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
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');
}
});
其他回答
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()
)
));
}
没有Jquery的现代方法是，当用户选择一个文件时，使用从＜input type＝“file”＞返回的FileList对象，然后使用Fetch发布包裹在FormData对象周围的FileList。
// The input DOM element // <input type="file">
const inputElement = document.querySelector('input[type=file]');
// Listen for a file submit from user
inputElement.addEventListener('change', () => {
const data = new FormData();
data.append('file', inputElement.files[0]);
data.append('imageName', 'flower');
// You can then post it to your server.
// Fetch can accept an object of type FormData on its body
fetch('/uploadImage', {
method: 'POST',
body: data
});
});
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
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>
