我正在使用jQuery和Ajax为我的表单提交数据和文件,但我不知道如何在一个表单中发送数据和文件?

我目前做的几乎相同的两个方法,但数据收集到数组的方式是不同的,数据使用.serialize();但是文件使用= new FormData($(this)[0]);

是否有可能结合这两种方法,以便通过Ajax以一种形式上传文件和数据?

数据jQuery, Ajax和html

$("form#data").submit(function(){

    var formData = $(this).serialize();

    $.ajax({
        url: window.location.pathname,
        type: 'POST',
        data: formData,
        async: false,
        success: function (data) {
            alert(data)
        },
        cache: false,
        contentType: false,
        processData: false
    });

    return false;
});

<form id="data" method="post">
    <input type="text" name="first" value="Bob" />
    <input type="text" name="middle" value="James" />
    <input type="text" name="last" value="Smith" />
    <button>Submit</button>
</form>

jQuery, Ajax和html文件

$("form#files").submit(function(){

    var formData = new FormData($(this)[0]);

    $.ajax({
        url: window.location.pathname,
        type: 'POST',
        data: formData,
        async: false,
        success: function (data) {
            alert(data)
        },
        cache: false,
        contentType: false,
        processData: false
    });

    return false;
});

<form id="files" method="post" enctype="multipart/form-data">
    <input name="image" type="file" />
    <button>Submit</button>
</form>

如何结合上述内容,以便通过Ajax以一种形式发送数据和文件?

我的目标是能够发送所有这些表单在一个帖子与Ajax,这是可能的吗?

<form id="datafiles" method="post" enctype="multipart/form-data">
    <input type="text" name="first" value="Bob" />
    <input type="text" name="middle" value="James" />
    <input type="text" name="last" value="Smith" />
    <input name="image" type="file" />
    <button>Submit</button>
</form>

当前回答

<form id="form" method="post" action="otherpage.php" enctype="multipart/form-data">
    <input type="text" name="first" value="Bob" />
    <input type="text" name="middle" value="James" />
    <input type="text" name="last" value="Smith" />
    <input name="image" type="file" />
    <button type='button' id='submit_btn'>Submit</button>
</form>

<script>
$(document).on("click", "#submit_btn", function (e) {
    //Prevent Instant Click  
    e.preventDefault();
    // Create an FormData object 
    var formData = $("#form").submit(function (e) {
        return;
    });
    //formData[0] contain form data only 
    // You can directly make object via using form id but it require all ajax operation inside $("form").submit(<!-- Ajax Here   -->)
    var formData = new FormData(formData[0]);
    $.ajax({
        url: $('#form').attr('action'),
        type: 'POST',
        data: formData,
        success: function (response) {
            console.log(response);
        },
        contentType: false,
        processData: false,
        cache: false
    });
    return false;
});
</script>

/ / / / / otherpage.php

<?php
    print_r($_FILES);
?>

其他回答

一个简单但更有效的方法: new FormData()本身就像一个容器(或袋子)。你可以把所有attr或文件本身。 你唯一需要附加属性file fileName eg:

let formData = new FormData()
formData.append('input', input.files[0], input.files[0].name)

然后传入AJAX request。例如:

    let formData = new FormData()
    var d = $('#fileid')[0].files[0]

    formData.append('fileid', d);
    formData.append('inputname', value);

    $.ajax({
        url: '/yourroute',
        method: 'POST',
        contentType: false,
        processData: false,
        data: formData,
        success: function(res){
            console.log('successfully')
        },
        error: function(){
            console.log('error')
        }
    })

你可以用FormData附加n个文件或数据。

如果你在Node.js中使用AJAX请求从Script.js文件到Route文件,请注意使用 要求的事情。主体用于访问数据(即文本) 要求的事情。访问文件的文件(如图像、视频等)

我在ASP中也遇到了同样的问题。Net MVC和HttpPostedFilebase,而不是在提交上使用表单,我需要使用按钮,点击,我需要做一些事情,然后如果都OK,提交表单,这是我如何让它工作

$(".submitbtn").on("click", function(e) {

    var form = $("#Form");

    // you can't pass Jquery form it has to be javascript form object
    var formData = new FormData(form[0]);

    //if you only need to upload files then 
    //Grab the File upload control and append each file manually to FormData
    //var files = form.find("#fileupload")[0].files;

    //$.each(files, function() {
    //  var file = $(this);
    //  formData.append(file[0].name, file[0]);
    //});

    if ($(form).valid()) {
        $.ajax({
            type: "POST",
            url: $(form).prop("action"),
            //dataType: 'json', //not sure but works for me without this
            data: formData,
            contentType: false, //this is requireded please see answers above
            processData: false, //this is requireded please see answers above
            //cache: false, //not sure but works for me without this
            error   : ErrorHandler,
            success : successHandler
        });
    }
});

这将比正确地填充你的MVC模型,请确保在你的模型,属性HttpPostedFileBase[]有相同的名称作为html中的输入控件的名称。

<input id="fileupload" type="file" name="UploadedFiles" multiple>

public class MyViewModel
{
    public HttpPostedFileBase[] UploadedFiles { get; set; }
}

另一种选择是使用iframe并将表单的目标设置为iframe。

你可以试试这个(它使用jQuery):

function ajax_form($form, on_complete)
{
    var iframe;

    if (!$form.attr('target'))
    {
        //create a unique iframe for the form
        iframe = $("<iframe></iframe>").attr('name', 'ajax_form_' + Math.floor(Math.random() * 999999)).hide().appendTo($('body'));
        $form.attr('target', iframe.attr('name'));
    }

    if (on_complete)
    {
        iframe = iframe || $('iframe[name="' + $form.attr('target') + '"]');
        iframe.load(function ()
        {
            //get the server response
            var response = iframe.contents().find('body').text();
            on_complete(response);
        });
    }
}

它适用于所有浏览器,您不需要序列化或准备数据。 一个缺点是你无法监控进程。

另外,至少对于chrome浏览器,请求不会出现在开发工具的“xhr”标签下,而是在“doc”下。

——DOT NET CORE MVC实现的解决方案—— 在看这个问题的时候,我想我应该正确的。net CORE实现,因为这个问题不是特定于任何后端语言。 这是独立实现的例子。 目标:-提交包括文件在内的表单字段,以及我们如何在后端单个模型中获得数据

HTML代码/视图代码- Views/Home/Index.cshtml

@{
    ViewData["Title"] = "Home Page";
}

<input type="file" id="FileUpload1" multiple />
<div>
    <label>Enter First Name :</label>
    <input type="text" id="nameText" maxlength="50" />

</div>
<input type="button" id="btnUpload" value="Submit Form with Files" />

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script>
    $(document).ready(function () {
        $('#btnUpload').click(function () {

            // Checking whether FormData is available in browser
            if (window.FormData !== undefined) {

                var fileUpload = $("#FileUpload1").get(0);
                var files = fileUpload.files;

                // Create FormData object
                var fileData = new FormData();

                // Looping over all files and add it to FormData object
                for (var i = 0; i < files.length; i++) {
                    fileData.append("files", files[i]);
                }
                // Adding one more key to FormData object
                fileData.append('FirstName', $("#nameText").val());

                $.ajax({
                    url: '/Home/UploadFiles',
                    type: "POST",
                    contentType: false, // Not to set any content header
                    processData: false, // Not to process data
                    data: fileData,
                    success: function (result) {
                        alert(result);
                    },
                    error: function (err) {
                        alert(err.statusText);
                    }
                });
            } else {
                alert("FormData is not supported.");
            }
        });
    });
</script>  

后台代码/控制器动作方法Controllers/ homeconcontroller .cs

public class HomeController : Controller
{
    private readonly ILogger<HomeController> _logger;
    private readonly IWebHostEnvironment _environment;

    public HomeController(ILogger<HomeController> logger, IWebHostEnvironment environment)
    {
        _logger = logger;
        _environment = environment;
    }

    public IActionResult Index()
    {
        return View();
    }

    public IActionResult Privacy()
    {
        return View();
    }

    [HttpPost]
    public async Task<IActionResult> UploadFiles(MyForm myForm)
    {
        var files = myForm.Files;
        // First Name 
        string name = myForm.FirstName;

        // check All files
        foreach (IFormFile source in files)
        {
            string filename = ContentDispositionHeaderValue.Parse(source.ContentDisposition).FileName.Trim('"');

            filename = this.EnsureCorrectFilename(filename);
            string fileWithPath = this.GetPathAndFilename(filename);
            // Create directory if not exist
            Directory.CreateDirectory(Path.GetDirectoryName(fileWithPath));

            using (FileStream output = System.IO.File.Create(fileWithPath))
                await source.CopyToAsync(output);
        }

        return Ok("Success");
    }

    [ResponseCache(Duration = 0, Location = ResponseCacheLocation.None, NoStore = true)]
    public IActionResult Error()
    {
        return View(new ErrorViewModel { RequestId = Activity.Current?.Id ?? HttpContext.TraceIdentifier });
    }

    public class MyForm
    {
        public string FirstName { get; set; }
        public IList<IFormFile> Files { get; set; }
    }

    private string EnsureCorrectFilename(string filename)
    {
        if (filename.Contains("\\"))
            filename = filename.Substring(filename.LastIndexOf("\\") + 1);

        return filename;
    }

    private string GetPathAndFilename(string filename)
    {
        return Path.Combine(_environment.ContentRootPath, "uploadedFiles", filename);
    }
}

完整源代码回购:https://github.com/rj-learning/DotNetCoreFileUpload

或更短:

$("form#data").submit(function() {
    var formData = new FormData(this);
    $.post($(this).attr("action"), formData, function() {
        // success    
    });
    return false;
});