js判断元素是否是disable状态

js判断元素是否是disable状态

1. jquery判断元素状态用$(select).prop(属性值) == true

• js判断button是否可以点击：

 ???//判断button是否为不可点击状态 ???if($("#buyButton").prop("disabled") == true){} ???//判断button是否为不可点击状态 ???if($("#buyButton").prop(‘disabled‘) != true){} ???/** ????* 购买按钮失效 ????*/ ???that.disableBuyButton = function(){ ???????if($("#buyButton").prop(‘disabled‘) != true){ ???????????$("#buyButton").prop(‘disabled‘, ‘disabled‘); ???????????$("#buyButton").css("background", "#8f8f96"); ???????} ???} ???/** ????* 购买按钮激活 ????*/ ???that.ableBuyButton = function(){ ???????if($("#buyButton").prop("disabled") == true){ ???????????$("#buyButton").removeProp("disabled"); ???????????$("#buyButton").css("background","#52a9e9"); ???????} ???}

js判断元素是否是disable状态

原文地址：https://www.cnblogs.com/alisleepy/p/9982296.html