How can i change behaviour of button if form submit was successful?
<form method="post" enctype="multipart/form-data" style="margin: 0;" id="frmGenerate">
<div class="clearfix">
<div class="input-group">
<input type="text" placeholder="Customer Name:" name="CustomerName">
<input type="text" placeholder="Generated Url:" name="CustomerUrl" readonly="readonly" />
</div>
</div>
<div class="clearfix">
<div class="input-group">
<button type="submit" class="btn">Generate</button>
</div>
</div>
</form>
And here is my JS code:
$('#frmGenerate').submit(function(e) {
var clientName = $(this).find('input[name="CustomerName"]').val();
$.ajax(
{
url: '@Url.Action("GenerateToken","Admin")',
type: 'GET',
data: { customerName: clientName },
success: function(response) {
if (response.result) {
$('#frmGenerate').find('input[name="CustomerUrl"]').val(response.message).select();
$('#frmGenerate').find('button.btn').text('Close');
} else {
alert(response.message);
}
}
});
e.preventDefault();
});
This is modal windows, which i open to generate token. When it's generated successful then i set generated token in modal input and change text of button to close, but i don't know how to change button's behaviour.
Better approach would be not to change existing button, but hide it and show another.
HTML:
<div class="clearfix">
<div class="input-group">
<button type="submit" class="btn">Generate</button>
<button type="button" class="btn btn-close hide">Close</button>
</div>
</div>
JS:
$('#frmGenerate').find('.btn').hide();
$('#frmGenerate').find('.btn-close').show();