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    <p>Note that doing an ajax post will not refresh or redirect the page like a classic post would.</p> <p>The view is not showing because there is no where to place it. You post successfully, and return a view successfully but then do nothing with it. If you want to show the returned view, then you have to append it to the dom somehow. Here is a common method:</p> <pre><code>&lt;div id="successDiv"&gt;&lt;/div&gt; &lt;script type="text/javascript" language="javascript"&gt; $(document).ready(function () { $("#HotelDDL").change(function () { var text = $("#HotelDDL option:selected").text(); var value = $("#HotelDDL option:selected").val(); alert("Selected text=" + text + " Selected value= " + value); $.post("/Home/SearchForRooms", { HotelDDL: $("#HotelDDL option:selected").val() }, function(result){//add callback for success - result is the returned view $("#successDiv").html(result);//place the view inside of the div }); }); }); &lt;/script&gt; </code></pre> <p><code></code></p> <hr> <p><strong>Comment Response</strong></p> <p>@TunAntun - You should use a classic post if you want the view to have its own page. There is no way to accomplish this from ajax. You could do this with javascript though</p> <pre><code> $.post("/Home/SearchForRooms", { HotelDDL: $("#HotelDDL option:selected").val() }, function(result){//add callback for success - result is the returned view $("#successDiv").html(result);//place the view inside of the div }); </code></pre> <p>would become</p> <pre><code> var postForm = document.createElement("form"); postForm.setAttribute("method","post"); postForm.setAttribute("action","/Home/SearchForRooms"); var postVal = document.createElement("input"); postVal.setAttribute("type","hidden"); postVal.setAttribute("name","HotelDDL"); postVal.setAttribute("value",$("#HotelDDL option:selected").val() ); postForm.appendChild(postVal); document.body.appendChild(postForm); $(postForm).submit(); </code></pre> <p>This will dynamically issue the post you wanted. Then in your controller it will properly redirect or return a view. It is highly suggested that when returning a view from a <code>[HttpPost]</code> that you use <code>RedirectToAction</code> and then return the view from a <code>[HttpGet]</code> method in order to prevent refresh from reposting old data.</p>
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    1. POASP.NET MVC 3 NOT showing appropriate view, action called using jquery
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    1. COIf I do place view inside <div id = "successDiv"><div/> then my Model is not working anymore and have to change it. But let's say I modified it. What about when I want View to have it's own page? Thanks
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    2. CO@TunAntun - See edits.
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    3. COok, can I just ask one more thing. I have two drop downs and one ok button. First DDL is filled normally from base and ajax is used to fill another DDL (the one in this question). Is there a better way to do this (2 dropdowns). I guess I made this much more complicated then needed.
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