语法错误，find()上无法识别的表达式

Question

我有饼干：

 var city_id_cookie_load = Cookies.get('city_id_cookie')

我需要这样做(或者类似的事情)

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

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

我该怎么做？

我试试看：

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');

但我错了：

未识别错误:语法错误，未识别表达式：$('#city-2')

Answer 1

看起来，您正在尝试在一个字符串中构建一个jQuery对象，这个字符串将永远无法工作。只需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');