Jquery和JS用外部变量获取Ajax返回的参数值的方法实例(超简单)_javascript技巧

var csj_data;

$.ajax
({
cache: false,
async: false,
type: 'post',
data: { aaa: "1" },
url: "../handle/Ladder_Fee_Code.ashx?ajaxaction=Select_FangAn",
success: function (data) {


csj_data = data;
alert(data);
}
});

alert(csj_data);