这个ajax调用永远不会在成功时创建警报,即使它已达到并在服务器端方法返回success = true。
@model IEnumerable<Test.Models.Task>
@Styles.Render("~/Content/Site.css")
@{
ViewBag.Title = "Index";
}
<h2>Index</h2>
<p>
@Html.ActionLink("Create New", "Create")
</p>
<div id ="alerts">
@Html.Action("_Tasks")
<script type="text/javascript">
$(document).ready(function poll() {
$.ajax({
type: 'GET',
cache: false,
url: '@Url.Action("TasksRefresh")',
dataType: "json",
complete: function () { setTimeout(poll, 10000); },
success: function (data) {
alert("Testing")
}
});
})();
</script>
@* <script type="text/javascript">
var alerts = '@ViewBag.Alerts';
@foreach (var i in alerts)
{
}
</script>*@
</div>
<table>
<tr>
<th>Category</th>
<th>Severity</th>
<th>Assigned to Role</th>
<th>Assigned To</th>
<th>Chart #</th>
<th>Note</th>
<th>Alert</th>
<th>Status</th>
<th>Creator By</th>
<th>Create Date</th>
<th>Due Date</th>
<th></th>
</tr>
@foreach (var item in Model) {
<tr>
<td>
@Html.DisplayFor(modelItem => item.LookupTaskCategory.CategoryName)
</td>
<td>
@Html.DisplayFor(modelItem => item.LookupTaskSeverity.SeverityName)
</td>
<td>
@Html.DisplayFor(modelItem => item.AssignedToRoleName)
</td>
<td>
@Html.DisplayFor(modelItem => item.AssignedToName)
</td>
<td>
@Html.DisplayFor(modelItem => item.Patient.ChartNo)
</td>
<td>
@Html.DisplayFor(modelItem => item.Note)
</td>
<td>
@Html.DisplayFor(modelItem => item.AlertFlag)
</td>
<td>
@Html.DisplayFor(modelItem => item.LookupTaskStatu.StatusName )
</td>
<td>
@Html.DisplayFor(modelItem => item.CreatedByName)
</td>
<td>
@Html.DisplayFor(modelItem => item.CreatedOnDate)
</td>
<td>
@Html.DisplayFor(modelItem => item.DueDate)
</td>
<td>
@Html.ActionLink("Edit", "Edit", new { id=item.Id }) |
@Html.ActionLink("Details", "Details", new { id=item.Id }) |
@Html.ActionLink("Delete", "Delete", new { id=item.Id })
</td>
</tr>
}
</table>
这是我控制器中的服务器端方法。我试图用ActionResult替换JsonResult,但它没有改变结果。
public JsonResult TasksRefresh()
{
//Testing to see if this return ever gets received by ajax.
return Json(new { success = true });
}
答案 0 :(得分:3)
您在服务器上遇到异常 - 尝试调试.NET代码,或使用浏览器工具观察您的响应,以查看它。
如果要在GET
方法上返回JSON对象,则需要在JsonRequestBehavior
调用中加入Json
参数,例如:
return Json(new { success = true }, JsonRequestBehavior.AllowGet);
修改强>
实际上,如果你在服务器上进行调试,看起来你看不到它 - 你必须在响应中看到它。显然,在Json
方法之后,异常会被进一步抛出。