如何在元素内到达子<a>元素</a>

Question

I have a structure like this

 <ul>
  <li>
    <a>
     <img />
     <span />
    </a>
  </li>
  <li> 
    ......
  </li>
  <li> 
    ......
  </li>
 </ul>

I am using jquery like this#

            $("ul#someid").find("li:last > a[name="some.....]

How can i access the img element and the span element within the "a" element like above.

i am using now like this

  $("ul#pushedProducts").find("li:last > a[name= <%=PushedProductsParametersMapper.PARAMS_PRODUCTS_INFO%>]").children('span').attr({name:'<%=PushedProductsParametersMapper.PARAMS_PRODUCTS_INFO%>' + pushedProductsTypesCount, id:'<%=PushedProductsParametersMapper.PARAMS_PRODUCTS_INFO%>' + pushedProductsTypesCount , style :'display:none;'});

Answer 1

If you want to select both, you can use .children() ^[docs] :

$("#someid").find("li:last > a[name=['someName']").children();

Otherwise, just use the child selector ^[docs] again, a > b > c ,
or the descendant selector ^[docs] , a > bc .

You can combine the selectors ^[docs] however you want to.

Answer 2

var $img = $("ul#someid").find('li:last > a[name="some....."]').children("img")

注意name旁边的引号

Answer 3

Asuming the A tag has an ID,

$('#id img')

and

$('#id span')

would be enough. Classes will work aswell using $('.class img') etc

Answer 4

$("ul#someid li:first a[name='somename'] img")...
$("ul#someid li:first a[name='somename'] span")...

Answer 5

$("ul#someid").find("li:last > a").children(); // will give you all children
$("ul#someid").find("li:last > a").find("img");
$("ul#someid").find("li:last > a").find("span");