Reputation: 198
In a View, there is a file upload control(excluded from below code), two buttons, one for saving and other for displaying the modal.
@using (Html.BeginForm("Display", "Home", FormMethod.Post,
new {enctype = "multipart/form-data"}))
{
<div id="myModal" class="modal fade" role="dialog">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-body">
</div>
</div>
</div>
</div><button id="btn1" name="update" class="btn btn-danger" type="submit">Save</button><br />
<button id="btn2" name="savenadd" class="btn btn-danger" data-toggle="modal" data-target="#myModal">Display Modal</button>
}
How do I prevent the page from posting back when btn2 is clicked? I don't mind posting back when modal is closed. Thanx
Upvotes: 2
Views: 202
Reputation:
The WC3 standards specify that the default type
for a <button>
in a <form>
is type="submit"
. To prevent the default behavior, make it type="button"
(note the name
attribute is not necessary)
<button id="btn2" type="button" class="btn btn-danger" data-toggle="modal" data-target="#myModal">Display Modal</button>
Alternatively, you could move the button outside the <form>
element.
Upvotes: 3