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2013-08-23T20:47:33.817
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<p>Although sending back a response and checking the data object in the callback (as explained above by @Richard and @Jatin) will work, I tend to follow the following pattern of rolling my own AjaxResult object, hydrating this object with the model state validity, the success state, as well as any data I need to pass back to the $.ajax success callback function. This enables a flexible and elegant solution in handling $.ajax requests. For example:</p> <pre><code>// custom ajax request result class public class AjaxResult { public bool IsValid { get; set; } public bool Success { get; set; } public string Message { get; set; } public Dictionary<string, object> Data { get; set; } public AjaxResult() { Data = new Dictionary<string, object>(); } public void AddData(string dataKey, object dataValue) { if (dataKey != null && !Data.ContainsKey(dataKey)) { Data.Add(dataKey, dataValue); } } } // controller public ActionResult NhpdAdminRejectAddress(NhpdAdminViewModel model) { AjaxResult result = new AjaxResult(); if (ModelState.IsValid) { // do something result.AddData("Code", -1); result.IsValid = true; result.Success = true; } return Json(result); } // ajax post request $.ajax({ type: "POST", url: '@Url.Action("NHPDAdminRejectAddress", "Home")', data: JSON.stringify(model), contentType: "application/json; charset=utf-8", dataType: "json", success: function (result) { if (result != null && result.Data != null && result.Data.Code != null) { int code = result.Data.Code; } }, error: function () { // handle error } }); </code></pre>
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1. POHow do I code an ajax function to receive a return code passed back from a controller method?
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