如何在javascript

Question

以下完整代码

我想在变量中添加一个数字到incrimented变量，但它们不会添加。相反，他们得到了连环。

这一行

countScroll-=parseInt(scrollWidth_g);

如果“scrollWidth_g”为300且“countScroll”为10，我得到10300，我想要310;

$(document).ready(function(e) {
    var countScroll = 10;
    $(document).on('click', '.scroller_g_nav', function(){
        var nav = $(this).attr('data-nav');
        var scrollWidth = $('.scroller_g').attr('data-scrollwidth');
        if (typeof scrollWidth == 'undefined'){
        var scrollWidth_g = 200;
        } else {
            var scrollWidth_g = scrollWidth;
            }

        if (nav == 'left'){         
                countScroll-=parseInt(scrollWidth_g);
                $('.scroller_g').animate({scrollLeft:countScroll},600);
            } else {
                countScroll+=scrollWidth_g;
                $('.scroller_g').animate({scrollLeft:countScroll},600);
        }
    });
});

Answer 1

在减法时，您不需要解析字符串编号（＆＃34; 123＆＃34;），但在添加数字时需要它。

Example: "12" - 1 = 11
         "12" + 1 = 121 //here you need to parse string to number to get correct result 
         parseInt("12") + 1 = 11 


It works : 

    if (nav == 'left'){         
                countScroll -= parseInt(scrollWidth_g);
                $('.scroller_g').animate({scrollLeft:countScroll},600);
            } else {
                countScroll += parseInt(scrollWidth_g);
                $('.scroller_g').animate({scrollLeft:countScroll},600);
    }