这是我的观点
<form action="@Url.Action("Index", "Home")" method="post" enctype="multipart/form-data">
@Html.AntiForgeryToken()
<label for="file">Filename:</label>
<input type="file" name="files" id="files" />
<input type="submit" name="submit" value="Upload" />
</form>
这是我的控制器
[HttpPost]
[ValidateAntiForgeryToken]
public ActionResult Index(IEnumerable<HttpPostedFileBase> files)
{
if (files != null)
{
foreach (var file in files)
{
try
{
if (file != null && file.ContentLength > 0)
{
var fileName = file.FileName;
var path = Path.Combine(Server.MapPath(@"\Upload"), fileName);
file.SaveAs(path);
ViewBag.Message = "File uploaded successfully";
}
}
catch (Exception ex)
{
ViewBag.Message = "ERROR:" + ex.Message.ToString();
}
}
}
return View();
}
问题是HttpPostedFileBase文件始终为null。我找不到问题。
答案 0 :(得分:1)
以下是如何使用表单onsubmit方法
的示例
您的HTML部分
<form id="formTest" method="post" enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="files" id="files" />
<input type="submit" name="submit" value="Upload" />
</form>
的 脚本 强>
<script type="text/javascript">
var form = document.getElementById('formTest').onsubmit = function (e) {
e.preventDefault();
var formdata = new FormData(); //FormData object
var fileInput = document.getElementById('files');
if (fileInput != "" && fileInput.files.length > 0) {
//Iterating through each files selected in fileInput
for (i = 0; i < fileInput.files.length; i++) {
//Appending each file to FormData object
formdata.append(fileInput.files[i].name, fileInput.files[i]);
}
//Creating an XMLHttpRequest and sending
var xhr = new XMLHttpRequest();
var url = '@Url.Action("Index","Home")';
xhr.open('POST', url);
xhr.send(formdata);
xhr.onreadystatechange = function () {
if (xhr.readyState == 4 && xhr.status == 200) {
var result = xhr.responseText;
}
}
return false;
}
}
</script>
的 C# 强>
public ActionResult Index()
{
if (Request.Files.Count > 0)
{
var file = Request.Files[0];
if (file != null && file.ContentLength > 0)
{
var fileName = Path.GetFileName(file.FileName);
var path = Path.Combine(Server.MapPath("~/Images/"), fileName);
file.SaveAs(path);
}
return View();
}
}
答案 1 :(得分:0)
您还可以使用Request.Files
处理文件:
public ActionResult Index()
{
if (Request.Files.Count > 0)
{
var file = Request.Files[0];
if (file != null && file.ContentLength > 0)
{
var fileName = Path.GetFileName(file.FileName);
var path = Path.Combine(Server.MapPath("~/Images/"), fileName);
file.SaveAs(path);
}
}
}
对于您的第二个问题,请尝试在Html.BeginForm
而不是form
之间使用它:
@using (Html.BeginForm("Index", "Home", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
@Html.AntiForgeryToken()
<label>Filename:</label>
<input type="file" name="file1"/>
<input type="submit" name="submit" value="Upload" />
}