Question

我想写一个优雅的解决方案，通过搜索选择相同的标签不同背景的列表

//i have this not editable code
var c1$ = $('#context1'),
    c2$ = $('#context2'),
    c3$ = $('#context3');

//I want to improve the code and 'only from here on
var selector = 'a.select';
c1$.find(selector).css('color','red');
c3$.find(selector).css('color','red');

我正在寻找任意数量的上下文的解决方案

Answer 1

您可以使用$.add()构建上下文，然后在选择器的context参数中将其用作单个jQuery对象：

var $foo = $("#foo"),
    $bar = $("#bar");

var $context = $foo.add($bar);

$("a.select", $context).css("color", "red");

Answer 2

为什么不加载上下文列表，然后在每个上下文中找到'a.select'？＃。

Answer 3

我绝望的不优雅的解决方案

var contexts = [$('#context1'),$('#context2'),$('#context3')];
//list of contexts in single variable(arbitrary number!)

var selector = 'a.select';
//same selector for all variable contexts

var cons$ = $([]);
//empty temporary selector

for(i in contexts)
    cons$ = cons$.add(contexts[i]);

var allmatches$ = cons$.find(selector);

//here processing allmatches$

allmatches$.css("color", "red"); 
//now apply css in one step

jquery用于多个上下文的相同选择器

3 个答案: