clydewinux clydewinux - 2 months ago 6
HTML Question

call the same jQuery function in multiple buttons

I am not really familiar with jQuery. I have this code that I downloaded to create a fade in/fade out popup form. Here's the code:

<script type='text/javascript'>
$(document).ready(function() {
$('#button').click(function(e) {
$('#modal').reveal({
animation: 'fade',
animationspeed: 150,
closeonbackgroundclick: true,
dismissmodalclass: 'close'
});
return false;
});
});
</script>


The code above executes when a button with an
id='button'
is clicked. Now I have multiple buttons, how can I call this function in all buttons? I tried setting the
id
of all buttons to
button
but only the first button works. Any help would be very much appreciated.
By the way I forgot to mention, in my .php file i have this codes:

for ($c=1;$c<=5;$c++){
echo "<input type='button' id='button'">
};


This php code will display 5 buttons with the same id which is 'button'. What I want to happen is when I click any of the 5 buttons the jQuery function will execute which is popping up a fade in/fade out form.

Answer

First solution :

function doClick(e) { 
   $('#modal').reveal({ 
     animation: 'fade',
     animationspeed: 150,
     closeonbackgroundclick: true, 
     dismissmodalclass: 'close'
   });
   return false;
}
$('#button1').click(doClick);
$('#button2').click(doClick);

Second solution :

Give a class "someClass" to all the involved buttons

<input type=button class=someClass ...

and do

$('.someClass').click(function(e) { 
...
});

Third solution :

Use the comma to separate ids :

$('#button1, #button2').click(function(e) { 
...
});

Generally, the best solution is the second one : it allows you to add buttons in your code without modifying the javascript part. If you add some of those buttons dynamically, you may even do

$(document).on('click', '.someClass', function(e) { 
...
});