jQuery: do I need to put calls to $("img").lazyload() in $(document).ready()?_问答_开发者

jQuery: do I need to put calls to $("img").lazyload() in $(document).ready()?

开发者 https://www.devze.com 2022-12-21 17:47 出处：网络

I am using lazy 开发者_如何学Pythonload jquery plugin. It tells me to put this in my code to activate it:

I am using lazy 开发者_如何学Pythonload jquery plugin.

It tells me to put this in my code to activate it:

$("img").lazyload();

Do I have to put this in $(document).ready()? or can I just put it at the very top of the page:

<head>
<script type="text/javascript">
$("img").lazyload();

$(document).ready...{}

</script>
</head>

The demo page puts it in the $(document).ready(), except it uses the shorthand $( ) function to do it.

(from the demo)

$( function() {
    $("img").lazyload({placeholder : "img/grey.gif"});
} );

(note that $() is an alias to the jQuery() function, which takes CSS selectors, HTML elements and also callbacks to run when the DOM loads)

If you don't put it in the ready() function, then it may only affect images that are earlier in the page than your script. And if you're placing scripts in the HEAD, that's none of the images.

So, unless the library is somehow using the .live() function of JQuery, you need to put it in ready().

The way I understand it is that if you put it in the $(document).ready(...) the script won't run until the DOM has loaded. If you just put it in <script></script> tags in the head then the page will have to wait for the script to complete before the remainder of the DOM can load.

If the script modifies the DOM this means it will run, do nothing, then the page will load (having not been affected by the $("img").lazyload() function.

You should put it in the $(document).ready().

You need to put it inside since it uses jQuery library.