我在Asp.net Mvc上传文件时遇到问题。首先,我应该使用Ajax传递上传文件值。
在javascript中我有模型我填充它,当我用调试器检查它是否正确填充对象,但当我将此模型发送到服务器(控制器)
httpPostedfileBase值始终为null。
我在谷歌搜索它,在一些帖子中我看到我不能使用文件上传器与Ajax,但在其他我看到我可以。
但我无法修复我的代码。
有我的Javascript代码。
$(document).ready(function () {
$('#btnUploadFile').on('click', function () {
var data= new FormData();
debugger;
var files = $("#fileUpload").get(0).files;
if (files.length > 0) {
data.append("UploadedImage", files[0]);
}
var ResturantSharingViewModel =
{
Type: $("#SharingTargetType").val(),
SharingTitle: $("#SharingTitle").val(),
content: $("#Content").val(),
ItemId : $("#ItemId").val(),
Photos: files[0]
};
$.ajax({
type: 'POST',
dataType: 'json',
contentType: 'application/json',
url: '<%= Url.Action("SaveOneDatabase")%>',
data: JSON.stringify(ResturantSharingViewModel),
success: function (result) {
var rs = result;
},
error: function () {
alert("Error loading data! Please try again.");
}
});
我的控制器public virtual bool SaveOneDatabase(ResturantSharingViewModel result)
我的ResturantSharingViewModel视图模型
public class ResturantSharingViewModel
{
public Guid SharingPremiumHistoryID { get; set; }
public string SharingTitle { get; set; }
public string Content { get; set; }
public DateTime AddedDate { get; set; }
public bool IsSubmit { get; set; }
public DateTime SubmitedDate { get; set; }
public IEnumerable<SelectListItem> SharingTypes { get; set; }
public IEnumerable<SelectListItem> SharingTargetType { get; set; }
public short Type { get; set; }
public Guid ItemId { get; set; }
public HttpPostedFileBase[] Photos { get; set; }
}
我的Html元素
<form enctype="multipart/form-data">
<article>
<%--<% =Html.BeginForm("Add","PremiumSharing") %>--%>
<hgroup class="radiogroup">
<h1>ارسال خبر</h1>
<%= Html.HiddenFor(model => model.SharingPremiumHistoryID) %>
<%= Html.HiddenFor(model => model.ItemId) %>
<div class="group">
<span> ارسال به </span>
<%= Html.DropDownListFor(model => model.SharingTargetType, Model.SharingTypes) %>
</div>
</hgroup>
<div class="newseditor">
<div class="input-form">
<%= Html.LabelFor(model => model.SharingTitle, "عنوان خبر") %>
<%= Html.TextBoxFor(model => model.SharingTitle) %>
</div>
<div class="input-form">
<%= Html.LabelFor(model => model.Content, "متن خبر") %>
<%= Html.TextAreaFor(model => model.Content) %>
</div>
<div><input id="fileUpload" type="file" />
</div>
<% if (ViewBag.IsInEditMode != null && !(bool)ViewBag.IsInEditMode)
{%>
<div class="input-form">
<%= Html.CheckBox("SendToInTheCity") %> ارسال در بخش «در شهر» فیدیلیو
</div>
<%} %>
<div class="input-submit">
<button name="post" id="btnUploadFile" onclick="uploadFile()" >ارسال خبر</button>
</div>
<br />
</div>
答案 0 :(得分:14)
首先,可以使用Ajax上传,重要的是你需要在表单上设置<form enctype="multipart/form-data"></form>
告诉它你的表单有文件上传输入。然后,您需要在控制器操作中接受HttpPostedFileBase
作为输入参数。
试试这个。 jquery上传代码示例。 (主要来自How can I upload files asynchronously?)
function uploadFile(uploadId) {
var formData = new FormData($('form')[0]);
$.ajax({
url: '<%= Url.Action("SaveOneDatabase")%>',
type: 'Post',
beforeSend: function(){},
success: function(result){
},
xhr: function() { // Custom XMLHttpRequest
var myXhr = $.ajaxSettings.xhr();
if(myXhr.upload) { // Check if upload property exists
// Progress code if you want
}
return myXhr;
},
error: function(){},
data: formData,
cache: false,
contentType: false,
processData: false
});
}
HTML表单需要此属性。看到这篇文章为什么你需要它 - &gt; What does enctype='multipart/form-data' mean?
enctype="multipart/form-data"
C#
[HttpPost]
public ActionResult SaveOneDatabase(HttpPostedFileBase file)
{
}
答案 1 :(得分:4)
查看:强>
<script/>
var add_photo_url = "@Url.Action("AddPhoto", "Gallery")";
var model = new FormData();
var i=0;//selected file index
model.append("File", files[i]);
model.append("Name", "test");
$.ajax({// and other parameter is set here
url: add_photo_url,
type: "POST",
data: model,
dataType: "json",
cache: false,
contentType: false,
processData: false
})
.always(function (result) {
});
</script>
查看型号:
public class PhotoAlbumViewModel {
public string Name { get; set; }
public HttpPostedFileBase File { get; set; }
}
<强>控制器:强>
public JsonResult AddPhoto(PhotoAlbumViewModel model) {
// var result =...
// and set your result;
return Json(result, JsonRequestBehavior.AllowGet);
}