在 :nth-child( X ) 中插入 VAL 不起作用
为什么我不能在底部使用 SlideNumber...？ 我尝试将 var SlideNumber = 3; 更改为 SlideNumber = 3; 但说实话，我在黑暗中摸索。
if(window.location.hash != '') {
var SlideNumber = 3;
$('.slideshow').cycle({
fx: 'blindY', // choose your transition type, ex: fade, scrollUp, shuffle, etc...
speed: 700,
cleartype: 1,
startingSlide: $(window.location.hash).index(),
timeout: 3000,
after: function(curr,next,opts) {
$("#menu ul li:nth-child(SlideNumber)").addClass("active");
ready = true;
}
});
先感谢您。
Why can't I use SlideNumber in the bottom...??
I tried to change var SlideNumber = 3; to SlideNumber = 3; but honestly, I'm fumbling in the dark.
if(window.location.hash != '') {
var SlideNumber = 3;
$('.slideshow').cycle({
fx: 'blindY', // choose your transition type, ex: fade, scrollUp, shuffle, etc...
speed: 700,
cleartype: 1,
startingSlide: $(window.location.hash).index(),
timeout: 3000,
after: function(curr,next,opts) {
$("#menu ul li:nth-child(SlideNumber)").addClass("active");
ready = true;
}
});
Thank you in advance.
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要添加，您可以使用
.eq来避免串联并稍微提高性能 过滤方法（采用从零开始的索引）：To add, you can avoid concatenation and crank up the performance a bit by using the
.eqfiltering method (which takes a zero-based index):要将
SlideNumber变量连接到选择器中，您需要分解字符串并将变量夹在+运算符之间：否则，您的伪类选择器将简单地显示为字符串
":nth-child(SlideNumber)"这并不完全符合您的意图。To concatenate your
SlideNumbervariable into your selector, you need to break up your string and sandwich your variable between+operators:Otherwise your pseudo-class selector will simply appear as the string
":nth-child(SlideNumber)"which isn't quite what you intend.