array($mimeType, $allowedMimeTypes))
throw new Exception('Only JPEG, PNG and GIFs are allowed');
// nice! it's a valid image
// get file extension (ex: jpg, png) not (.jpg)
$fileExtention = strtolower(pathinfo($image['name'] ,PATHINFO_EXTENSION));
// create random name for your image
$fileName = round(microtime(true)) . mt_rand() . '.' . $fileExtention; // anyfilename.jpg
// Create the path starting from DOCUMENT ROOT of your website
$path = '/examples/image-upload/images/' . $fileName;
// file path in the computer - where to save it
$destination = $_SERVER['DOCUMENT_ROOT'] . $path;
if (!move_uploaded_file($image['tmp_name'], $destination))
throw new Exception('Error in moving the uploaded file');
// create the url
$protocol = stripos($_SERVER['SERVER_PROTOCOL'],'https') === true ? 'https://' : 'http://';
$domain = $protocol . $_SERVER['SERVER_NAME'];
$url = $domain . $path;
$stmt = $mysqli -> prepare('INSERT INTO image_uploads (url) VALUES (?)');
if (
$stmt &&
$stmt -> bind_param('s', $url) &&
$stmt -> execute()
) {
exit(
json_encode(
array(
'status' => true,
'url' => $url
)
)
);
} else
throw new Exception('Error in saving into the database');
} catch (Exception $e) {
exit(json_encode(
array (
'status' => false,
'error' => $e -> getMessage()
)
));
}
示例：如果使用jQuery，您可以轻松地上传文件。这是一个小而强大的jQuery插件，http://jquery.malsup.com/form/.
实例
var $bar = $('.ProgressBar');
$('.Form').ajaxForm({
dataType: 'json',
beforeSend: function(xhr) {
var percentVal = '0%';
$bar.width(percentVal);
},
uploadProgress: function(event, position, total, percentComplete) {
var percentVal = percentComplete + '%';
$bar.width(percentVal)
},
success: function(response) {
// Response
}
});
我希望这会有帮助
这是我的解决方案。
<form enctype="multipart/form-data">
<div class="form-group">
<label class="control-label col-md-2" for="apta_Description">Description</label>
<div class="col-md-10">
<input class="form-control text-box single-line" id="apta_Description" name="apta_Description" type="text" value="">
</div>
</div>
<input name="file" type="file" />
<input type="button" value="Upload" />
</form>
和js
<script>
$(':button').click(function () {
var formData = new FormData($('form')[0]);
$.ajax({
url: '@Url.Action("Save", "Home")',
type: 'POST',
success: completeHandler,
data: formData,
cache: false,
contentType: false,
processData: false
});
});
function completeHandler() {
alert(":)");
}
</script>
控制器
[HttpPost]
public ActionResult Save(string apta_Description, HttpPostedFileBase file)
{
[...]
}
