jQuery Counter add number to class

Question

I'm have a for loop on jQuery and I want my counter to add a class each time with the number of the array it's coming from. But if I write:.addClass("delay"+counter++) that obviously won't work. Would there be another way of going about this.

For example this is what I have so far

            $(document).ready(function() {
                var counter = 1;
                $.getJSON('http://www.passionla.com/app/blog-data.php', function(data) {
                    $.each(data, function(key, value) {
                        $('.blog').append("<div class='post animated slideInRight "+ counter++ +"' style=\"background:url('"+value.header+"') no-repeat; background-size: 100% 100%\"><div class='color'><div class='wrapper'><h1>"+value.title+"</h1></div></div></div>");
                    });
                }); 
            });

It wouldn't work but regardless .addClass typo*

NooBskie
– NooBskie

2015-09-11 03:16:09 +00:00
Commented Sep 11, 2015 at 3:16 — NooBskie
– NooBskie, Commented Sep 11, 2015 at 3:16
Please add more code

Damian0o
– Damian0o

2015-09-11 03:16:29 +00:00
Commented Sep 11, 2015 at 3:16 — Damian0o
– Damian0o, Commented Sep 11, 2015 at 3:16
Make a codepen/jsfiddle of what you have tried at least

NooBskie
– NooBskie

2015-09-11 03:17:23 +00:00
Commented Sep 11, 2015 at 3:17 — NooBskie
– NooBskie, Commented Sep 11, 2015 at 3:17
what do you mean by won't work....

Arun P Johny
– Arun P Johny

2015-09-11 03:17:41 +00:00
Commented Sep 11, 2015 at 3:17 — Arun P Johny
– Arun P Johny, Commented Sep 11, 2015 at 3:17
try .addClass("delay"+ ( counter++ ) )

Tamil Selvan C
– Tamil Selvan C

2015-09-11 03:19:04 +00:00
Commented Sep 11, 2015 at 3:19 — Tamil Selvan C
– Tamil Selvan C, Commented Sep 11, 2015 at 3:19

Derek Story · Accepted Answer · 2015-09-11 03:24:37Z

2

Its probably easier to increase the counter as a variable and then attach the variable to the class:

JS Fiddle

$('div').each(function(i) {
    var $this = $(this); 
    var newClass = "delay" + i++;
    $this.addClass(newClass);
});

But you would use the .each() function on your returned data instead of div.

edited Sep 11, 2015 at 3:24

answered Sep 11, 2015 at 3:19

Derek Story

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