By the plugin documentation, you can destroy the validator instance with the method:
validator.destroy();
Where validator
is the name of the plugin instance you started. Example:
var validator = $("SELETOR DO FORMULÁRIO").validate();
Then your event change
would be:
$('#pessoaNatureza').on('change', function (e) {
validator.destroy();
$(this).closest('form').submit();
});
The validator will be deactivated before Submit and the form will be submitted in the event change
dropdown.
Plugin documentation
Example:
var validator = $("#commentForm").validate();
$('#pessoaNatureza').on('change', function (e) {
validator.destroy();
$(this).closest('form').submit();
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://cdn.jsdelivr.net/jquery.validation/1.16.0/jquery.validate.min.js"></script>
<form class="cmxform" id="commentForm" method="get" action="./">
<p>
<select id="pessoaNatureza">
<option value="">Selecione...</option>
<option value="1">1</option>
<option value="2">2</option>
</select>
</p>
<p>
<input id="cname" name="name" minlength="2" type="text" required>
</p>
<button>Salvar</button>
</form>
You can pass a parameter executes validations and do within the Submit method a validation for if the execute validations is true execute, if it does not pass directly to the http request.
$('#pessoaNatureza').on('change', function (e) { 
 $(this).closest('form').submit(false);
});
– Lucas Brogni
Only in all cases I have to give a Submit, only when it is by the change event, it does not validate the fields, that is, it must send the form without validating...
– Master JR