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    copied!<p>You have to manually fire the validator on your form that was loaded dynamically into your html page.</p> <p>try this:</p> <p>in your view use <code>Ajax.ActionLink</code> to load the content of the partial view into your dialog container to avoid unnecessary JavaScript.</p> <pre><code>@Ajax.ActionLink("AjaxLink", "EditPerson", new { PersonID = model.Id_Person }, new AjaxOptions { UpdateTargetId = "myModalDialog", HttpMethod = "Post",OnSuccess="OpenDialog(myModalDialog)" }) &lt;div id="myModalDialog" title="" style="display: none"&gt; &lt;/div&gt; </code></pre> <p>in you JS file do this</p> <p></p> <pre><code>function OpenDialog(DialogContainerID) { var $DialogContainer = $('#' + DialogContainerID); var $jQval = $.validator; //This is the validator $jQval.unobtrusive.parse($DialogContainer); // and here is where you set it up. $DialogContainer.modal(); var $form = $DialogContainer.find("form"); $.validator.unobtrusive.parse($form); $form.on("submit", function (event) { var $form = $(this); //Function is defined later... submitAsyncForm($form, function (data) { $DialogContainer.modal("hide"); window.location.href = window.location.href; }, function (xhr, ajaxOptions, thrownError) { console.log(xhr.responseText); $("body").html(xhr.responseText); }); event.preventDefault(); }); } //This is the function that will submit the form using ajax and check for validation errors before that. function submitAsyncForm($formToSubmit, fnSuccess, fnError) { if (!$formToSubmit.valid()) return false; $.ajax({ type: $formToSubmit.attr('method'), url: $formToSubmit.attr('action'), data: $formToSubmit.serialize(), success: fnSuccess, error: fnError }); } </code></pre>
 

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