我能够上传文件并将其保存到目录中。这都很好。我需要在我的数据库中输入有关该文件的信息。到目前为止,我不确定在这种特殊情况下如何将值从视图传递给控制器。我试图将它作为方法参数传递,但没有发布该值。
这是我的剃刀形式:
@using (Html.BeginForm("AjaxUpload", "Cases", FormMethod.Post, new { enctype = "multipart/form-data", id = "ajaxUploadForm" }))
{
<fieldset>
<legend>Upload a file</legend>
<label>File to Upload: <input type="file" name="file" />(100MB max size)</label>
<input id="ajaxUploadButton" type="submit" value="Submit" />
</fieldset>
}
<div id="attachments">
@Html.Partial("_AttachmentList", Model.Attachments)
</div>
这是我的 jQuery 来 ajaxify 表单:
$(function () {
$('#ajaxUploadForm').ajaxForm({
iframe: true,
dataType: "json",
beforeSubmit: function () {
$('#ajaxUploadForm').block({ message: '<h1><img src="/Content/images/busy.gif" /> Uploading file...</h1>' });
},
success: function (result) {
$('#ajaxUploadForm').unblock();
$('#ajaxUploadForm').resetForm();
$.growlUI(null, result.message);
//$('#attachments').html(result);
},
error: function (xhr, textStatus, errorThrown) {
$('#ajaxUploadForm').unblock();
$('#ajaxUploadForm').resetForm();
$.growlUI(null, 'Error uploading file');
}
});
});
这是控制器方法:
public FileUpload AjaxUpload(HttpPostedFileBase file, int cid)
{
file.SaveAs(Server.MapPath("~/Uploads/" + file.FileName));
var attach = new Attachment { CasesID = cid, FileName = file.FileName, FileType = file.ContentType, FilePath = "Demo", AttachmentDate = DateTime.Now, Description = "test" };
db.Attachments.Add(attach);
db.SaveChanges();
//TODO change this to return a partial view
return new FileUpload { Data = new { message = string.Format("{0} uploaded successfully.", System.IO.Path.GetFileName(file.FileName)) } };
}
我希望将 cid 传递给此方法,以便我可以将记录插入数据库。