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");
        }
    }

本文来自博客园,作者:alisleepy,转载请注明原文链接:https://www.cnblogs.com/alisleepy/p/9982296.html

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