HTTPPostedFileBase 在控制器中始终为 null。请审查并让我知道我错在哪里。
控制器开机自检方法
public ActionResult EditProfile(Contact model, HttpPostedFileBase picture, string currentPassword, string CurrentPasswordQ, string newPassword, string loginPwd, string currentPinQ, string newPin, int? selectedQuestion, string answer, bool pwdChange = false, bool questionChange = false, bool pinChange = false)
{
我的表单页眉
@using (Html.BeginForm("EditProfile", "CompanyAdmin", FormMethod.Post, new { enctype = "multipart/form-data", @data_ajax = "false" }))
{
和我的文件输入
<tr>
<td class="label_form_div">
<label>Profile Picture</label>
</td>
<td>
<input type="file" name="picture" />
</td>
</tr>
请查看,看看是否可以找到问题所在。 谢谢
看来您的代码是正确的,无论如何请在控制器中尝试这种方式。它可能会对您有所帮助:
public ActionResult EditProfile(Contact model, string currentPassword, string CurrentPasswordQ, string newPassword, string loginPwd, string currentPinQ, string newPin, int? selectedQuestion, string answer, bool pwdChange = false, bool questionChange = false, bool pinChange = false)
{
if (Request.Files != null && Request.Files.Count > 0)
{
HttpPostedFileBase file = Request.Files[0];
if (file != null && file.ContentLength > 0)
{
//other logic
}
}
}
来源: https://stackoverflow.com/a/32219011/3397630
请按照以下步骤操作。
步骤 :- 1
@using (Html.BeginForm("FileUpload", "Home", FormMethod.Post,
new { enctype = "multipart/form-data" }))
{
<label for="file">Upload Image:</label>
<input type="file" name="file" id="file" style="width: 100%;" />
<input type="submit" value="Upload" class="submit" />
}
步骤 :- 2
public ActionResult FileUpload(HttpPostedFileBase file)
{
if (file != null)
{
string pic = System.IO.Path.GetFileName(file.FileName);
string path = System.IO.Path.Combine(
Server.MapPath("~/images/profile"), pic);
// file is uploaded
file.SaveAs(path);
// save the image path path to the database or you can send image
// directly to database
// in-case if you want to store byte[] ie. for DB
using (MemoryStream ms = new MemoryStream())
{
file.InputStream.CopyTo(ms);
byte[] array = ms.GetBuffer();
}
}
// after successfully uploading redirect the user
return RedirectToAction("actionname", "controller name");
}