jquery第n个子选择器问题

发布于 2024-11-02 03:22:33 字数 128 浏览 1 评论 0原文

我想在 div 中选择图像。我想要的图像编号为 2、5、8、11 等。

$('.thediv img:nth-child(3n+1)')..

不适合我，我错过了什么吗？谢谢

原文

I want to select images in a div. The images i want are number 2,5,8,11 etc.

$('.thediv img:nth-child(3n+1)')..

Did not work for me, did i miss something? Thanks

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

童话 2024-11-09 03:22:33

您的公式 3n + 1 不起作用，因为这些是将为以下 n 值选择的元素：

3(0) + 1 = 0 + 1 = 1
3(1) + 1 = 3 + 1 = 4
3(2) + 1 = 6 + 1 = 7
3(3) + 1 = 9 + 1 = 10
...

显然，这些不是第二个、第五个、第 8 个、第 11 个...选定的元素。它们中的每一个都相差 1。您需要使用公式 3n + 2 来代替，因此将选择这些元素：

3(0) + 2 = 0 + 2 = 2
3(1) + 2 = 3 + 2 = 5
3(2) + 2 = 6 + 2 = 8
3(3) + 2 = 9 + 2 = 11
...

并且因为您在评论中说每个 img位于 a 中，:nth-child() 伪类应附加到 a，然后选择 img ：

$('.thediv a:nth-child(3n+2) img')

Your formula 3n + 1 doesn't work because these are the elements that will be selected for the following values of n:

3(0) + 1 = 0 + 1 = 1
3(1) + 1 = 3 + 1 = 4
3(2) + 1 = 6 + 1 = 7
3(3) + 1 = 9 + 1 = 10
...

Clearly, these aren't the 2nd, 5th, 8th, 11th ... elements selected. Each of them is off by 1. You need to use the formula 3n + 2 instead, so these elements will be selected:

3(0) + 2 = 0 + 2 = 2
3(1) + 2 = 3 + 2 = 5
3(2) + 2 = 6 + 2 = 8
3(3) + 2 = 9 + 2 = 11
...

And since you said in your comment that each img is in an a, the :nth-child() pseudo-class should be attached to a, then you select the img: