jQuery圆形滑块 [英] jQuery rounded slider
本文介绍了jQuery圆形滑块的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想像下图所示做一个圆形滑块。
jQuery能够做到这一点吗?
我知道直滑块如何工作,但我想制作一个HTML5圆形滑块。
以下是我在网上找到的
$(function(){
var $ circle = $('#circle'),
$ handler = $('#handler'),
$ p = $('#test'),
handlerW2 = $ handler.width()/ 2,
rad = $ circle.width()/ 2,
offs = $ circle.offset (),
elPos = {x:offs.left,y:offs.top},
mHold = 0,
PI2 = Math.PI / 180;
$ handler。 mousedown(function(){mHold = 1;});
$(document).mousemove(function(e){
if(mHold){
var mPos = {x:e.pageX-elPos.x,y:e.pageY-elPos.y},
atan = Math.atan2(mPos.x-rad,mPos.y-rad),
deg = -atan / PI2 + 180,
perc =(deg * 100/360)| 0,
X = Math.round(rad * Math.sin(deg * PI2)),
Y = Math.round(rad * -Math.cos(deg * PI2));
$ handler.css({left:X + rad-handlerW2,top:Y + rad-handlerW2,transform:'rotate('+ deg +'deg)'});
}
})。mouseup(function(){mHold = 0;});
});
I would like to do a rounded slider like the image below. Is jQuery able to do this?
I know how a straight slider works but I would like to make a HTML5 rounded slider. Here is what I found online http://jsfiddle.net/XdvNg/1/ - but I dont know how to get the slider value one the user lets go
解决方案
Here is what I came up with:
jsBin demo
$(function () {
var $circle = $('#circle'),
$handler = $('#handler'),
$p = $('#test'),
handlerW2 = $handler.width()/2,
rad = $circle.width()/2,
offs = $circle.offset(),
elPos = {x:offs.left, y:offs.top},
mHold = 0,
PI2 = Math.PI/180;
$handler.mousedown(function() { mHold = 1; });
$(document).mousemove(function(e) {
if (mHold) {
var mPos = {x:e.pageX-elPos.x, y:e.pageY-elPos.y},
atan = Math.atan2(mPos.x-rad, mPos.y-rad),
deg = -atan/PI2+180,
perc = (deg*100/360)|0,
X = Math.round(rad* Math.sin(deg*PI2)),
Y = Math.round(rad* -Math.cos(deg*PI2));
$handler.css({left:X+rad-handlerW2, top:Y+rad-handlerW2, transform:'rotate('+deg+'deg)'});
}
}).mouseup(function() { mHold = 0; });
});
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