I am using jquery form plug-in for posting my form ajax way.
Post is working fine with default button.But it is not working when I am using custom bootstrap button as below . Please let me know what i am missing here?
http://jquery.malsup.com/form/#validation
Working case :
<input name="Update" value="Update" type="submit">
Not working case:
<a class="btn btn-sm btn-bitbucket" name="Update" value="Update" type="submit">
<i class="demo-icon icon-ccw fa-fw" style="font-size:1.3em;"> </i> Update
</a>
I've just removed my comment as any of these should work:
<input name="Update" value="Update" type="submit" class="btn btn-sm btn-bitbucket">
<button class="btn btn-sm btn-bitbucket" name="Update" type="submit">
<i class="demo-icon icon-ccw fa-fw" style="font-size:1.3em;"></i> Update
</button>
An anchor can't (by default) be used to submit the form, hence why it isn't working. However Bootstrap allows you to use the button classes on input and button tags.
Sorry for the ghost comment now!
Related
I have 3 buttons on my toolbar (delete,block,unblock). Сan i change form action dynamically
<form action="/users/groupUnblock" method="POST">
<button type="submit" class="btn btn-danger btn-sm">Delete</button>
</form>
Tried to write a func,but its not working
If you don't want to use JavaScript at all, this can be done with pure HTML5. You can specify the action of each button with the formaction attribute.
Example:
<form>
<button type="submit" formaction="http://firsttarget.com">Submit to first</button>
<button type="submit" formaction="http://secondtarget.com">Submit to second</button>
</form>
Read more about this attribute here
How can I add action = "/#" to a button without using <form>?
For example, I want to do
<button type="submit" action = "/logout" class="btn btn-default">Logout</button>
instead of:
<form action = "/logout">
<button type="submit" class="btn btn-default">Logout</button>
</form>
Here is the current code:
<div class="navbar-form navbar-right btn-group">
<button type="button" class="btn btn-default">Score</button>
<button type="button" class="btn btn-default" href="#menu-toggle" id="menu-toggle">Favorites</button>
<button type="submit" class="btn btn-default">Logout</button>
</div>
You can't use action attribute on a button it only works with HTML form. type="submit" is used on a form element to submit the form data. An element with this type will appear as a button by default.
If you don't want to use HTML form you can use anchor tag to send user on any page. You can also do it with JavaScript or jQuery
I am using jquery file-upload plugin in backbone js.
I have more than one add-files button on a same page. Each button comes form their respective backbone views.
**View**
render: ->
$(#el).html(#template(model: #model.attributes ))
#
**Template**
<form id="fileupload" class="fileupload" action="some_action" method="POST" enctype="multipart/form-data">
<span class="btn btn-xs btn-success fileinput-button">
<i class="glyphicon glyphicon-plus"></i>
<span>Add files...</span>
<input type="file" class="select_file" name="files" multiple>
</span>
<button type="submit" class="btn btn-xs btn-primary start">
<i class="glyphicon glyphicon-upload"></i>
<span>Start upload</span>
</button>
</form>
The Add files button works for any one form but not for the others
What can be the reason and how to solve it?
Try this jsFiddle.
My guess is, you might have initialized fileupload like below in your code.
$("#fileupload").fileupload(....)
If you have done like that it should be the issue. You should initialize fileupload on each form.
Note:
I found this jsFiddle on google search and edited it for your requirement.
I have been trying to put a font awesome or glyphicon (or even an image) inside a submit button. This quickly falls apart, and I need to implement another option. I have seen a few ways of doing this, but I don't know which is the "right" or better way.
The three contenders I've seen are:
divs
links
buttons
Which is the best way to get an image / icon inside a "submit" button?
Wouldn't all three need a JS component?
This is how bootstrap add glyphicon to button object:
<button type="button" class="btn btn-default btn-lg">
<span class="glyphicon glyphicon-star" aria-hidden="true"></span> Star
</button>
Bootstrap button with glyphicon
The button option does not need JS. I have tested it and you can see a working example using Font Awesome here https://jsfiddle.net/mikhailjan/sf9s28et/5/ or just see the code below:
<form action="add_person.php">
First name:<br>
<input type="text" name="firstname" value="Mickey">
<br>
Last name:<br>
<input type="text" name="lastname" value="Mouse">
<br><br>
<button type="submit" class="button">
<i class="fa fa-user-plus"></i> Add Person
</button>
</form>
Jonathan Anctil is correct, the button needs to have type="submit" for the form to work normally.
I have an ecommerce website that uses jQuery 1.9.1 and Bootstrap 2.3.2
I'd like to prevent customers from double-submitting orders on accident (hitting submit twice).
Any ideas on how to do this?
My input submit is simply:
<input type="submit" class="btn btn-orange pull-right" value="Place Order">
Is there anything I can do to prevent this from occurring? I don't want to hinder other customers, I just don't want folks to submit, wait, get impatient, submit again, and they are double charged for their order. That's sloppy.
Thanks.
Here is a little trick:
<input type="submit" class="btn btn-orange pull-right" value="Place Order" onClick="this.disabled=1">
Disable the button when form is submitted. You can do something like this:
$('.pull-right').click(function() {
// Code to submit the form
$(this).attr('disabled', true);
});
This SO Question might help
How-to-prevent-calling-of-en-event-handler-twice-on-fast-clicks
You can do it like this:
<input type="submit" class="btn btn-orange pull-right" value="Place Order" onclick="$(this).attr('disabled', true);">
Some times you may want to also run a function like: OnSubmit function (or a validation) you can do this:
<input type="submit" class="btn btn-orange pull-right" value="Place Order" onclick="this.disabled=true;OnSubmit(this.parentNode);" />