将变量与jquery选择器一起使用 [英] using variable with jquery selector

问题描述

我有以下几行代码，我想使用visibleoffer变量作为选择器.如何连接它?

I have following lines of code and I want to use visibleoffer variable as selector. How can I concatenate it?

var visibleoffer = '.' + obj.parents('tbody').attr('providername') + ' .visible-offers'; //visibleoffer = .Txy .visible-offers
$(obj).parents('.Provider').find(".Offers:not(.HideByProviderFilter,.HideByMainFilter,.HideByProviderFilter.HideByMainFilter,+ visibleoffer)").length;

它变成:

$(obj).parents('.Provider').find(".Offers:not(.HideByProviderFilter,.HideByMainFilter,.HideByProviderFilter.HideByMainFilter,.Txy .visible-offers)").length;

推荐答案

在构建选择器字符串时，您需要连接变量:

You need to concatenate variable while building selector string:

$(obj).parents('.Provider').find(".Offers:not(.HideByProviderFilter,.HideByMainFilter,.HideByProviderFilter.HideByMainFilter," + visibleoffer +")").length;

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