我正在开发一个 ASP.NET MVC 网站，其中有一个表单，可以使用表单标记上的 multipart/form data enctype 选项上传文件。像这样：
<form enctype="multipart/form-data" method="post" action='<%= Url.Action("Post","Entries",new {id=ViewData.Model.MemberDetermination.DeterminationMemberID}) %>'>
我该如何编写代码，以实现 ASP.NET MVC Ajax 表单提交呢？
例如：
@using (Ajax.BeginForm(YourMethod, YourController, new { id= Model.Id }, new AjaxOptions {//needed options }, new { enctype = "multipart/form-data" }))
{
<input type="file" id="image" name="image" />
<input type="submit" value="Modify" />
}
步骤2：拦截请求并将其发送到服务器
<script type="text/javascript">
$(function() {
$("#form0").submit(function(event) {
var dataString;
event.preventDefault();
var action = $("#form0").attr("action");
if ($("#form0").attr("enctype") == "multipart/form-data") {
//this only works in some browsers.
//purpose? to submit files over ajax. because screw iframes.
//also, we need to call .get(0) on the jQuery element to turn it into a regular DOM element so that FormData can use it.
dataString = new FormData($("#form0").get(0));
contentType = false;
processData = false;
} else {
// regular form, do your own thing if you need it
}
$.ajax({
type: "POST",
url: action,
data: dataString,
dataType: "json", //change to your own, else read my note above on enabling the JsonValueProviderFactory in MVC
contentType: contentType,
processData: processData,
success: function(data) {
//BTW, data is one of the worst names you can make for a variable
//handleSuccessFunctionHERE(data);
},
error: function(jqXHR, textStatus, errorThrown) {
//do your own thing
alert("fail");
}
});
}); //end .submit()
});
</script>
步骤3：因为您进行了ajax调用，所以您可能想要替换一些multipart/form-data的图像或其他内容。
例如：
handleSuccessFunctionHERE(data)
{
$.ajax({
type: "GET",
url: "/Profile/GetImageModified",
data: {},
dataType: "text",
success: function (MSG) {
$("#imageUploaded").attr("src", "data:image/gif;base64,"+msg);
},
error: function (msg) {
alert(msg);
}
});
}
MSG变量是一个base64加密的字符串。在我的情况下，它是图像的源。
通过这种方式，我成功地更改了个人资料图片，之后图片立即更新。
同时，请确保在Application_Start（global.asax）中添加以下代码：
ValueProviderFactories.Factories.Add(new JsonValueProviderFactory());
非常好用吧？
P.S.: 这个解决方案可行，所以不要犹豫，可以询问更多细节。
我遇到了这个小技巧，它可以很好地解决这个问题。
window.addEventListener("submit", function (e) {
var form = e.target;
if (form.getAttribute("enctype") === "multipart/form-data") {
if (form.dataset.ajax) {
e.preventDefault();
e.stopImmediatePropagation();
var xhr = new XMLHttpRequest();
xhr.open(form.method, form.action);
xhr.onreadystatechange = function () {
if (xhr.readyState == 4 && xhr.status == 200) {
if (form.dataset.ajaxUpdate) {
var updateTarget = document.querySelector(form.dataset.ajaxUpdate);
if (updateTarget) {
updateTarget.innerHTML = xhr.responseText;
}
}
}
};
xhr.send(new FormData(form));
}
}
}, true);
Use:
AjaxHelper.BeginForm("Post", "Entries", new {id=ViewData.Model.MemberDetermination.DeterminationMemberID}, new AjaxOptions(){/*some options*/}, new {enctype="multipart/form-data"})
但是在第二种情况下，我不确定它是否有效。
以下是我使用并且可行的代码！！这是@James 'Fluffy' Burton方案的复制。我仅仅对他的答案进行了改进，以便新手在MVC上能够快速理解后果。
以下是我的视图：
@using (Ajax.BeginForm("FileUploader", null, new AjaxOptions { HttpMethod = "POST", UpdateTargetId = "AjaxUpdatePanel" }, new { enctype = "multipart/form-data", id = "frmUploader" })){
<div id="AjaxUpdatePanel">
<div class="form-group">
<input type="file" id="dataFile" name="upload" />
</div>
<div class="form-group">
<input type="submit" value="Upload" class="btn btn-default" id="btnUpload"/>
</div>
</div>}
<script>
window.addEventListener("submit", function (e) {
var form = e.target;
if (form.getAttribute("enctype") === "multipart/form-data") {
if (form.dataset.ajax) {
e.preventDefault();
e.stopImmediatePropagation();
var xhr = new XMLHttpRequest();
xhr.open(form.method, form.action);
xhr.onreadystatechange = function () {
if (xhr.readyState == 4 && xhr.status == 200) {
if (form.dataset.ajaxUpdate) {
var updateTarget = document.querySelector(form.dataset.ajaxUpdate);
if (updateTarget) {
updateTarget.innerHTML = xhr.responseText;
}
}
}
};
xhr.send(new FormData(form));
}
}
}, true);
[HttpPost]
public JsonResult FileUploader(HttpPostedFileBase upload)
{
if (ModelState.IsValid)
{
if (upload != null && upload.ContentLength > 0)
{
if (upload.FileName.EndsWith(".csv"))
{
Stream stream = upload.InputStream;
DataTable csvTable = new DataTable();
using (CsvReader csvReader = new CsvReader(new StreamReader(stream), true))
{
csvTable.Load(csvReader);
}
}
else
{
return Json(new { dataerror = true, errormsg = "This file format is not supported" });
}
}
else
{
return Json(new { dataerror = true, errormsg = "Please Upload Your file" });
}
}
return Json(new { result = true });
}
对于那些在MVC中使用@Ajax.BeginForm进行multipart enctypes / file uploads仍有问题的人
在由@Ajax.BeginForm助手生成的表单元素上运行“检查元素”工具会发现，该助手不可思议地覆盖了指定的控制器参数。如果您为部分Postback实现了一个单独的控制器，那么就会出现这种情况。
修复问题的快速方法是将html action属性值明确指定为/<yourcontrollername>/<youractionname>。
@using (Ajax.BeginForm("", "", new AjaxOptions() { HttpMethod = "POST", UpdateTargetId = "<TargetElementId>", InsertionMode = InsertionMode.Replace }, new { enctype = "multipart/form-data", action = "/<Controller>/<Action>" }))
实际上我自己回答了这个问题...
<% using (Ajax.BeginForm("Post", "Entries", new { id = ViewData.Model.MemberDetermination.DeterminationMemberID }, new AjaxOptions { UpdateTargetId = "dc_goal_placeholder" }, new { enctype = "multipart/form-data" }))
OnSuccess AjaxOption并/或在控制器中使用Request.IsAjaxRequest()来检查请求类型，即...@using (Ajax.BeginForm("FileUploader", null, new AjaxOptions { HttpMethod = "POST", UpdateTargetId = "elementToUpdate", OnSuccess = "mySuccessFuntion(returnedData)", OnFailure = "myFailureFuntion(returnedData)"}, new { enctype = "multipart/form-data" }))
然后您可以使用以下代码（我修改了@James 'Fluffy' Burton的答案）。如果可能的话，这也将把响应文本转换为JSON对象（如果您不想要，可以省略此步骤）。
<script>
if(typeof window.FormData === 'undefined') {
alert("This browser doesn't support HTML5 file uploads!");
}
window.addEventListener("submit", function (e) {
var form = e.target;
if (form.getAttribute("enctype") === "multipart/form-data") {
if (form.dataset.ajax) {
e.preventDefault();
e.stopImmediatePropagation();
var xhr = new XMLHttpRequest();
xhr.open(form.method, form.action);
xhr.setRequestHeader("x-Requested-With", "XMLHttpRequest"); // this allows 'Request.IsAjaxRequest()' to work in the controller code
xhr.onreadystatechange = function () {
if (xhr.readyState === XMLHttpRequest.DONE && xhr.status === 200) {
var returnedData; //this variable needs to be named the same as the parameter in the function call specified for the AjaxOptions.OnSuccess
try {
returnedData = JSON.parse(xhr.responseText); //I also want my returned data to be parsed if it is a JSON object
}catch(e){
returnedData = xhr.responseText;
}
if (form.dataset.ajaxSuccess) {
eval(form.dataset.ajaxSuccess); //converts function text to real function and executes (not very safe though)
}
else if (form.dataset.ajaxFailure) {
eval(form.dataset.ajaxFailure);
}
if (form.dataset.ajaxUpdate) {
var updateTarget = document.querySelector(form.dataset.ajaxUpdate);
if (updateTarget) {
updateTarget.innerHTML = data;
}
}
}
};
xhr.send(new FormData(form));
}
}
}, true);
</script>
注意：我使用javascript函数eval()将字符串转换为函数……如果有更好的解决方案，请留言评论。
我还使用了JQuery JSON.parse()，因此这不是一个纯javascript解决方案，但它并不是脚本运行所必须的，因此可以删除。
我将Brad Larson的答案与Amirhossein Mehrvarzi的答案结合起来，因为Brad的答案没有提供任何处理响应的方法，而Amirhossein的答案会导致2次postback。 在发送之前，我只是添加了($('#formBacklink').valid())来调用模型验证。
window.addEventListener("submit", function (e) {
if ($('#formBacklink').valid()) {
var form = e.target;
if (form.getAttribute("enctype") === "multipart/form-data") {
if (form.dataset.ajax) {
e.preventDefault();
e.stopImmediatePropagation();
var dataString;
event.preventDefault();
var action = $("#formBacklink").attr("action");
if ($("#formBacklink").attr("enctype") == "multipart/form-data") {
//this only works in some browsers.
//purpose? to submit files over ajax. because screw iframes.
//also, we need to call .get(0) on the jQuery element to turn it into a regular DOM element so that FormData can use it.
dataString = new FormData($("#formBacklink").get(0));
contentType = false;
processData = false;
} else {
// regular form, do your own thing if you need it
}
$.ajax({
type: "POST",
url: action,
data: dataString,
dataType: "json", //change to your own, else read my note above on enabling the JsonValueProviderFactory in MVC
contentType: contentType,
processData: processData,
success: function (data) {
//BTW, data is one of the worst names you can make for a variable
//handleSuccessFunctionHERE(data);
},
error: function (jqXHR, textStatus, errorThrown) {
//do your own thing
}
});
}
}
}
}, true);
Ajax.BegineForm() 适用于多部分表单数据，下面是相同功能的工作代码示例：
视图：
@using(Ajax.BeginForm("UploadFile","MyPOC",
new AjaxOptions {
HttpMethod = "POST"
},
new
{
enctype = "multipart/form-data"
}))
{
<input type="file" name="files" id="fileUploaderControl" />
<input type="submit" value="Upload" id="btnFileUpload" />
}
控制器操作方法：
public void UploadFile(IEnumerable<HttpPostedFileBase> files)
{
HttpPostedFileBase file = files.FirstOrDefault(); //Attach a debugger here and check whether you are getting your file on server side or null.
if (file != null && file.ContentLength > 0)
{
//Do other validations before saving the file
//Save File
file.SaveAs(path);
}
}
顺便提一下，确保文件上传控件的“name”属性和传递给Action方法UploadFile()的参数名称相同（在本例中为“files”）。
return RedirectToAction(YourAction); }- Demian Flavius