您可以使用JavaScript或jQuery进行异步多文件上传，而无需使用任何插件。您还可以在进度控件中显示文件上载的实时进度。我遇到了两个不错的链接-

带有进度条的基于ASP.NET Web表单的多文件上载功能jQuery中基于ASP.NET MVC的多文件上载

服务器端语言是C#，但您可以进行一些修改，使其与其他语言（如PHP）一起使用。

文件上载ASP.NET核心MVC：

在html中的View create file upload控件中：

<form method="post" asp-action="Add" enctype="multipart/form-data">
    <input type="file" multiple name="mediaUpload" />
    <button type="submit">Submit</button>
</form>

现在在控制器中创建动作方法：

[HttpPost]
public async Task<IActionResult> Add(IFormFile[] mediaUpload)
{
    //looping through all the files
    foreach (IFormFile file in mediaUpload)
    {
        //saving the files
        string path = Path.Combine(hostingEnvironment.WebRootPath, "some-folder-path"); 
        using (var stream = new FileStream(path, FileMode.Create))
        {
            await file.CopyToAsync(stream);
        }
    }
}

hostingEnvironment变量的类型为IHostingEnvironment，可以使用依赖注入将其注入控制器，例如：

private IHostingEnvironment hostingEnvironment;
public MediaController(IHostingEnvironment environment)
{
    hostingEnvironment = environment;
}

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。

注意：此答案已过时，现在可以使用XHR上载文件。

不能使用XMLHttpRequest（Ajax）上载文件。可以使用iframe或Flash模拟效果。优秀的jQuery表单插件，通过iframe发布文件以获得效果。

jQueryUploadify是我以前用来上传文件的另一个好插件。JavaScript代码如下所示：code。但是，新版本在Internet Explorer中不起作用。

$('#file_upload').uploadify({
    'swf': '/public/js/uploadify.swf',
    'uploader': '/Upload.ashx?formGuid=' + $('#formGuid').val(),
    'cancelImg': '/public/images/uploadify-cancel.png',
    'multi': true,
    'onQueueComplete': function (queueData) {
        // ...
    },
    'onUploadStart': function (file) {
        // ...
    }
});

我做了大量的搜索，我找到了另一种不用任何插件、只使用ajax上传文件的解决方案。解决方案如下：

$(document).ready(function () {
    $('#btn_Upload').live('click', AjaxFileUpload);
});

function AjaxFileUpload() {
    var fileInput = document.getElementById("#Uploader");
    var file = fileInput.files[0];
    var fd = new FormData();
    fd.append("files", file);
    var xhr = new XMLHttpRequest();
    xhr.open("POST", 'Uploader.ashx');
    xhr.onreadystatechange = function () {
        if (xhr.readyState == 4) {
             alert('success');
        }
        else if (uploadResult == 'success')
            alert('error');
    };
    xhr.send(fd);
}

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

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()
        )
    ));
}