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
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；）
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
});
});
使用HTML5，您可以使用Ajax和jQuery上传文件。不仅如此，您还可以进行文件验证（名称、大小和MIME类型）或使用HTML5进度标记（或div）处理进度事件。最近我不得不做一个文件上传器，但我不想使用Flash或Iframes或插件，经过一些研究，我想出了解决方案。
HTML：
<form enctype="multipart/form-data">
<input name="file" type="file" />
<input type="button" value="Upload" />
</form>
<progress></progress>
首先，如果需要，可以进行一些验证。例如，在文件的.on（'change'）事件中：
$(':file').on('change', function () {
var file = this.files[0];
if (file.size > 1024) {
alert('max upload size is 1k');
}
// Also see .name, .type
});
现在，单击按钮即可提交$.ajax（）：
$(':button').on('click', function () {
$.ajax({
// Your server script to process the upload
url: 'upload.php',
type: 'POST',
// Form data
data: new FormData($('form')[0]),
// Tell jQuery not to process data or worry about content-type
// You *must* include these options!
cache: false,
contentType: false,
processData: false,
// Custom XMLHttpRequest
xhr: function () {
var myXhr = $.ajaxSettings.xhr();
if (myXhr.upload) {
// For handling the progress of the upload
myXhr.upload.addEventListener('progress', function (e) {
if (e.lengthComputable) {
$('progress').attr({
value: e.loaded,
max: e.total,
});
}
}, false);
}
return myXhr;
}
});
});
正如你所看到的，通过HTML5（以及一些研究）文件上传不仅变得可能，而且非常容易。尝试使用Google Chrome，因为示例中的一些HTML5组件在每个浏览器中都不可用。
为此，我建议使用Fine Uploader插件。您的JavaScript代码为：
$(document).ready(function() {
$("#uploadbutton").jsupload({
action: "addFile.do",
onComplete: function(response){
alert( "server response: " + response);
}
});
});
在此处查找“异步处理文件的上载过程”：https://developer.mozilla.org/en-US/docs/Using_files_from_web_applications
链接中的示例
<?php
if (isset($_FILES['myFile'])) {
// Example:
move_uploaded_file($_FILES['myFile']['tmp_name'], "uploads/" . $_FILES['myFile']['name']);
exit;
}
?><!DOCTYPE html>
<html>
<head>
<title>dnd binary upload</title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<script type="text/javascript">
function sendFile(file) {
var uri = "/index.php";
var xhr = new XMLHttpRequest();
var fd = new FormData();
xhr.open("POST", uri, true);
xhr.onreadystatechange = function() {
if (xhr.readyState == 4 && xhr.status == 200) {
// Handle response.
alert(xhr.responseText); // handle response.
}
};
fd.append('myFile', file);
// Initiate a multipart/form-data upload
xhr.send(fd);
}
window.onload = function() {
var dropzone = document.getElementById("dropzone");
dropzone.ondragover = dropzone.ondragenter = function(event) {
event.stopPropagation();
event.preventDefault();
}
dropzone.ondrop = function(event) {
event.stopPropagation();
event.preventDefault();
var filesArray = event.dataTransfer.files;
for (var i=0; i<filesArray.length; i++) {
sendFile(filesArray[i]);
}
}
}
</script>
</head>
<body>
<div>
<div id="dropzone" style="margin:30px; width:500px; height:300px; border:1px dotted grey;">Drag & drop your file here...</div>
</div>
</body>
</html>
