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这里是我的Web方法Web方法返回OK,但火失败功能
[HttpGet]
public ActionResult EditEmp(int? id)
{
if (id == null)
{
return new HttpStatusCodeResult(HttpStatusCode.BadRequest);
}
Employee Emp = db.Employees.Find(id);
if (Emp == null)
{
return HttpNotFound();
}
ViewBag.dept_id = new SelectList(db.Departments, "dept_id", "dept_name", Emp.dept_id);
return PartialView("_EditEmp", Emp);
}
,这里是Ajax调用
$.ajax({
type: "GET",
url: '/Employee/EditEmp',
data: { id: idp },
dataType: "json",
success: function (result) {
alert(result);
$('#editid').html(result);
},
error: function (result) {
alert("FAILED : " + result.status + ' ' + result.statusText);
}
});
它给了我result.status = 200和result.statusText = OK但它火灾事件
非常感谢你,你让我 – freelancer
天乐意提供帮助。 –