Reputation: 149
I have this code:
$(document).on('submit','#roomsend',function(e) {
var form = $('#roomsend');
var data = form.serialize();
$.post('php/roomlist.php', data, function(response) {
console.log(response);
$('#power').replaceWith(response);
});
return false;
});
This works with a Submit button inside the form. But now I have a button outside the form:
<button class="btn btn-success" id="roomsendbutton"><i class="icon-ok"></i> Save</button>
How must I must be the code, that when i clicked the button the button all actions will be work and the form submits?
Upvotes: 0
Views: 9491
Reputation: 95054
If you're using HTML5, you can use the new form attribute:
<button form="roomsend" type="submit" class="btn btn-success" id="roomsendbutton"><i class="icon-ok"></i> Save</button>
Should probably look up browser support for it though.
Upvotes: 4
Reputation: 4712
$(document).on('click','#roomsendbutton',function(e) {
var form = $('#roomsend');
var data = form.serialize();
$.post('php/roomlist.php', data, function(response) {
console.log(response);
$('#power').replaceWith(response);
});
return false;
});
Just bind the click event on the button.
Upvotes: 0
Reputation: 620
Just do something like this in your jQuery..
$('#roomsendbutton').click(function(){
$(your form name).trigger('submit');
});
Upvotes: 0
Reputation:
Use .trigger()
method:
$('#roomsendbutton').on('click', function () {
$('#roomsend').trigger('submit');
});
References:
.trigger()
- jQuery API DocumentationUpvotes: 2