通过类名称在Jquery选择器中选择元素并将其转换回Jquery对象
[英]Select element in Jquery selector by class name and convert it back to Jquery object
问题描述
I don't understand why when i looking for how to get an element in list of element selected by class name like in traditional JS, I've always seen complicated answers . 
我不明白为什么当我寻找如何像传统JS一样在类名选择的元素列表中获取元素时，总是会看到复杂的答案。
document.getElementsByClassName('anyclass')[1]
so, i have found out myself that i cant do that (may be its the wrong approach) 所以，我发现自己做不到（可能是错误的方法）
$('.anyclass')[1]
but i get a DOM element! 但是我得到了一个DOM元素！ so logically i tried 所以我在逻辑上尝试过
$('.anyclass')[1][0]
and it doesnt work 'TypeError: $(...)[0][0] is undefined' Anyone can explain why ? 并且不起作用'TypeError：$（...）[0] [0]未定义'任何人都可以解释为什么？ thank you! 谢谢！
3 个解决方案
解决方案1
1 2018-01-04 11:40:46
var elm = document.createElement("div");
var jelm = $(elm); //convert to jQuery Element
var htmlElm = jelm[0]; //convert to HTML Element
解决方案2
1 已采纳 2018-01-04 11:42:04
I think you need .eq(index) 我认为你需要.eq(index)
var secondElement = $('.anyClass').eq(1); //jQuery object var domElement = secondElement[0]; //DOM element console.log(secondElement, domElement)
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div class='anyClass'>1</div> <div class='anyClass'>2</div>
解决方案3
0 2018-01-04 13:01:30
The :first pseudo-class is equivalent to :eq( 0 ) . :first伪类等效于:eq( 0 ) 。 It could also be written as :lt( 1 ) . 它也可以写成:lt( 1 ) 。 While this matches only a single element, :first-child can match more than one: One for each parent. 虽然这仅匹配一个元素，但：first-child可以匹配多个：每个父对象一个。
here 这里
$('.anyclass:first)
Quick example: 快速示例：
<!DOCTYPE html> <html> <head> <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script> <script> $(document).ready(function(){ $($(".t")[1]).css("background-color", "yellow"); }); </script> </head> <body> <p class=t>This 1.</p> <p class=t>This 2.</p> <p class=t>This 3.</p> </body> </html>
' “
$('.anyclass')[0].attr("src"); you can use like this .
after your comment: $(".anyclass")[1] is a DOM element not a jquery object. 在您的评论之后： $(".anyclass")[1]是DOM元素，而不是jquery对象。 Simply wrap it as jquery $($("td")[1]).width() 只需将其包装为jquery $($("td")[1]).width()
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