Why is my javaScript not working locally or on server, but working on jsFiddle

问题

This is a link to the working jsfiddle http://jsfiddle.net/akshaytyagi/SD66b/

and following is the code that I am trying to run on my website ( same as the one jsFiddle )

I have tried on two computers. What am I doing wrong?

<html>
<head>

<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>

<script type="text/javascript">

$(document).ready(function() {

var tips = [
    "creative",
    "innovative",
    "awesome",
    "amazing",
    "social"
    ];

setInterval(function() {
    var i = Math.round((Math.random()) * tips.length);
    if (i == tips.length)--i;

    $('#tip').slideUp(500, function() {
        var $this = $(this);
        $this.html(tips[i]);
        $this.toggleClass('first second');
        $this.slideDown(500);
    });

}, 3 *1000);

});
</script>
</head>

<body>
<div style=" background-position:center; background-repeat:no-repeat; background-color:#c84d5f; height:500px">
<div class="thousand">
<div style="font-size:72px; font-family:Museo; padding-top:100px; padding-left:auto; padding-right:auto; color:#FFF;">The <span id="tip">creative</span><br />brand.
</div>
</div>
</div>


</body>
</html>

回答1:

I copy and pasted your HTML, and after }, 3 * 1000); there is a special char.

Delete that whole line (}, 3 * 1000);) and re-type it.

See:

As andyb has commented, if you're loading the file locally your jquery url wont work. You can either change it to http:// or upload your file somewhere.

回答2:

You need to put the script which access DOM element in $(document).ready to make sure the elements are ready before they are accessed.

$(document).ready(function(){

})

Edit based on comments

Change

<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2jquery.min.js"></script>

回答3:

Although the right answer was already given, I've taken the liberty to fix your markup. And may I suggest you use proper CSS instead on inline styling? It makes your code much more readable and separates markup and design as you should,

<html><head>
    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
    <script>
    var tips = [
        'creative',
        'innovative',
        'awesome',
        'amazing',
        'social'
    ];

    setInterval(function() {
        var i = Math.round((Math.random()) * tips.length);
        if (i == tips.length)--i;

        $('#tip').slideUp(500, function() {
            var $this = $(this);
            $this.html(tips[i]);
            $this.toggleClass('first second');
            $this.slideDown(500);
        });
    }, 3000);
    </script>
</head><body>
    <div style="background-position:center; background-repeat:no-repeat; background-color:#c84d5f; height:500px">
        <div class="thousand">
            <div style="font-size:72px; font-family:Museo; padding-top:100px; padding-left:auto; padding-right:auto; color:#FFF;">
                The <span id="tip">creative</span><br />brand.
            </div>
        </div>
    </div>
</body></html>

回答4:

The problem making it work, locally, was that // links do not get resolved to http:// ( src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js instead of just src="//ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js")

[ Thanks to @andyb : I was wondering why Google had improper code on their site. ]

来源：https://stackoverflow.com/questions/13309903/why-is-my-javascript-not-working-locally-or-on-server-but-working-on-jsfiddle

标签

javascript

jquery

jsfiddle