1

I currently have this code which checks the margin-left of the navigation, and scrolls it left accordingly. However, this method is slow, so I would like to know how to change it into a for loop or something of the sort to catch every multiple of 200 then animate the navigation accordingly.
This is my code currently, as you can tell not very efficient if this needs to go up to -5000 ect.
EDIT
These are my functions, but they do not work

    function goleft(){
    var nav = $("#innernavigation");
    var navmargin = parseInt(nav.css("margin-left"), 10);
    if (navmargin != 0)
    {
        //time!
        if (navmargin > -nav.width() && navmargin < -200)
        {   
            nav.animate({'marginLeft':'-' + ((Math.floor(navmargin / 200)-1) * 200) + 'px'});
        }
        if (navmargin > 0)
        {
            nav.css({'marginLeft':'0px'});
        }
    }
}
function goright(){
    var nav = $("#innernavigation");
    var navmargin = parseInt(nav.css("margin-left"), 10);
    if (navmargin != -nav.width())
    {
        //time!
        if (navmargin > 0 && navmargin < -nav.width())
        {   
            nav.animate({'marginLeft':'-' + ((Math.floor(navmargin / 200)-1) * 200) + 'px'});
        }
        if (navmargin < -nav.width())
        {
            nav.css({'marginLeft':'-' + nav.width()});
        }
    }
}
4

3 回答 3

2

You can do something like this in single if statement.

if (navmargin > -5000 && navmargin < -200)
{
    nav.animate({'marginLeft': (-((Math.floor(navmargin / 200)-1) * 200) + 'px'});
}
于 2013-10-05T16:55:19.473 回答
0 
nav.animate({'marginLeft': (Math.floor(Math.abs(navmargin) / 200) * -200) + 'px'})
于 2013-10-05T17:04:02.170 回答
0

Just a fast guess (not tested yet):

if (navmargin<-200 && navmargin>-5000) {
    nav.animate({'marginLeft':((navmargin%200-1)*-200)+'px'});
}
于 2013-10-05T17:07:12.823 回答