CODEWITHSUNDEEP
jqueryfunctionvar
Scott Pedder
Scott Pedder

Reputation: 57

Combine two or more functions

I have 10 images each with a class 1-10 how can i make the function work for every class? Works fine with class="1". Just not sure how to add them together?

<body>
<div>
    <img src="img/13-01_basic_hover.jpg" class="1" width="220" height="184" title="huge, gigantic">
    <img src="img/13-01_basic_hover.jpg" class="2" width="220" height="184" title="Number 2">
</div>
<div class="project"></div>
<script>
var title = $(".1, .2").attr("title");
$(".project").text(title);
$(document).ready(function () {
    $(".project").fadeTo(0,0);
    $(".1, .2").hover(
        function () {
            $(".project").stop(true).fadeTo("normal",1);
        }, 
        function () {
            $(".project").fadeTo("normal",0);
        }
    );       
});
</script>
</body>

Upvotes: 0

Views: 93

Answers (2)

Selvakumar Arumugam
Selvakumar Arumugam

Reputation: 79840

Just add them as a comma seperated value,

$(".1, .2, .3, .4, .5, .6, .7, .8, .9, .10").hover(
  function () {
    $(".project")
        .text(this.title)
        .stop(true).fadeTo("normal",1);
  }, 
  function () {
    $(".project").fadeTo("normal",0);
  }
);

Or add another class to those elements and group them and then you can,

$(".grouped").hover(
  function () {
     $(".project")
        .text(this.title)
        .stop(true).fadeTo("normal",1);
  }, 
  function () {
   $(".project").fadeTo("normal",0);
  }
);

Upvotes: 2

Diodeus - James MacFarlane
Diodeus - James MacFarlane

Reputation: 114417

Class names starting with numbers are invalid. You should change this.

Class names are usually used to select groups of items anyway. Just add a second class name for all your items and use that instead of .1.

Upvotes: 1

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