colorbox - 如何使两种类型的jquery脚本不冲突

Question

我不是 jQuery 人，但愿我是。我没有处理任何冲突，但它仍然一次只允许一个脚本工作，而不是同时工作。我正在使用滚动到jquery脚本和colorbox(如lightbox)这是我的代码。任何帮助都很棒。

    <!--jquery scroll-->

<script src="js_scroll/jquery.js"></script>
<script src="js_scroll/jquery.scrollTo.js"></script>
<script src="js_scroll/jquery.nav.js"></script>
<script>

$(document).ready(function() {$('#nav').onePageNav();});

</script>


<!--colorbox-->
<link rel="stylesheet" href="colorbox.css" />
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.0/jquery.min.js"></script>
    <script src="colorbox/jquery.colorbox.js"></script>
    <script>
        $(document).ready(function(){
            //Examples of how to assign the ColorBox event to elements
            $(".group1").colorbox({rel:'group1'});
            $(".group2").colorbox({rel:'group2', transition:"fade"});
            $(".group3").colorbox({rel:'group3', transition:"none", width:"75%", height:"75%"});
            $(".group4").colorbox({rel:'group4', slideshow:true});
            $(".ajax").colorbox();
            $(".youtube").colorbox({iframe:true, innerWidth:425, innerHeight:344});
            $(".iframe").colorbox({iframe:true, width:"80%", height:"80%"});
            $(".inline").colorbox({inline:true, width:"50%"});
            $(".callbacks").colorbox({
                onOpen:function(){ alert('onOpen: colorbox is about to open'); },
                onLoad:function(){ alert('onLoad: colorbox has started to load the targeted content'); },
                onComplete:function(){ alert('onComplete: colorbox has displayed the loaded content'); },
                onCleanup:function(){ alert('onCleanup: colorbox has begun the close process'); },
                onClosed:function(){ alert('onClosed: colorbox has completely closed'); }
            });

            //Example of preserving a JavaScript event for inline calls.
            $("#click").click(function(){ 
                $('#click').css({"background-color":"#f00", "color":"#fff", "cursor":"inherit"}).text("Open this window again and this message will still be here.");
                return false;
            });
        });
    </script>

</body>

Answer 1

如果你想加载多个 jQuery 库，就是这样。

<script src='jquery-1.3.2.js'></script>
<script>
var $j132 = jQuery.noConflict();
</script>

<script src='jquery-1.4.2.js'></script>
<script>
var $j142 = jQuery.noConflict();
</script>

现在使用 $j132() 进行 jQuery 1.3.2 操作，使用 $j142() 进行 jQuery 1.4.2