如何使用JQuery定位变量的孩子

Question

如何定位post_ad_cloned的输入孩子？

var post_ad = $('.duplicate_name'); 
var post_ad_cloned = post_ad.clone().removeClass('duplicate_name').appendTo('#sss').show(); 
$("post_ad_cloned input").css('color', 'red'); 

Answer 1

var post_ad = $('.duplicate_name'); 
var post_ad_cloned = post_ad.clone().removeClass('duplicate_name').appendTo('#sss').show(); 
$("input", post_ad_cloned).css('color', 'red'); 

或：

var post_ad = $('.duplicate_name'); 
var post_ad_cloned = post_ad.clone() 
    .removeClass('duplicate_name') 
    .appendTo('#sss') 
    .show() 
    .children('input') 
    .css('color', 'red');