语法错误,find()上无法识别的表达式

问题描述 投票:2回答:1

我有饼干:

 var city_id_cookie_load = Cookies.get('city_id_cookie')

我需要这样做(或类似的东西)

var $cities = $('#cities');

$cities.on('click', 'a', function() {}).find(city_id_cookie_load).trigger('click');

我怎样才能做到这一点?

我试试这个:

var city_id_cookie_load = '$(' + "'" + '#' + Cookies.get('city_id_cookie') + "'" + ')';

$cities.on('click', 'a', function() 
$cities.on('click', 'a', function() {}).find(city_id_cookie_load).trigger('click');

但我得到错误:

未捕获错误:语法错误,无法识别的表达式:$('#city-2')

jquery
1个回答
0
投票

看起来你正在尝试在一个永远不会工作的字符串中构建一个jQuery对象。尝试在字符串中构建选择器:

var city_id_cookie_load =  '#' + Cookies.get('city_id_cookie')

$cities.on('click', 'a', function() {
  // your click handler logic...
}).find(city_id_cookie_load).trigger('click');
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