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
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()
)
));
}
其他回答
您可以通过JavaScript使用更新的Fetch API。这样地：
function uploadButtonCLicked(){
var input = document.querySelector('input[type="file"]')
fetch('/url', {
method: 'POST',
body: input.files[0]
}).then(res => res.json()) // you can do something with response
.catch(error => console.error('Error:', error))
.then(response => console.log('Success:', response));
}
优点：所有现代浏览器都支持Fetch API，因此您不必导入任何内容。此外，请注意，fetch（）返回Promise，然后使用.then（..代码处理响应..）异步处理Promise。
你也可以考虑使用类似的https://uppy.io.
它可以在不离开页面的情况下进行文件上传，并提供一些奖励，如拖放、在浏览器崩溃/网络不稳定的情况下恢复上传，以及从例如Instagram导入。它是开源的，不依赖于jQuery/React/Angular/Vue，但可以与它一起使用。免责声明：作为它的创建者，我有偏见；）
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)
{
[...]
}
Try
异步函数saveFile（）{let formData=新formData（）；formData.append（“file”，file.files[0]）；wait-fetch（'addFile.do'，｛method:“POST”，body:formData｝）；alert（“上传的数据：”）；}<span>文件</span><input type=“file”id=“file”name=“file”size=“10”/><input type=“button”value=“Upload”onclick=“saveFile（）”/>
content-type='multipart/form-data'由浏览器自动设置，文件名也自动添加到文件名FormData参数中（服务器可以轻松读取）。下面是一个更为成熟的错误处理和json添加示例
异步函数saveFile（inp）{让用户＝{name:'john'，年龄：34}；let formData=新formData（）；let photo=inp.files[0]；formData.append（“照片”，照片）；formData.append（“用户”，JSON.stringify（用户））；尝试{let r=等待获取（'/upload/image'，｛method:“POST”，body:formData｝）；console.log（'HTTP响应代码：'，r.status）；警报（“成功”）；}捕获（e）{console.log（'休斯顿我们有问题…：'，e）；}}<input-type=“file”onchange=“saveFile（this）”><br><br>在选择文件之前，打开chrome控制台>网络选项卡以查看请求详细信息。<br><br><small>因为在本例中，我们将请求发送到https://stacksnippets.net/upload/image响应代码当然是404</小>
如果使用承诺使用哪种ajax，并检查文件是否有效并保存在后端，那么您可以在用户浏览页面时在前面使用一些动画。
您甚至可以使用递归方法使其并行上传或堆叠
