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home / developersection / forums / how do you handle model validation errors in an ajax form submission without reloading the page?
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Anubhav Sharma
Anubhav Sharma

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The Anubhav portal was launched in March 2015 at the behest of the Hon'ble Prime Minister for retiring government officials to leave a record of their experiences while in Govt service .

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2 Answers

Termo Infinito
Termo Infinito 01 Aug 2025
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O Termo Infinito é um verdadeiro desafio mental com palavras, ótimo para quem busca atividades que combinem entretenimento e estímulo cognitivo.

Ravi Vishwakarma
Ravi Vishwakarma 28 Jul 2025
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Handling model validation errors in an AJAX form in ASP.NET MVC:

1. View (Main Form)

Use Ajax.BeginForm with UpdateTargetId:

<div id="formContainer">
    @Html.Partial("_FormPartial", Model)
</div>

2. Partial View: _FormPartial.cshtml

@model MyViewModel
@using (Ajax.BeginForm("SubmitForm", "Home", new AjaxOptions {
    UpdateTargetId = "formContainer", InsertionMode = InsertionMode.Replace, HttpMethod = "POST"
}))
{
    @Html.AntiForgeryToken()
    @Html.ValidationSummary(true)
    @Html.TextBoxFor(m => m.Name)
    @Html.ValidationMessageFor(m => m.Name)
    <input type="submit" value="Submit" />
}

3. Controller

[HttpPost]
public ActionResult SubmitForm(MyViewModel model)
{
    if (!ModelState.IsValid)
        return PartialView("_FormPartial", model); // return form with errors

    return PartialView("_SuccessPartial", model);  // return success view
}

4. Scripts

Make sure these are included:

<script src="jquery.js"></script>
<script src="jquery.validate.js"></script>
<script src="jquery.validate.unobtrusive.js"></script>
<script src="jquery.unobtrusive-ajax.js"></script>

5. Re-enable Validation After AJAX

$(document).ajaxComplete(function () {
    $.validator.unobtrusive.parse("#formContainer");
});
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