some times we see that the post is not work, like as button or other. we can manage it using form and script to serialize the form to reach our controller action. define your action path, form id and method type in your form.
and script
<form id="subscriptionForm" action="/Category/Create" method="post">
<div class="form-horizontal">
<h4>PIS_tblCategory</h4>
<hr />
@Html.ValidationSummary(true)
<div class="form-group">
<div class="form-group">
@Html.LabelFor(model => model.CategoryName, new { @class = "control-label col-md-2" })
<div class="col-md-10">
@Html.EditorFor(model => model.CategoryName)
@Html.ValidationMessageFor(model => model.CategoryName)
</div>
</div>
<div class="form-group">
@Html.LabelFor(model => model.GCompanyID, new { @class = "control-label col-md-2" })
<div class="col-md-10">
@Html.EditorFor(model => model.GCompanyID)
@Html.ValidationMessageFor(model => model.GCompanyID)
</div>
</div>
<div class="form-group">
<div class="col-md-offset-2 col-md-10">
<input type="submit" value="Create" class="btn btn-default" id="Save" />
</div>
</div>
</div>
@*}*@
<div>
@Html.ActionLink("Back to List", "Index")
</div>
</form>
and script
<script type="text/javascript">
$('#Save').click(function () {
var form = $("#subscriptionForm");
var url = form.attr("action");
var formData = form.serialize();
$.post(url, formData, function (data) {
$("#msg").html(data);
});
})
</script>
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