我有一个包含ajax表单的局部视图.视图只是添加和更新用户信息.
控制器将相同的部分视图发送回ActionResult.
我想要做的是在事务不成功时显示错误消息.但是它仍然应该发回部分视图,这次只发送一条消息.
如何实现？
码：
ManageUsers.cshtml
<div id="details">
@{
Html.RenderPartial("AddModifyUserPartialView");
}
</div>
@{
Html.RenderPartial("ListUsersPartialView");
}
AddModifyUserPartialView.cshtml
@using (Ajax.BeginForm("AddModifyUser","Account",FormMethod.Post,new AjaxOptions()
{
UpdateTargetId = "details",OnFailure= "handleError",OnSuccess="handleSuccess"
},new { id = "useragentform",enctype = "multipart/form-data" }))
{
//form fields here
<input type="submit" id="savebutton" name="savebutton" value="Add New User" />
}
另外在局部视图中：
function handleError(ajaxContext) {
var response = ajaxContext.get_response();
var statusCode = response.get_statusCode();
alert(statusCode);
}
账户管理员
try
{
SecurityManager.AddUpdateUserAgent(ua);
}
catch (Exception ex)
{
//how do I send the message back along with the partial view???
}
return PartialView("AddModifyUserPartialView");
解决方法
解决这个问题的两个部分,创建一个新的异常,让我们用你的消息称它为StatusException,并在你发现正常异常时抛出它：
try
{
SecurityManager.AddUpdateUserAgent(ua);
}
catch (Exception ex)
{
throw new StatusException("Your error message here")
}
return PartialView("AddModifyUserPartialView");
覆盖Controller :: OnException并通过将其设置为handling来处理异常,将错误代码设置为500,将HttpContext.Response.StatusDescription设置为StatusException消息.例如：
protected override void OnException(ExceptionContext filterContext)
{
if (filterContext.Exception == null) return;
Type exceptionType = filterContext.Exception.GetType();
if (exceptionType == typeof(StatusException))
{
filterContext.ExceptionHandled = true;
filterContext.HttpContext.Response.Clear();
filterContext.HttpContext.Response.ContentEncoding = Encoding.UTF8;
filterContext.HttpContext.Response.HeaderEncoding = Encoding.UTF8;
filterContext.HttpContext.Response.TrySkipIisCustomErrors = true;
filterContext.HttpContext.Response.StatusCode = 500;
filterContext.HttpContext.Response.StatusDescription = filterContext.Exception.Message;
}
}
然后,在Ajax.BeginForm的OnFailure处理程序中,显示错误参数：
function handleError(data){
//display data.errorThrown,data.statusCode,etc...
}
通过在OnException覆盖中将错误代码设置为500,AjaxForm将检测错误并跳转到您的处理程序.我们也在覆盖中设置StatusDescription,以便在handleError回调中提供该消息.
