本方法定义在jQuery.prototype
addClass: function( value ) { var classNames, i, l, elem, setClass, c, cl; if ( jQuery.isFunction( value ) ) { return this.each(function( j ) { jQuery( this ).addClass( value.call(this, j, this.className) ); }); } if ( value && typeof value === "string" ) { classNames = value.split( rspace ); for ( i = 0, l = this.length; i < l; i++ ) { elem = this[ i ]; if ( elem.nodeType === 1 ) { if ( !elem.className && classNames.length === 1 ) { elem.className = value; } else { setClass = " " + elem.className + " "; for ( c = 0, cl = classNames.length; c < cl; c++ ) { if ( !~setClass.indexOf( " " + classNames[ c ] + " " ) ) { setClass += classNames[ c ] + " "; } } elem.className = jQuery.trim( setClass ); } } } } return this; }
发现一个转整型的好方法: ~~x
不论x是什么,基本都不会报错,实在不行就返回0,比如~~1.5,返回1
jQuery这里也用了~,这是一种取巧,即indexOf如果返回-1,~运算完正好为0,可以进行if判断,关于单个~运算,我试了几个数,感觉公式就是 -1 * ( 原值 + 1 )
if ( !~setClass.indexOf( " " + classNames[ c ] + " " ) ) { setClass += classNames[ c ] + " "; }
我写一个通用的:
function addClass( el, className ) { var classNames = className.split(/\s/), curClass = trim(el.className), itemClass; if( !curClass && classNames.length === 1 ) {//如果el的className是空的,并且只传了一个className this.className = className; } else { for( var i = 0, len = classNames.length; i < len; i++ ){ curClass = ' ' + curClass + ' '; //这是为了进行下面的匹配 itemClass = classNames[i]; if( !~curClass.indexOf( ' ' + itemClass + ' ' )){ curClass += classNames[i] + ' '; } } el.className = curClass; } }